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checking for palindromes in js [duplicate]

I have the following:
function checkPalindrom(palindrom)
{
for( var i = palindrom.length; i > 0; i-- )
{
if( palindrom[i] = palindrom.charAt(palindrom.length)-1 )
{
document.write('the word is palindrome.');
}else{
document.write('the word is not palindrome!');
}
}
}
checkPalindrom('wordthatwillbechecked');
What is wrong with my code? I want to check if the word is a palindrome.

Maybe I will suggest alternative solution:
function checkPalindrom (str) {
return str == str.split('').reverse().join('');
}
UPD. Keep in mind however that this is pretty much "cheating" approach, a demonstration of smart usage of language features, but not the most practical algorithm (time O(n), space O(n)). For real life application or coding interview you should definitely use loop solution. The one posted by Jason Sebring in this thread is both simple and efficient (time O(n), space O(1)).

25x faster than the standard answer
function isPalindrome(s,i) {
return (i=i||0)<0||i>=s.length>>1||s[i]==s[s.length-1-i]&&isPalindrome(s,++i);
}
use like:
isPalindrome('racecar');
as it defines "i" itself
Fiddle: http://jsfiddle.net/namcx0yf/9/
This is ~25 times faster than the standard answer below.
function checkPalindrome(str) {
return str == str.split('').reverse().join('');
}
Fiddle: http://jsfiddle.net/t0zfjfab/2/
View console for performance results.
Although the solution is difficult to read and maintain, I would recommend understanding it to demonstrate non-branching with recursion and bit shifting to impress your next interviewer.
explained
The || and && are used for control flow like "if" "else". If something left of || is true, it just exits with true. If something is false left of || it must continue. If something left of && is false, it exits as false, if something left of a && is true, it must continue. This is considered "non-branching" as it does not need if-else interupts, rather its just evaluated.
1. Used an initializer not requiring "i" to be defined as an argument. Assigns "i" to itself if defined, otherwise initialize to 0. Always is false so next OR condition is always evaluated.
(i = i || 0) < 0
2. Checks if "i" went half way but skips checking middle odd char. Bit shifted here is like division by 2 but to lowest even neighbor division by 2 result. If true then assumes palindrome since its already done. If false evaluates next OR condition.
i >= s.length >> 1
3. Compares from beginning char and end char according to "i" eventually to meet as neighbors or neighbor to middle char. If false exits and assumes NOT palindrome. If true continues on to next AND condition.
s[i] == s[s.length-1-i]
4. Calls itself again for recursion passing the original string as "s". Since "i" is defined for sure at this point, it is pre-incremented to continue checking the string's position. Returns boolean value indicating if palindrome.
isPalindrome(s,++i)
BUT...
A simple for loop is still about twice as fast as my fancy answer (aka KISS principle)
function fastestIsPalindrome(str) {
var len = Math.floor(str.length / 2);
for (var i = 0; i < len; i++)
if (str[i] !== str[str.length - i - 1])
return false;
return true;
}
http://jsfiddle.net/6L953awz/1/

The logic here is not quite correct, you need to check every letter to determine if the word is a palindrome. Currently, you print multiple times. What about doing something like:
function checkPalindrome(word) {
var l = word.length;
for (var i = 0; i < l / 2; i++) {
if (word.charAt(i) !== word.charAt(l - 1 - i)) {
return false;
}
}
return true;
}
if (checkPalindrome("1122332211")) {
document.write("The word is a palindrome");
} else {
document.write("The word is NOT a palindrome");
}
Which should print that it IS indeed a palindrome.

First problem
= is assign
== is compare
Second problem, Your logic here is wrong
palindrom.charAt(palindrom.length)-1
You are subtracting one from the charAt and not the length.
Third problem, it still will be wrong since you are not reducing the length by i.

It works to me
function palindrome(str) {
/* remove special characters, spaces and make lowercase*/
var removeChar = str.replace(/[^A-Z0-9]/ig, "").toLowerCase();
/* reverse removeChar for comparison*/
var checkPalindrome = removeChar.split('').reverse().join('');
/* Check to see if str is a Palindrome*/
return (removeChar === checkPalindrome);
}

As a much clearer recursive function: http://jsfiddle.net/dmz2x117/
function isPalindrome(letters) {
var characters = letters.split(''),
firstLetter = characters.shift(),
lastLetter = characters.pop();
if (firstLetter !== lastLetter) {
return false;
}
if (characters.length < 2) {
return true;
}
return isPalindrome(characters.join(''));
}

SHORTEST CODE (31 chars)(ES6):
p=s=>s==[...s].reverse().join``
p('racecar'); //true
Keep in mind short code isn't necessarily the best. Readability and efficiency can matter more.

At least three things:
You are trying to test for equality with =, which is used for setting. You need to test with == or ===. (Probably the latter, if you don't have a reason for the former.)
You are reporting results after checking each character. But you don't know the results until you've checked enough characters.
You double-check each character-pair, as you really only need to check if, say first === last and not also if last === first.

function checkPalindrom(palindrom)
{
var flag = true;
var j = 0;
for( var i = palindrom.length-1; i > palindrom.length / 2; i-- )
{
if( palindrom[i] != palindrom[j] )
{
flag = false;
break; // why this? It'll exit the loop at once when there is a mismatch.
}
j++;
}
if( flag ) {
document.write('the word is palindrome.');
}
else {
document.write('the word is not palindrome.');
}
}
checkPalindrom('wordthatwillbechecked');
Why am I printing the result outside the loop? Otherwise, for each match in the word, it'll print "is or is not pallindrome" rather than checking the whole word.
EDIT: Updated with changes and a fix suggested by Basemm.

I've added some more to the above functions, to check strings like, "Go hang a salami, I'm a lasagna hog".
function checkPalindrom(str) {
var str = str.replace(/[^a-zA-Z0-9]+/gi, '').toLowerCase();
return str == str.split('').reverse().join('');
}
Thanks

The most important thing to do when solving a Technical Test is Don't use shortcut methods -- they want to see how you think algorithmically! Not your use of methods.
Here is one that I came up with (45 minutes after I blew the test). There are a couple optimizations to make though. When writing any algorithm, its best to assume false and alter the logic if its looking to be true.
isPalindrome():
Basically, to make this run in O(N) (linear) complexity you want to have 2 iterators whose vectors point towards each other. Meaning, one iterator that starts at the beginning and one that starts at the end, each traveling inward. You could have the iterators traverse the whole array and use a condition to break/return once they meet in the middle, but it may save some work to only give each iterator a half-length by default.
for loops seem to force the use of more checks, so I used while loops - which I'm less comfortable with.
Here's the code:
/**
* TODO: If func counts out, let it return 0
* * Assume !isPalindrome (invert logic)
*/
function isPalindrome(S){
var s = S
, len = s.length
, mid = len/2;
, i = 0, j = len-1;
while(i<mid){
var l = s.charAt(i);
while(j>=mid){
var r = s.charAt(j);
if(l === r){
console.log('#while *', i, l, '...', j, r);
--j;
break;
}
console.log('#while !', i, l, '...', j, r);
return 0;
}
++i;
}
return 1;
}
var nooe = solution('neveroddoreven'); // even char length
var kayak = solution('kayak'); // odd char length
var kayaks = solution('kayaks');
console.log('#isPalindrome', nooe, kayak, kayaks);
Notice that if the loops count out, it returns true. All the logic should be inverted so that it by default returns false. I also used one short cut method String.prototype.charAt(n), but I felt OK with this as every language natively supports this method.

function palindromCheck(str) {
var palinArr, i,
palindrom = [],
palinArr = str.split(/[\s!.?,;:'"-()]/ig);
for (i = 0; i < palinArr.length; i++) {
if (palinArr[i].toLowerCase() === palinArr[i].split('').reverse().join('').toLowerCase() &&
palinArr[i] !== '') {
palindrom.push(palinArr[i]);
}
}
return palindrom.join(', ');
}
console.log(palindromCheck('There is a man, his name! was Bob.')); //a, Bob
Finds and upper to lower case. Split string into array, I don't know why a few white spaces remain, but I wanted to catch and single letters.

= in palindrom[i] = palindrom.charAt(palindrom.length)-1 should be == or ===
palindrom.charAt(palindrom.length)-1 should be palindrom.charAt(palindrom.length - i)

Sharing my fast variant which also support spaces
function isPalindrom(str) {
var ia = 0;
var ib = str.length - 1;
do {
if (str[ia] === str[ib]) continue;
// if spaces skip & retry
if (str[ia] === ' ' && ib++) continue;
if (str[ib] === ' ' && ia--) continue;
return false;
} while (++ia < --ib);
return true;
}
var palindrom="never odd or even";
var res = isPalindrom(palindrom);
document.getElementById('check').innerHTML ='"'+ palindrom + '"'+" checked to be :" +res;
<span id="check" />

Some above short anwsers is good, but it's not easy for understand, I suggest one more way:
function checkPalindrome(inputString) {
if(inputString.length == 1){
return true;
}else{
var i = 0;
var j = inputString.length -1;
while(i < j){
if(inputString[i] != inputString[j]){
return false;
}
i++;
j--;
}
}
return true;
}
I compare each character, i start form left, j start from right, until their index is not valid (i<j).
It's also working in any languages

One more solution with ES6
isPalin = str => [...str].every((c, i) => c === str[str.length-1-i]);

You can try the following
function checkPalindrom (str) {
str = str.toLowerCase();
return str == str.split('').reverse().join('');
}
if(checkPalindrom('Racecar')) {
console.log('Palindrome');
} else {
console.log('Not Palindrome');
}

function checkPalindrom(palindrom)
{
palindrom= palindrom.toLowerCase();
var flag = true;
var j;
j = (palindrom.length) -1 ;
//console.log(j);
var cnt = j / 2;
//console.log(cnt);
for( i = 0; i < cnt+1 ; i++,j-- )
{
console.log("J is => "+j);
console.log(palindrom[i] + "<==>" + palindrom[j]);
if( palindrom[i] != palindrom[j] )
{
flag = false;
break;
}
}
if( flag ) {
console.log('the word is palindrome.');
}
else {
console.log('the word is not palindrome.');
}
}
checkPalindrom('Avid diva');

I'm wondering why nobody suggested this:
ES6:
// "aba" -> true
// "acb" -> false
// "aa" -> true
// "abba" -> true
// "s" -> true
isPalindrom = (str = "") => {
if (str[0] === str[str.length - 1]) {
return str.length <= 1 ? true : isPalindrom(str.slice(1, -1))
}
return false;
}
alert(["aba", "acb", "aa", "abba", "s"].map((e, i) => isPalindrom(e)).join())
ES5:
// "aba" -> true
// "acb" -> false
// "aa" -> true
// "abba" -> true
// "s" -> true
function isPalindrom(str) => {
var str = typeof str !== "string" ? "" : str;
if (str[0] === str[str.length - 1]) {
return str.length <= 1 ? true : isPalindrom(str.slice(1, -1))
}
return false;
}
alert(["aba", "acb", "aa", "abba", "s"].map(function (e, i) {
return isPalindrom(e);
}).join());

Recursive Method:
var low;
var high;
var A = "abcdcba";
function palindrome(A , low, high){
A = A.split('');
if((low > high) || (low == high)){
return true;
}
if(A[low] === A[high]){
A = A.join('');
low = low + 1;
high = high - 1;
return palindrome(A , low, high);
}
else{
return "not a palindrome";
}
}
palindrome(A, 0, A.length-1);

I thought I'd share my own solution:
function palindrome(string){
var reverseString = '';
for(var k in string){
reverseString += string[(string.length - k) - 1];
}
if(string === reverseString){
console.log('Hey there palindrome');
}else{
console.log('You are not a palindrome');
}
}
palindrome('ana');
Hope will help someone.

I found this on an interview site:
Write an efficient function that checks whether any permutation of an
input string is a palindrome. You can ignore punctuation, we only care
about the characters.
Playing around with it I came up with this ugly piece of code :)
function checkIfPalindrome(text) {
var found = {};
var foundOne = 0;
text = text.replace(/[^a-z0-9]/gi, '').toLowerCase();
for (var i = 0; i < text.length; i++) {
if (found[text[i]]) {
found[text[i]]++;
} else {
found[text[i]] = 1;
}
}
for (var x in found) {
if (found[x] === 1) {
foundOne++;
if (foundOne > 1) {
return false;
}
}
}
for (var x in found) {
if (found[x] > 2 && found[x] % 2 && foundOne) {
return false;
}
}
return true;
}
Just leaving it here for posterity.

How about this, using a simple flag
function checkPalindrom(str){
var flag = true;
for( var i = 0; i <= str.length-1; i++){
if( str[i] !== str[str.length - i-1]){
flag = false;
}
}
if(flag == false){
console.log('the word is not a palindrome!');
}
else{
console.log('the word is a palindrome!');
}
}
checkPalindrom('abcdcba');

(JavaScript) Using regexp, this checks for alphanumeric palindrome and disregards space and punctuation.
function palindrome(str) {
str = str.match(/[A-Za-z0-9]/gi).join("").toLowerCase();
// (/[A-Za-z0-9]/gi) above makes str alphanumeric
for(var i = 0; i < Math.floor(str.length/2); i++) { //only need to run for half the string length
if(str.charAt(i) !== str.charAt(str.length-i-1)) { // uses !== to compare characters one-by-one from the beginning and end
return "Try again.";
}
}
return "Palindrome!";
}
palindrome("A man, a plan, a canal. Panama.");
//palindrome("4_2 (: /-\ :) 2-4"); // This solution would also work on something like this.

`
function checkPalindrome (str) {
var str = str.toLowerCase();
var original = str.split(' ').join('');
var reversed = original.split(' ').reverse().join('');
return (original === reversed);
}
`

This avoids regex while also dealing with strings that have spaces and uppercase...
function isPalindrome(str) {
str = str.split("");
var str2 = str.filter(function(x){
if(x !== ' ' && x !== ',') {
return x;
}
});
return console.log(str2.join('').toLowerCase()) == console.log(str2.reverse().join('').toLowerCase());
};
isPalindrome("A car, a man, a maraca"); //true

function myPolidrome(polidrome){
var string=polidrome.split('').join(',');
for(var i=0;i<string.length;i++){
if(string.length==1){
console.log("is polidrome");
}else if(string[i]!=string.charAt(string.length-1)){
console.log("is not polidrome");
break;
}else{
return myPolidrome(polidrome.substring(1,polidrome.length-1));
}
}
}
myPolidrome("asasdsdsa");

Thought I will share my solution using Array.prototype.filter(). filter()
filters the array based on boolean values the function returns.
var inputArray=["","a","ab","aba","abab","ababa"]
var outputArray=inputArray.filter(function isPalindrome(x){
if (x.length<2) return true;
var y=x.split("").reverse().join("");
return x==y;
})
console.log(outputArray);

This worked for me.
var number = 8008
number = number + "";
numberreverse = number.split("").reverse().join('');
console.log ("The number if reversed is: " +numberreverse);
if (number == numberreverse)
console.log("Yes, this is a palindrome");
else
console.log("Nope! It isnt a palindrome");

Here is a solution that works even if the string contains non-alphanumeric characters.
function isPalindrome(str) {
str = str.toLowerCase().replace(/\W+|_/g, '');
return str == str.split('').reverse().join('');
}

Floating point comparison with zero at end [duplicate]

Here is the software version number:
"1.0", "1.0.1", "2.0", "2.0.0.1", "2.0.1"
How can I compare this?
Assume the correct order is:
"1.0", "1.0.1", "2.0", "2.0.0.1", "2.0.1"
The idea is simple...:
Read the first digit, than, the second, after that the third...
But I can't convert the version number to float number...
You also can see the version number like this:
"1.0.0.0", "1.0.1.0", "2.0.0.0", "2.0.0.1", "2.0.1.0"
And this is clearer to see what is the idea behind...
But, how can I convert it into a computer program?

semver
The semantic version parser used by npm.
$ npm install semver
var semver = require('semver');
semver.diff('3.4.5', '4.3.7') //'major'
semver.diff('3.4.5', '3.3.7') //'minor'
semver.gte('3.4.8', '3.4.7') //true
semver.ltr('3.4.8', '3.4.7') //false
semver.valid('1.2.3') // '1.2.3'
semver.valid('a.b.c') // null
semver.clean(' =v1.2.3 ') // '1.2.3'
semver.satisfies('1.2.3', '1.x || >=2.5.0 || 5.0.0 - 7.2.3') // true
semver.gt('1.2.3', '9.8.7') // false
semver.lt('1.2.3', '9.8.7') // true
var versions = [ '1.2.3', '3.4.5', '1.0.2' ]
var max = versions.sort(semver.rcompare)[0]
var min = versions.sort(semver.compare)[0]
var max = semver.maxSatisfying(versions, '*')
Semantic Versioning Link : https://www.npmjs.com/package/semver#prerelease-identifiers

The basic idea to make this comparison would be to use Array.split to get arrays of parts from the input strings and then compare pairs of parts from the two arrays; if the parts are not equal we know which version is smaller.
There are a few of important details to keep in mind:
How should the parts in each pair be compared? The question wants to compare numerically, but what if we have version strings that are not made up of just digits (e.g. "1.0a")?
What should happen if one version string has more parts than the other? Most likely "1.0" should be considered less than "1.0.1", but what about "1.0.0"?
Here's the code for an implementation that you can use directly (gist with documentation):
function versionCompare(v1, v2, options) {
var lexicographical = options && options.lexicographical,
zeroExtend = options && options.zeroExtend,
v1parts = v1.split('.'),
v2parts = v2.split('.');
function isValidPart(x) {
return (lexicographical ? /^\d+[A-Za-z]*$/ : /^\d+$/).test(x);
}
if (!v1parts.every(isValidPart) || !v2parts.every(isValidPart)) {
return NaN;
}
if (zeroExtend) {
while (v1parts.length < v2parts.length) v1parts.push("0");
while (v2parts.length < v1parts.length) v2parts.push("0");
}
if (!lexicographical) {
v1parts = v1parts.map(Number);
v2parts = v2parts.map(Number);
}
for (var i = 0; i < v1parts.length; ++i) {
if (v2parts.length == i) {
return 1;
}
if (v1parts[i] == v2parts[i]) {
continue;
}
else if (v1parts[i] > v2parts[i]) {
return 1;
}
else {
return -1;
}
}
if (v1parts.length != v2parts.length) {
return -1;
}
return 0;
}
This version compares parts naturally, does not accept character suffixes and considers "1.7" to be smaller than "1.7.0". The comparison mode can be changed to lexicographical and shorter version strings can be automatically zero-padded using the optional third argument.
There is a JSFiddle that runs "unit tests" here; it is a slightly expanded version of ripper234's work (thank you).
Important note: This code uses Array.map and Array.every, which means that it will not run in IE versions earlier than 9. If you need to support those you will have to provide polyfills for the missing methods.

The simplest is to use localeCompare :
a.localeCompare(b, undefined, { numeric: true, sensitivity: 'base' })
This will return:
0: version strings are equal
1: version a is greater than b
-1: version b is greater than a

// Return 1 if a > b
// Return -1 if a < b
// Return 0 if a == b
function compare(a, b) {
if (a === b) {
return 0;
}
var a_components = a.split(".");
var b_components = b.split(".");
var len = Math.min(a_components.length, b_components.length);
// loop while the components are equal
for (var i = 0; i < len; i++) {
// A bigger than B
if (parseInt(a_components[i]) > parseInt(b_components[i])) {
return 1;
}
// B bigger than A
if (parseInt(a_components[i]) < parseInt(b_components[i])) {
return -1;
}
}
// If one's a prefix of the other, the longer one is greater.
if (a_components.length > b_components.length) {
return 1;
}
if (a_components.length < b_components.length) {
return -1;
}
// Otherwise they are the same.
return 0;
}
console.log(compare("1", "2"));
console.log(compare("2", "1"));
console.log(compare("1.0", "1.0"));
console.log(compare("2.0", "1.0"));
console.log(compare("1.0", "2.0"));
console.log(compare("1.0.1", "1.0"));

This very small, yet very fast compare function takes version numbers of any length and any number size per segment.
Return values:
- a number < 0 if a < b
- a number > 0 if a > b
- 0 if a = b
So you can use it as compare function for Array.sort();
EDIT: Bugfixed Version stripping trailing zeros to recognize "1" and "1.0.0" as equal
function cmpVersions (a, b) {
var i, diff;
var regExStrip0 = /(\.0+)+$/;
var segmentsA = a.replace(regExStrip0, '').split('.');
var segmentsB = b.replace(regExStrip0, '').split('.');
var l = Math.min(segmentsA.length, segmentsB.length);
for (i = 0; i < l; i++) {
diff = parseInt(segmentsA[i], 10) - parseInt(segmentsB[i], 10);
if (diff) {
return diff;
}
}
return segmentsA.length - segmentsB.length;
}
// TEST
console.log(
['2.5.10.4159',
'1.0.0',
'0.5',
'0.4.1',
'1',
'1.1',
'0.0.0',
'2.5.0',
'2',
'0.0',
'2.5.10',
'10.5',
'1.25.4',
'1.2.15'].sort(cmpVersions));
// Result:
// ["0.0.0", "0.0", "0.4.1", "0.5", "1.0.0", "1", "1.1", "1.2.15", "1.25.4", "2", "2.5.0", "2.5.10", "2.5.10.4159", "10.5"]

Simple and short function:
function isNewerVersion (oldVer, newVer) {
const oldParts = oldVer.split('.')
const newParts = newVer.split('.')
for (var i = 0; i < newParts.length; i++) {
const a = ~~newParts[i] // parse int
const b = ~~oldParts[i] // parse int
if (a > b) return true
if (a < b) return false
}
return false
}
Tests:
isNewerVersion('1.0', '2.0') // true
isNewerVersion('1.0', '1.0.1') // true
isNewerVersion('1.0.1', '1.0.10') // true
isNewerVersion('1.0.1', '1.0.1') // false
isNewerVersion('2.0', '1.0') // false
isNewerVersion('2', '1.0') // false
isNewerVersion('2.0.0.0.0.1', '2.1') // true
isNewerVersion('2.0.0.0.0.1', '2.0') // false

Taken from http://java.com/js/deployJava.js:
// return true if 'installed' (considered as a JRE version string) is
// greater than or equal to 'required' (again, a JRE version string).
compareVersions: function (installed, required) {
var a = installed.split('.');
var b = required.split('.');
for (var i = 0; i < a.length; ++i) {
a[i] = Number(a[i]);
}
for (var i = 0; i < b.length; ++i) {
b[i] = Number(b[i]);
}
if (a.length == 2) {
a[2] = 0;
}
if (a[0] > b[0]) return true;
if (a[0] < b[0]) return false;
if (a[1] > b[1]) return true;
if (a[1] < b[1]) return false;
if (a[2] > b[2]) return true;
if (a[2] < b[2]) return false;
return true;
}

Here is another short version that works with any number of sub versions, padded zeros and even numbers with letters (1.0.0b3)
const compareVer = ((prep, repl) =>
{
prep = t => ("" + t)
//treat non-numerical characters as lower version
//replacing them with a negative number based on charcode of first character
.replace(/[^0-9\.]+/g, c => "." + (c.replace(/[\W_]+/, "").toLowerCase().charCodeAt(0) - 65536) + ".")
//remove trailing "." and "0" if followed by non-numerical characters (1.0.0b);
.replace(/(?:\.0+)*(\.-[0-9]+)(\.[0-9]+)?\.*$/g, "$1$2")
.split('.');
return (a, b, c, i, r) =>
{
a = prep(a);
b = prep(b);
for (i = 0, r = 0, c = Math.max(a.length, b.length); !r && i++ < c;)
{
r = -1 * ((a[i] = ~~a[i]) < (b[i] = ~~b[i])) + (a[i] > b[i]);
}
return r;
}
})();
Function returns:
0 if a = b
1 if a > b
-1 if a < b
1.0 = 1.0.0.0.0.0
1.0 < 1.0.1
1.0b1 < 1.0
1.0b = 1.0b
1.1 > 1.0.1b
1.1alpha < 1.1beta
1.1rc1 > 1.1beta
1.1rc1 < 1.1rc2
1.1.0a1 < 1.1a2
1.1.0a10 > 1.1.0a1
1.1.0alpha = 1.1a
1.1.0alpha2 < 1.1b1
1.0001 > 1.00000.1.0.0.0.01
/*use strict*/
const compareVer = ((prep, repl) =>
{
prep = t => ("" + t)
//treat non-numerical characters as lower version
//replacing them with a negative number based on charcode of first character
.replace(/[^0-9\.]+/g, c => "." + (c.replace(/[\W_]+/, "").toLowerCase().charCodeAt(0) - 65536) + ".")
//remove trailing "." and "0" if followed by non-numerical characters (1.0.0b);
.replace(/(?:\.0+)*(\.-[0-9]+)(\.[0-9]+)?\.*$/g, "$1$2")
.split('.');
return (a, b, c, i, r) =>
{
a = prep(a);
b = prep(b);
for (i = 0, r = 0, c = Math.max(a.length, b.length); !r && i++ < c;)
{
r = -1 * ((a[i] = ~~a[i]) < (b[i] = ~~b[i])) + (a[i] > b[i]);
}
return r;
}
})();
//examples
let list = [
["1.0", "1.0.0.0.0.0"],
["1.0", "1.0.1"],
["1.0b1", "1.0"],
["1.0b", "1.0b"],
["1.1", "1.0.1b"],
["1.1alpha", "1.1beta"],
["1.1rc1", "1.1beta"],
["1.1rc1", "1.1rc2"],
["1.1.0a1", "1.1a2"],
["1.1.0a10", "1.1.0a1"],
["1.1.0alpha", "1.1a"],
["1.1.0alpha2", "1.1b1"],
["1.0001", "1.00000.1.0.0.0.01"]
]
for(let i = 0; i < list.length; i++)
{
console.log( list[i][0] + " " + "<=>"[compareVer(list[i][0], list[i][1]) + 1] + " " + list[i][1] );
}
https://jsfiddle.net/vanowm/p7uvtbor/

Couldn't find a function doing what I wanted here. So I wrote my own. This is my contribution. I hope someone find it useful.
Pros:
Handles version strings of arbitrary length. '1' or '1.1.1.1.1'.
Defaults each value to 0 if not specified. Just because a string is longer doesn't mean it's a bigger version. ('1' should be the same as '1.0' and '1.0.0.0'.)
Compare numbers not strings. ('3'<'21' should be true. Not false.)
Don't waste time on useless compares in the loop. (Comparing for ==)
You can choose your own comparator.
Cons:
It does not handle letters in the version string. (I don't know how that would even work?)
My code, similar to the accepted answer by Jon:
function compareVersions(v1, comparator, v2) {
"use strict";
var comparator = comparator == '=' ? '==' : comparator;
if(['==','===','<','<=','>','>=','!=','!=='].indexOf(comparator) == -1) {
throw new Error('Invalid comparator. ' + comparator);
}
var v1parts = v1.split('.'), v2parts = v2.split('.');
var maxLen = Math.max(v1parts.length, v2parts.length);
var part1, part2;
var cmp = 0;
for(var i = 0; i < maxLen && !cmp; i++) {
part1 = parseInt(v1parts[i], 10) || 0;
part2 = parseInt(v2parts[i], 10) || 0;
if(part1 < part2)
cmp = 1;
if(part1 > part2)
cmp = -1;
}
return eval('0' + comparator + cmp);
}
Examples:
compareVersions('1.2.0', '==', '1.2'); // true
compareVersions('00001', '==', '1.0.0'); // true
compareVersions('1.2.0', '<=', '1.2'); // true
compareVersions('2.2.0', '<=', '1.2'); // false

2017 answer:
v1 = '20.0.12';
v2 = '3.123.12';
compareVersions(v1,v2)
// return positive: v1 > v2, zero:v1 == v2, negative: v1 < v2
function compareVersions(v1, v2) {
v1= v1.split('.')
v2= v2.split('.')
var len = Math.max(v1.length,v2.length)
/*default is true*/
for( let i=0; i < len; i++)
v1 = Number(v1[i] || 0);
v2 = Number(v2[i] || 0);
if (v1 !== v2) return v1 - v2 ;
i++;
}
return 0;
}
Simplest code for modern browsers:
function compareVersion2(ver1, ver2) {
ver1 = ver1.split('.').map( s => s.padStart(10) ).join('.');
ver2 = ver2.split('.').map( s => s.padStart(10) ).join('.');
return ver1 <= ver2;
}
The idea here is to compare numbers but in the form of string. to make the comparison work the two strings must be at the same length. so:
"123" > "99" become "123" > "099"
padding the short number "fix" the comparison
Here I padding each part with zeros to lengths of 10. then just use simple string compare for the answer
Example :
var ver1 = '0.2.10', ver2=`0.10.2`
//become
ver1 = '0000000000.0000000002.0000000010'
ver2 = '0000000000.0000000010.0000000002'
// then it easy to see that
ver1 <= ver2 // true

I faced the similar issue, and I had already created a solution for it. Feel free to give it a try.
It returns 0 for equal, 1 if the version is greater and -1 if it is less
function compareVersion(currentVersion, minVersion) {
let current = currentVersion.replace(/\./g," .").split(' ').map(x=>parseFloat(x,10))
let min = minVersion.replace(/\./g," .").split(' ').map(x=>parseFloat(x,10))
for(let i = 0; i < Math.max(current.length, min.length); i++) {
if((current[i] || 0) < (min[i] || 0)) {
return -1
} else if ((current[i] || 0) > (min[i] || 0)) {
return 1
}
}
return 0
}
console.log(compareVersion("81.0.1212.121","80.4.1121.121"));
console.log(compareVersion("81.0.1212.121","80.4.9921.121"));
console.log(compareVersion("80.0.1212.121","80.4.9921.121"));
console.log(compareVersion("4.4.0","4.4.1"));
console.log(compareVersion("5.24","5.2"));
console.log(compareVersion("4.1","4.1.2"));
console.log(compareVersion("4.1.2","4.1"));
console.log(compareVersion("4.4.4.4","4.4.4.4.4"));
console.log(compareVersion("4.4.4.4.4.4","4.4.4.4.4"));
console.log(compareVersion("0","1"));
console.log(compareVersion("1","1"));
console.log(compareVersion("1","1.0.00000.0000"));
console.log(compareVersion("","1"));
console.log(compareVersion("10.0.1","10.1"));

Although this question already has a lot of answers, each one promotes their own backyard-brewn solution, whilst we have a whole ecosystem of (battle-)tested libraries for this.
A quick search on NPM, GitHub, X will give us some lovely libs, and I'd want to run through some:
semver-compare is a great lightweight (~230 bytes) lib that's especially useful if you want to sort by version numbers, as the library's exposed method returns -1, 0 or 1 appropriately.
The core of the library:
module.exports = function cmp (a, b) {
var pa = a.split('.');
var pb = b.split('.');
for (var i = 0; i < 3; i++) {
var na = Number(pa[i]);
var nb = Number(pb[i]);
if (na > nb) return 1;
if (nb > na) return -1;
if (!isNaN(na) && isNaN(nb)) return 1;
if (isNaN(na) && !isNaN(nb)) return -1;
}
return 0;
};
compare-semver is rather hefty in size (~4.4 kB gzipped), but allows for some nice unique comparisons like to find the minimum/maximum of a stack of versions or to find out if the provided version is unique or less than anything else in a collection of versions.
compare-versions is another small library (~630 bytes gzipped) and follows the spec nicely, meaning you can compare versions with alpha/beta flags and even wildcards (like for minor/patch versions: 1.0.x or 1.0.*)
The point being: there's not always a need to copy-paste code from Stack Overflow, if you can find decent, (unit-)tested versions via your package manager of choice.

Forgive me if this idea already been visited in a link I have not seen.
I have had some success with conversion of the parts into a weighted sum like so:
partSum = this.major * Math.Pow(10,9);
partSum += this.minor * Math.Pow(10, 6);
partSum += this.revision * Math.Pow(10, 3);
partSum += this.build * Math.Pow(10, 0);
Which made comparisons very easy (comparing a double).
Our version fields are never more than 4 digits.
7.10.2.184 -> 7010002184.0
7.11.0.1385 -> 7011001385.0
I hope this helps someone, as the multiple conditionals seem a bit overkill.

We can now use Intl.Collator API now to create numeric comparators. Browser support is pretty decent, but not supported in Node.js at the time of writing.
const semverCompare = new Intl.Collator("en", { numeric: true }).compare;
const versions = ['1.0.1', '1.10.2', '1.1.1', '1.10.1', '1.5.10', '2.10.0', '2.0.1'];
console.log(versions.sort(semverCompare))
const example2 = ["1.0", "1.0.1", "2.0", "2.0.0.1", "2.0.1"];
console.log(example2.sort(semverCompare))

A dead simple way:
function compareVer(previousVersion, currentVersion) {
try {
const [prevMajor, prevMinor = 0, prevPatch = 0] = previousVersion.split('.').map(Number);
const [curMajor, curMinor = 0, curPatch = 0] = currentVersion.split('.').map(Number);
if (curMajor > prevMajor) {
return 'major update';
}
if (curMajor < prevMajor) {
return 'major downgrade';
}
if (curMinor > prevMinor) {
return 'minor update';
}
if (curMinor < prevMinor) {
return 'minor downgrade';
}
if (curPatch > prevPatch) {
return 'patch update';
}
if (curPatch < prevPatch) {
return 'patch downgrade';
}
return 'same version';
} catch (e) {
return 'invalid format';
}
}
Output:
compareVer("3.1", "3.1.1") // patch update
compareVer("3.1.1", "3.2") // minor update
compareVer("2.1.1", "1.1.1") // major downgrade
compareVer("1.1.1", "1.1.1") // same version

Check the function version_compare() from the php.js project. It's is similar to PHP's version_compare().
You can simply use it like this:
version_compare('2.0', '2.0.0.1', '<');
// returns true

My less verbose answer than most of the answers here
/**
* Compare two semver versions. Returns true if version A is greater than
* version B
* #param {string} versionA
* #param {string} versionB
* #returns {boolean}
*/
export const semverGreaterThan = function(versionA, versionB){
var versionsA = versionA.split(/\./g),
versionsB = versionB.split(/\./g)
while (versionsA.length || versionsB.length) {
var a = Number(versionsA.shift()), b = Number(versionsB.shift())
if (a == b)
continue
return (a > b || isNaN(b))
}
return false
}

You could use String#localeCompare with options
sensitivity
Which differences in the strings should lead to non-zero result values. Possible values are:
"base": Only strings that differ in base letters compare as unequal. Examples: a ≠ b, a = á, a = A.
"accent": Only strings that differ in base letters or accents and other diacritic marks compare as unequal. Examples: a ≠ b, a ≠ á, a = A.
"case": Only strings that differ in base letters or case compare as unequal. Examples: a ≠ b, a = á, a ≠ A.
"variant": Strings that differ in base letters, accents and other diacritic marks, or case compare as unequal. Other differences may also be taken into consideration. Examples: a ≠ b, a ≠ á, a ≠ A.
The default is "variant" for usage "sort"; it's locale dependent for usage "search".
numeric
Whether numeric collation should be used, such that "1" < "2" < "10". Possible values are true and false; the default is false. This option can be set through an options property or through a Unicode extension key; if both are provided, the options property takes precedence. Implementations are not required to support this property.
var versions = ["2.0.1", "2.0", "1.0", "1.0.1", "2.0.0.1"];
versions.sort((a, b) => a.localeCompare(b, undefined, { numeric: true, sensitivity: 'base' }));
console.log(versions);

The (most of the time) correct JavaScript answer in 2020
Both Nina Scholz in March 2020 and Sid Vishnoi in April 2020 post the modern answer:
var versions = ["2.0.1", "2.0", "1.0", "1.0.1", "2.0.0.1"];
versions.sort((a, b) =>
a.localeCompare(b, undefined, { numeric: true, sensitivity: 'base' })
);
console.log(versions);
localCompare has been around for some time
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Intl/Collator/Collator
But what about 1.0a and 1.0.1
localCompare doesn't solve that, still returns 1.0.1 , 1.0a
Michael Deal in his (longish &complex) solution already cracked that in 2013
He converts Numbers to another Base, so they can be sorted better
His answer got me thinking...
666 - Don't think in numbers - 999
Sorting is alphanumeric, based on the ASCII values, so let's (ab)use ASCII as the "base"
My solution is to convert 1.0.2.1 to b.a.c.b to bacb , and then sort
This solves 1.1 vs. 1.0.0.0.1 with: bb vs. baaab
And immediately solves the 1.0a and 1.0.1 sorting problem with notation: baa and bab
Conversion is done with:
const str = s => s.match(/(\d+)|[a-z]/g)
.map(c => c == ~~c ? String.fromCharCode(97 + c) : c);
= Calculate ASCII value for 0...999 Numbers, otherwise concat letter
1.0a >>> [ "1" , "0" , "a" ] >>> [ "b" , "a" , "a" ]
For comparison sake there is no need to concatenate it to one string with .join("")
Oneliner
const sortVersions=(x,v=s=>s.match(/(\d+)|[a-z]/g)
.map(c=>c==~~c?String.fromCharCode(97+c):c))
=>x.sort((a,b)=>v(b)<v(a)?1:-1)
Test snippet:
function log(label,val){
document.body.append(label,String(val).replace(/,/g," - "),document.createElement("BR"));
}
let v = ["1.90.1", "1.9.1", "1.89", "1.090", "1.2", "1.0a", "1.0.1", "1.10", "1.0.0a"];
log('not sorted input :',v);
v.sort((a, b) => a.localeCompare(b,undefined,{numeric:true,sensitivity:'base' }));
log(' locale Compare :', v); // 1.0a AFTER 1.0.1
const str = s => s.match(/(\d+)|[a-z]/g)
.map(c => c == ~~c ? String.fromCharCode(97 + c) : c);
const versionCompare = (a, b) => {
a = str(a);
b = str(b);
return b < a ? 1 : a == b ? 0 : -1;
}
v.sort(versionCompare);
log('versionCompare:', v);
Note how 1.090 is sorted in both results.
My code will not solve the 001.012.001 notation mentioned in one answer, but the localeCompare gets that part of the challenge right.
You could combine the two methods:
sort with .localCompare OR versionCompare when there is a letter involved
Final JavaScript solution
const sortVersions = (
x,
v = s => s.match(/[a-z]|\d+/g).map(c => c==~~c ? String.fromCharCode(97 + c) : c)
) => x.sort((a, b) => (a + b).match(/[a-z]/)
? v(b) < v(a) ? 1 : -1
: a.localeCompare(b, 0, {numeric: true}))
let v=["1.90.1","1.090","1.0a","1.0.1","1.0.0a","1.0.0b","1.0.0.1"];
console.log(sortVersions(v));

Few lines of code and good if you don't want to allow letters or symbols. This works if you control the versioning scheme and it's not something a 3rd party provides.
// we presume all versions are of this format "1.4" or "1.10.2.3", without letters
// returns: 1 (bigger), 0 (same), -1 (smaller)
function versionCompare (v1, v2) {
const v1Parts = v1.split('.')
const v2Parts = v2.split('.')
const length = Math.max(v1Parts.length, v2Parts.length)
for (let i = 0; i < length; i++) {
const value = (parseInt(v1Parts[i]) || 0) - (parseInt(v2Parts[i]) || 0)
if (value < 0) return -1
if (value > 0) return 1
}
return 0
}
console.log(versionCompare('1.2.0', '1.2.4') === -1)
console.log(versionCompare('1.2', '1.2.0') === 0)
console.log(versionCompare('1.2', '1') === 1)
console.log(versionCompare('1.2.10', '1.2.1') === 1)
console.log(versionCompare('1.2.134230', '1.2.2') === 1)
console.log(versionCompare('1.2.134230', '1.3.0.1.2.3.1') === -1)

You can use a JavaScript localeCompare method:
a.localeCompare(b, undefined, { numeric: true })
Here is an example:
"1.1".localeCompare("2.1.1", undefined, { numeric: true }) => -1
"1.0.0".localeCompare("1.0", undefined, { numeric: true }) => 1
"1.0.0".localeCompare("1.0.0", undefined, { numeric: true }) => 0

// Returns true if v1 is bigger than v2, and false if otherwise.
function isNewerThan(v1, v2) {
v1=v1.split('.');
v2=v2.split('.');
for(var i = 0; i<Math.max(v1.length,v2.length); i++){
if(v1[i] == undefined) return false; // If there is no digit, v2 is automatically bigger
if(v2[i] == undefined) return true; // if there is no digit, v1 is automatically bigger
if(v1[i] > v2[i]) return true;
if(v1[i] < v2[i]) return false;
}
return false; // Returns false if they are equal
}

The idea is to compare two versions and know which is the biggest. We delete "." and we compare each position of the vector with the other.
// Return 1 if a > b
// Return -1 if a < b
// Return 0 if a == b
function compareVersions(a_components, b_components) {
if (a_components === b_components) {
return 0;
}
var partsNumberA = a_components.split(".");
var partsNumberB = b_components.split(".");
for (var i = 0; i < partsNumberA.length; i++) {
var valueA = parseInt(partsNumberA[i]);
var valueB = parseInt(partsNumberB[i]);
// A bigger than B
if (valueA > valueB || isNaN(valueB)) {
return 1;
}
// B bigger than A
if (valueA < valueB) {
return -1;
}
}
}

The replace() function only replaces the first occurence in the string. So, lets replace the . with ,. Afterwards delete all . and make the , to . again and parse it to float.
for(i=0; i<versions.length; i++) {
v = versions[i].replace('.', ',');
v = v.replace(/\./g, '');
versions[i] = parseFloat(v.replace(',', '.'));
}
finally, sort it:
versions.sort();

Check out this blog post. This function works for numeric version numbers.
function compVersions(strV1, strV2) {
var nRes = 0
, parts1 = strV1.split('.')
, parts2 = strV2.split('.')
, nLen = Math.max(parts1.length, parts2.length);
for (var i = 0; i < nLen; i++) {
var nP1 = (i < parts1.length) ? parseInt(parts1[i], 10) : 0
, nP2 = (i < parts2.length) ? parseInt(parts2[i], 10) : 0;
if (isNaN(nP1)) { nP1 = 0; }
if (isNaN(nP2)) { nP2 = 0; }
if (nP1 != nP2) {
nRes = (nP1 > nP2) ? 1 : -1;
break;
}
}
return nRes;
};
compVersions('10', '10.0'); // 0
compVersions('10.1', '10.01.0'); // 0
compVersions('10.0.1', '10.0'); // 1
compVersions('10.0.1', '10.1'); // -1

If, for example, we want to check if the current jQuery version is less than 1.8, parseFloat($.ui.version) < 1.8 ) would give a wrong result if version is "1.10.1", since parseFloat("1.10.1") returns 1.1.
A string compare would also go wrong, since "1.8" < "1.10" evaluates to false.
So we need a test like this
if(versionCompare($.ui.version, "1.8") < 0){
alert("please update jQuery");
}
The following function handles this correctly:
/** Compare two dotted version strings (like '10.2.3').
* #returns {Integer} 0: v1 == v2, -1: v1 < v2, 1: v1 > v2
*/
function versionCompare(v1, v2) {
var v1parts = ("" + v1).split("."),
v2parts = ("" + v2).split("."),
minLength = Math.min(v1parts.length, v2parts.length),
p1, p2, i;
// Compare tuple pair-by-pair.
for(i = 0; i < minLength; i++) {
// Convert to integer if possible, because "8" > "10".
p1 = parseInt(v1parts[i], 10);
p2 = parseInt(v2parts[i], 10);
if (isNaN(p1)){ p1 = v1parts[i]; }
if (isNaN(p2)){ p2 = v2parts[i]; }
if (p1 == p2) {
continue;
}else if (p1 > p2) {
return 1;
}else if (p1 < p2) {
return -1;
}
// one operand is NaN
return NaN;
}
// The longer tuple is always considered 'greater'
if (v1parts.length === v2parts.length) {
return 0;
}
return (v1parts.length < v2parts.length) ? -1 : 1;
}
Here are some examples:
// compare dotted version strings
console.assert(versionCompare("1.8", "1.8.1") < 0);
console.assert(versionCompare("1.8.3", "1.8.1") > 0);
console.assert(versionCompare("1.8", "1.10") < 0);
console.assert(versionCompare("1.10.1", "1.10.1") === 0);
// Longer is considered 'greater'
console.assert(versionCompare("1.10.1.0", "1.10.1") > 0);
console.assert(versionCompare("1.10.1", "1.10.1.0") < 0);
// Strings pairs are accepted
console.assert(versionCompare("1.x", "1.x") === 0);
// Mixed int/string pairs return NaN
console.assert(isNaN(versionCompare("1.8", "1.x")));
//works with plain numbers
console.assert(versionCompare("4", 3) > 0);
See here for a live sample and test suite:
http://jsfiddle.net/mar10/8KjvP/

This is a neat trick. If you are dealing with numeric values, between a specific range of values, you can assign a value to each level of the version object. For instance "largestValue" is set to 0xFF here, which creates a very "IP" sort of look to your versioning.
This also handles alpha-numeric versioning (i.e. 1.2a < 1.2b)
// The version compare function
function compareVersion(data0, data1, levels) {
function getVersionHash(version) {
var value = 0;
version = version.split(".").map(function (a) {
var n = parseInt(a);
var letter = a.replace(n, "");
if (letter) {
return n + letter[0].charCodeAt() / 0xFF;
} else {
return n;
}
});
for (var i = 0; i < version.length; ++i) {
if (levels === i) break;
value += version[i] / 0xFF * Math.pow(0xFF, levels - i + 1);
}
return value;
};
var v1 = getVersionHash(data0);
var v2 = getVersionHash(data1);
return v1 === v2 ? -1 : v1 > v2 ? 0 : 1;
};
// Returns 0 or 1, correlating to input A and input B
// Direct match returns -1
var version = compareVersion("1.254.253", "1.254.253a", 3);

I made this based on Kons idea, and optimized it for Java version "1.7.0_45". It's just a function meant to convert a version string to a float. This is the function:
function parseVersionFloat(versionString) {
var versionArray = ("" + versionString)
.replace("_", ".")
.replace(/[^0-9.]/g, "")
.split("."),
sum = 0;
for (var i = 0; i < versionArray.length; ++i) {
sum += Number(versionArray[i]) / Math.pow(10, i * 3);
}
console.log(versionString + " -> " + sum);
return sum;
}
String "1.7.0_45" is converted to 1.0070000450000001 and this is good enough for a normal comparison. Error explained here: How to deal with floating point number precision in JavaScript?. If need more then 3 digits on any part you can change the divider Math.pow(10, i * 3);.
Output will look like this:
1.7.0_45 > 1.007000045
ver 1.7.build_45 > 1.007000045
1.234.567.890 > 1.23456789

Here's a coffeescript implementation suitable for use with Array.sort inspired by other answers here:
# Returns > 0 if v1 > v2 and < 0 if v1 < v2 and 0 if v1 == v2
compareVersions = (v1, v2) ->
v1Parts = v1.split('.')
v2Parts = v2.split('.')
minLength = Math.min(v1Parts.length, v2Parts.length)
if minLength > 0
for idx in [0..minLength - 1]
diff = Number(v1Parts[idx]) - Number(v2Parts[idx])
return diff unless diff is 0
return v1Parts.length - v2Parts.length

I wrote a node module for sorting versions, you can find it here: version-sort
Features:
no limit of sequences '1.0.1.5.53.54654.114.1.154.45' works
no limit of sequence length: '1.1546515465451654654654654138754431574364321353734' works
can sort objects by version (see README)
stages (like alpha, beta, rc1, rc2)
Do not hesitate to open an issue if you need an other feature.

Palindrome check in Javascript

I have the following:
function checkPalindrom(palindrom)
{
for( var i = palindrom.length; i > 0; i-- )
{
if( palindrom[i] = palindrom.charAt(palindrom.length)-1 )
{
document.write('the word is palindrome.');
}else{
document.write('the word is not palindrome!');
}
}
}
checkPalindrom('wordthatwillbechecked');
What is wrong with my code? I want to check if the word is a palindrome.

Maybe I will suggest alternative solution:
function checkPalindrom (str) {
return str == str.split('').reverse().join('');
}
UPD. Keep in mind however that this is pretty much "cheating" approach, a demonstration of smart usage of language features, but not the most practical algorithm (time O(n), space O(n)). For real life application or coding interview you should definitely use loop solution. The one posted by Jason Sebring in this thread is both simple and efficient (time O(n), space O(1)).

25x faster than the standard answer
function isPalindrome(s,i) {
return (i=i||0)<0||i>=s.length>>1||s[i]==s[s.length-1-i]&&isPalindrome(s,++i);
}
use like:
isPalindrome('racecar');
as it defines "i" itself
Fiddle: http://jsfiddle.net/namcx0yf/9/
This is ~25 times faster than the standard answer below.
function checkPalindrome(str) {
return str == str.split('').reverse().join('');
}
Fiddle: http://jsfiddle.net/t0zfjfab/2/
View console for performance results.
Although the solution is difficult to read and maintain, I would recommend understanding it to demonstrate non-branching with recursion and bit shifting to impress your next interviewer.
explained
The || and && are used for control flow like "if" "else". If something left of || is true, it just exits with true. If something is false left of || it must continue. If something left of && is false, it exits as false, if something left of a && is true, it must continue. This is considered "non-branching" as it does not need if-else interupts, rather its just evaluated.
1. Used an initializer not requiring "i" to be defined as an argument. Assigns "i" to itself if defined, otherwise initialize to 0. Always is false so next OR condition is always evaluated.
(i = i || 0) < 0
2. Checks if "i" went half way but skips checking middle odd char. Bit shifted here is like division by 2 but to lowest even neighbor division by 2 result. If true then assumes palindrome since its already done. If false evaluates next OR condition.
i >= s.length >> 1
3. Compares from beginning char and end char according to "i" eventually to meet as neighbors or neighbor to middle char. If false exits and assumes NOT palindrome. If true continues on to next AND condition.
s[i] == s[s.length-1-i]
4. Calls itself again for recursion passing the original string as "s". Since "i" is defined for sure at this point, it is pre-incremented to continue checking the string's position. Returns boolean value indicating if palindrome.
isPalindrome(s,++i)
BUT...
A simple for loop is still about twice as fast as my fancy answer (aka KISS principle)
function fastestIsPalindrome(str) {
var len = Math.floor(str.length / 2);
for (var i = 0; i < len; i++)
if (str[i] !== str[str.length - i - 1])
return false;
return true;
}
http://jsfiddle.net/6L953awz/1/

The logic here is not quite correct, you need to check every letter to determine if the word is a palindrome. Currently, you print multiple times. What about doing something like:
function checkPalindrome(word) {
var l = word.length;
for (var i = 0; i < l / 2; i++) {
if (word.charAt(i) !== word.charAt(l - 1 - i)) {
return false;
}
}
return true;
}
if (checkPalindrome("1122332211")) {
document.write("The word is a palindrome");
} else {
document.write("The word is NOT a palindrome");
}
Which should print that it IS indeed a palindrome.

First problem
= is assign
== is compare
Second problem, Your logic here is wrong
palindrom.charAt(palindrom.length)-1
You are subtracting one from the charAt and not the length.
Third problem, it still will be wrong since you are not reducing the length by i.

It works to me
function palindrome(str) {
/* remove special characters, spaces and make lowercase*/
var removeChar = str.replace(/[^A-Z0-9]/ig, "").toLowerCase();
/* reverse removeChar for comparison*/
var checkPalindrome = removeChar.split('').reverse().join('');
/* Check to see if str is a Palindrome*/
return (removeChar === checkPalindrome);
}

As a much clearer recursive function: http://jsfiddle.net/dmz2x117/
function isPalindrome(letters) {
var characters = letters.split(''),
firstLetter = characters.shift(),
lastLetter = characters.pop();
if (firstLetter !== lastLetter) {
return false;
}
if (characters.length < 2) {
return true;
}
return isPalindrome(characters.join(''));
}

SHORTEST CODE (31 chars)(ES6):
p=s=>s==[...s].reverse().join``
p('racecar'); //true
Keep in mind short code isn't necessarily the best. Readability and efficiency can matter more.

At least three things:
You are trying to test for equality with =, which is used for setting. You need to test with == or ===. (Probably the latter, if you don't have a reason for the former.)
You are reporting results after checking each character. But you don't know the results until you've checked enough characters.
You double-check each character-pair, as you really only need to check if, say first === last and not also if last === first.

function checkPalindrom(palindrom)
{
var flag = true;
var j = 0;
for( var i = palindrom.length-1; i > palindrom.length / 2; i-- )
{
if( palindrom[i] != palindrom[j] )
{
flag = false;
break; // why this? It'll exit the loop at once when there is a mismatch.
}
j++;
}
if( flag ) {
document.write('the word is palindrome.');
}
else {
document.write('the word is not palindrome.');
}
}
checkPalindrom('wordthatwillbechecked');
Why am I printing the result outside the loop? Otherwise, for each match in the word, it'll print "is or is not pallindrome" rather than checking the whole word.
EDIT: Updated with changes and a fix suggested by Basemm.

I've added some more to the above functions, to check strings like, "Go hang a salami, I'm a lasagna hog".
function checkPalindrom(str) {
var str = str.replace(/[^a-zA-Z0-9]+/gi, '').toLowerCase();
return str == str.split('').reverse().join('');
}
Thanks

The most important thing to do when solving a Technical Test is Don't use shortcut methods -- they want to see how you think algorithmically! Not your use of methods.
Here is one that I came up with (45 minutes after I blew the test). There are a couple optimizations to make though. When writing any algorithm, its best to assume false and alter the logic if its looking to be true.
isPalindrome():
Basically, to make this run in O(N) (linear) complexity you want to have 2 iterators whose vectors point towards each other. Meaning, one iterator that starts at the beginning and one that starts at the end, each traveling inward. You could have the iterators traverse the whole array and use a condition to break/return once they meet in the middle, but it may save some work to only give each iterator a half-length by default.
for loops seem to force the use of more checks, so I used while loops - which I'm less comfortable with.
Here's the code:
/**
* TODO: If func counts out, let it return 0
* * Assume !isPalindrome (invert logic)
*/
function isPalindrome(S){
var s = S
, len = s.length
, mid = len/2;
, i = 0, j = len-1;
while(i<mid){
var l = s.charAt(i);
while(j>=mid){
var r = s.charAt(j);
if(l === r){
console.log('#while *', i, l, '...', j, r);
--j;
break;
}
console.log('#while !', i, l, '...', j, r);
return 0;
}
++i;
}
return 1;
}
var nooe = solution('neveroddoreven'); // even char length
var kayak = solution('kayak'); // odd char length
var kayaks = solution('kayaks');
console.log('#isPalindrome', nooe, kayak, kayaks);
Notice that if the loops count out, it returns true. All the logic should be inverted so that it by default returns false. I also used one short cut method String.prototype.charAt(n), but I felt OK with this as every language natively supports this method.

function palindromCheck(str) {
var palinArr, i,
palindrom = [],
palinArr = str.split(/[\s!.?,;:'"-()]/ig);
for (i = 0; i < palinArr.length; i++) {
if (palinArr[i].toLowerCase() === palinArr[i].split('').reverse().join('').toLowerCase() &&
palinArr[i] !== '') {
palindrom.push(palinArr[i]);
}
}
return palindrom.join(', ');
}
console.log(palindromCheck('There is a man, his name! was Bob.')); //a, Bob
Finds and upper to lower case. Split string into array, I don't know why a few white spaces remain, but I wanted to catch and single letters.

= in palindrom[i] = palindrom.charAt(palindrom.length)-1 should be == or ===
palindrom.charAt(palindrom.length)-1 should be palindrom.charAt(palindrom.length - i)

Sharing my fast variant which also support spaces
function isPalindrom(str) {
var ia = 0;
var ib = str.length - 1;
do {
if (str[ia] === str[ib]) continue;
// if spaces skip & retry
if (str[ia] === ' ' && ib++) continue;
if (str[ib] === ' ' && ia--) continue;
return false;
} while (++ia < --ib);
return true;
}
var palindrom="never odd or even";
var res = isPalindrom(palindrom);
document.getElementById('check').innerHTML ='"'+ palindrom + '"'+" checked to be :" +res;
<span id="check" />

Some above short anwsers is good, but it's not easy for understand, I suggest one more way:
function checkPalindrome(inputString) {
if(inputString.length == 1){
return true;
}else{
var i = 0;
var j = inputString.length -1;
while(i < j){
if(inputString[i] != inputString[j]){
return false;
}
i++;
j--;
}
}
return true;
}
I compare each character, i start form left, j start from right, until their index is not valid (i<j).
It's also working in any languages

One more solution with ES6
isPalin = str => [...str].every((c, i) => c === str[str.length-1-i]);

You can try the following
function checkPalindrom (str) {
str = str.toLowerCase();
return str == str.split('').reverse().join('');
}
if(checkPalindrom('Racecar')) {
console.log('Palindrome');
} else {
console.log('Not Palindrome');
}

function checkPalindrom(palindrom)
{
palindrom= palindrom.toLowerCase();
var flag = true;
var j;
j = (palindrom.length) -1 ;
//console.log(j);
var cnt = j / 2;
//console.log(cnt);
for( i = 0; i < cnt+1 ; i++,j-- )
{
console.log("J is => "+j);
console.log(palindrom[i] + "<==>" + palindrom[j]);
if( palindrom[i] != palindrom[j] )
{
flag = false;
break;
}
}
if( flag ) {
console.log('the word is palindrome.');
}
else {
console.log('the word is not palindrome.');
}
}
checkPalindrom('Avid diva');

I'm wondering why nobody suggested this:
ES6:
// "aba" -> true
// "acb" -> false
// "aa" -> true
// "abba" -> true
// "s" -> true
isPalindrom = (str = "") => {
if (str[0] === str[str.length - 1]) {
return str.length <= 1 ? true : isPalindrom(str.slice(1, -1))
}
return false;
}
alert(["aba", "acb", "aa", "abba", "s"].map((e, i) => isPalindrom(e)).join())
ES5:
// "aba" -> true
// "acb" -> false
// "aa" -> true
// "abba" -> true
// "s" -> true
function isPalindrom(str) => {
var str = typeof str !== "string" ? "" : str;
if (str[0] === str[str.length - 1]) {
return str.length <= 1 ? true : isPalindrom(str.slice(1, -1))
}
return false;
}
alert(["aba", "acb", "aa", "abba", "s"].map(function (e, i) {
return isPalindrom(e);
}).join());

Recursive Method:
var low;
var high;
var A = "abcdcba";
function palindrome(A , low, high){
A = A.split('');
if((low > high) || (low == high)){
return true;
}
if(A[low] === A[high]){
A = A.join('');
low = low + 1;
high = high - 1;
return palindrome(A , low, high);
}
else{
return "not a palindrome";
}
}
palindrome(A, 0, A.length-1);

I thought I'd share my own solution:
function palindrome(string){
var reverseString = '';
for(var k in string){
reverseString += string[(string.length - k) - 1];
}
if(string === reverseString){
console.log('Hey there palindrome');
}else{
console.log('You are not a palindrome');
}
}
palindrome('ana');
Hope will help someone.

I found this on an interview site:
Write an efficient function that checks whether any permutation of an
input string is a palindrome. You can ignore punctuation, we only care
about the characters.
Playing around with it I came up with this ugly piece of code :)
function checkIfPalindrome(text) {
var found = {};
var foundOne = 0;
text = text.replace(/[^a-z0-9]/gi, '').toLowerCase();
for (var i = 0; i < text.length; i++) {
if (found[text[i]]) {
found[text[i]]++;
} else {
found[text[i]] = 1;
}
}
for (var x in found) {
if (found[x] === 1) {
foundOne++;
if (foundOne > 1) {
return false;
}
}
}
for (var x in found) {
if (found[x] > 2 && found[x] % 2 && foundOne) {
return false;
}
}
return true;
}
Just leaving it here for posterity.

How about this, using a simple flag
function checkPalindrom(str){
var flag = true;
for( var i = 0; i <= str.length-1; i++){
if( str[i] !== str[str.length - i-1]){
flag = false;
}
}
if(flag == false){
console.log('the word is not a palindrome!');
}
else{
console.log('the word is a palindrome!');
}
}
checkPalindrom('abcdcba');

(JavaScript) Using regexp, this checks for alphanumeric palindrome and disregards space and punctuation.
function palindrome(str) {
str = str.match(/[A-Za-z0-9]/gi).join("").toLowerCase();
// (/[A-Za-z0-9]/gi) above makes str alphanumeric
for(var i = 0; i < Math.floor(str.length/2); i++) { //only need to run for half the string length
if(str.charAt(i) !== str.charAt(str.length-i-1)) { // uses !== to compare characters one-by-one from the beginning and end
return "Try again.";
}
}
return "Palindrome!";
}
palindrome("A man, a plan, a canal. Panama.");
//palindrome("4_2 (: /-\ :) 2-4"); // This solution would also work on something like this.

`
function checkPalindrome (str) {
var str = str.toLowerCase();
var original = str.split(' ').join('');
var reversed = original.split(' ').reverse().join('');
return (original === reversed);
}
`

This avoids regex while also dealing with strings that have spaces and uppercase...
function isPalindrome(str) {
str = str.split("");
var str2 = str.filter(function(x){
if(x !== ' ' && x !== ',') {
return x;
}
});
return console.log(str2.join('').toLowerCase()) == console.log(str2.reverse().join('').toLowerCase());
};
isPalindrome("A car, a man, a maraca"); //true

function myPolidrome(polidrome){
var string=polidrome.split('').join(',');
for(var i=0;i<string.length;i++){
if(string.length==1){
console.log("is polidrome");
}else if(string[i]!=string.charAt(string.length-1)){
console.log("is not polidrome");
break;
}else{
return myPolidrome(polidrome.substring(1,polidrome.length-1));
}
}
}
myPolidrome("asasdsdsa");

Thought I will share my solution using Array.prototype.filter(). filter()
filters the array based on boolean values the function returns.
var inputArray=["","a","ab","aba","abab","ababa"]
var outputArray=inputArray.filter(function isPalindrome(x){
if (x.length<2) return true;
var y=x.split("").reverse().join("");
return x==y;
})
console.log(outputArray);

This worked for me.
var number = 8008
number = number + "";
numberreverse = number.split("").reverse().join('');
console.log ("The number if reversed is: " +numberreverse);
if (number == numberreverse)
console.log("Yes, this is a palindrome");
else
console.log("Nope! It isnt a palindrome");

Here is a solution that works even if the string contains non-alphanumeric characters.
function isPalindrome(str) {
str = str.toLowerCase().replace(/\W+|_/g, '');
return str == str.split('').reverse().join('');
}

How can I compare software version number using JavaScript? (only numbers)

Here is the software version number:
"1.0", "1.0.1", "2.0", "2.0.0.1", "2.0.1"
How can I compare this?
Assume the correct order is:
"1.0", "1.0.1", "2.0", "2.0.0.1", "2.0.1"
The idea is simple...:
Read the first digit, than, the second, after that the third...
But I can't convert the version number to float number...
You also can see the version number like this:
"1.0.0.0", "1.0.1.0", "2.0.0.0", "2.0.0.1", "2.0.1.0"
And this is clearer to see what is the idea behind...
But, how can I convert it into a computer program?

semver
The semantic version parser used by npm.
$ npm install semver
var semver = require('semver');
semver.diff('3.4.5', '4.3.7') //'major'
semver.diff('3.4.5', '3.3.7') //'minor'
semver.gte('3.4.8', '3.4.7') //true
semver.ltr('3.4.8', '3.4.7') //false
semver.valid('1.2.3') // '1.2.3'
semver.valid('a.b.c') // null
semver.clean(' =v1.2.3 ') // '1.2.3'
semver.satisfies('1.2.3', '1.x || >=2.5.0 || 5.0.0 - 7.2.3') // true
semver.gt('1.2.3', '9.8.7') // false
semver.lt('1.2.3', '9.8.7') // true
var versions = [ '1.2.3', '3.4.5', '1.0.2' ]
var max = versions.sort(semver.rcompare)[0]
var min = versions.sort(semver.compare)[0]
var max = semver.maxSatisfying(versions, '*')
Semantic Versioning Link : https://www.npmjs.com/package/semver#prerelease-identifiers

The basic idea to make this comparison would be to use Array.split to get arrays of parts from the input strings and then compare pairs of parts from the two arrays; if the parts are not equal we know which version is smaller.
There are a few of important details to keep in mind:
How should the parts in each pair be compared? The question wants to compare numerically, but what if we have version strings that are not made up of just digits (e.g. "1.0a")?
What should happen if one version string has more parts than the other? Most likely "1.0" should be considered less than "1.0.1", but what about "1.0.0"?
Here's the code for an implementation that you can use directly (gist with documentation):
function versionCompare(v1, v2, options) {
var lexicographical = options && options.lexicographical,
zeroExtend = options && options.zeroExtend,
v1parts = v1.split('.'),
v2parts = v2.split('.');
function isValidPart(x) {
return (lexicographical ? /^\d+[A-Za-z]*$/ : /^\d+$/).test(x);
}
if (!v1parts.every(isValidPart) || !v2parts.every(isValidPart)) {
return NaN;
}
if (zeroExtend) {
while (v1parts.length < v2parts.length) v1parts.push("0");
while (v2parts.length < v1parts.length) v2parts.push("0");
}
if (!lexicographical) {
v1parts = v1parts.map(Number);
v2parts = v2parts.map(Number);
}
for (var i = 0; i < v1parts.length; ++i) {
if (v2parts.length == i) {
return 1;
}
if (v1parts[i] == v2parts[i]) {
continue;
}
else if (v1parts[i] > v2parts[i]) {
return 1;
}
else {
return -1;
}
}
if (v1parts.length != v2parts.length) {
return -1;
}
return 0;
}
This version compares parts naturally, does not accept character suffixes and considers "1.7" to be smaller than "1.7.0". The comparison mode can be changed to lexicographical and shorter version strings can be automatically zero-padded using the optional third argument.
There is a JSFiddle that runs "unit tests" here; it is a slightly expanded version of ripper234's work (thank you).
Important note: This code uses Array.map and Array.every, which means that it will not run in IE versions earlier than 9. If you need to support those you will have to provide polyfills for the missing methods.

The simplest is to use localeCompare :
a.localeCompare(b, undefined, { numeric: true, sensitivity: 'base' })
This will return:
0: version strings are equal
1: version a is greater than b
-1: version b is greater than a

// Return 1 if a > b
// Return -1 if a < b
// Return 0 if a == b
function compare(a, b) {
if (a === b) {
return 0;
}
var a_components = a.split(".");
var b_components = b.split(".");
var len = Math.min(a_components.length, b_components.length);
// loop while the components are equal
for (var i = 0; i < len; i++) {
// A bigger than B
if (parseInt(a_components[i]) > parseInt(b_components[i])) {
return 1;
}
// B bigger than A
if (parseInt(a_components[i]) < parseInt(b_components[i])) {
return -1;
}
}
// If one's a prefix of the other, the longer one is greater.
if (a_components.length > b_components.length) {
return 1;
}
if (a_components.length < b_components.length) {
return -1;
}
// Otherwise they are the same.
return 0;
}
console.log(compare("1", "2"));
console.log(compare("2", "1"));
console.log(compare("1.0", "1.0"));
console.log(compare("2.0", "1.0"));
console.log(compare("1.0", "2.0"));
console.log(compare("1.0.1", "1.0"));

This very small, yet very fast compare function takes version numbers of any length and any number size per segment.
Return values:
- a number < 0 if a < b
- a number > 0 if a > b
- 0 if a = b
So you can use it as compare function for Array.sort();
EDIT: Bugfixed Version stripping trailing zeros to recognize "1" and "1.0.0" as equal
function cmpVersions (a, b) {
var i, diff;
var regExStrip0 = /(\.0+)+$/;
var segmentsA = a.replace(regExStrip0, '').split('.');
var segmentsB = b.replace(regExStrip0, '').split('.');
var l = Math.min(segmentsA.length, segmentsB.length);
for (i = 0; i < l; i++) {
diff = parseInt(segmentsA[i], 10) - parseInt(segmentsB[i], 10);
if (diff) {
return diff;
}
}
return segmentsA.length - segmentsB.length;
}
// TEST
console.log(
['2.5.10.4159',
'1.0.0',
'0.5',
'0.4.1',
'1',
'1.1',
'0.0.0',
'2.5.0',
'2',
'0.0',
'2.5.10',
'10.5',
'1.25.4',
'1.2.15'].sort(cmpVersions));
// Result:
// ["0.0.0", "0.0", "0.4.1", "0.5", "1.0.0", "1", "1.1", "1.2.15", "1.25.4", "2", "2.5.0", "2.5.10", "2.5.10.4159", "10.5"]

Simple and short function:
function isNewerVersion (oldVer, newVer) {
const oldParts = oldVer.split('.')
const newParts = newVer.split('.')
for (var i = 0; i < newParts.length; i++) {
const a = ~~newParts[i] // parse int
const b = ~~oldParts[i] // parse int
if (a > b) return true
if (a < b) return false
}
return false
}
Tests:
isNewerVersion('1.0', '2.0') // true
isNewerVersion('1.0', '1.0.1') // true
isNewerVersion('1.0.1', '1.0.10') // true
isNewerVersion('1.0.1', '1.0.1') // false
isNewerVersion('2.0', '1.0') // false
isNewerVersion('2', '1.0') // false
isNewerVersion('2.0.0.0.0.1', '2.1') // true
isNewerVersion('2.0.0.0.0.1', '2.0') // false

Taken from http://java.com/js/deployJava.js:
// return true if 'installed' (considered as a JRE version string) is
// greater than or equal to 'required' (again, a JRE version string).
compareVersions: function (installed, required) {
var a = installed.split('.');
var b = required.split('.');
for (var i = 0; i < a.length; ++i) {
a[i] = Number(a[i]);
}
for (var i = 0; i < b.length; ++i) {
b[i] = Number(b[i]);
}
if (a.length == 2) {
a[2] = 0;
}
if (a[0] > b[0]) return true;
if (a[0] < b[0]) return false;
if (a[1] > b[1]) return true;
if (a[1] < b[1]) return false;
if (a[2] > b[2]) return true;
if (a[2] < b[2]) return false;
return true;
}

Here is another short version that works with any number of sub versions, padded zeros and even numbers with letters (1.0.0b3)
const compareVer = ((prep, repl) =>
{
prep = t => ("" + t)
//treat non-numerical characters as lower version
//replacing them with a negative number based on charcode of first character
.replace(/[^0-9\.]+/g, c => "." + (c.replace(/[\W_]+/, "").toLowerCase().charCodeAt(0) - 65536) + ".")
//remove trailing "." and "0" if followed by non-numerical characters (1.0.0b);
.replace(/(?:\.0+)*(\.-[0-9]+)(\.[0-9]+)?\.*$/g, "$1$2")
.split('.');
return (a, b, c, i, r) =>
{
a = prep(a);
b = prep(b);
for (i = 0, r = 0, c = Math.max(a.length, b.length); !r && i++ < c;)
{
r = -1 * ((a[i] = ~~a[i]) < (b[i] = ~~b[i])) + (a[i] > b[i]);
}
return r;
}
})();
Function returns:
0 if a = b
1 if a > b
-1 if a < b
1.0 = 1.0.0.0.0.0
1.0 < 1.0.1
1.0b1 < 1.0
1.0b = 1.0b
1.1 > 1.0.1b
1.1alpha < 1.1beta
1.1rc1 > 1.1beta
1.1rc1 < 1.1rc2
1.1.0a1 < 1.1a2
1.1.0a10 > 1.1.0a1
1.1.0alpha = 1.1a
1.1.0alpha2 < 1.1b1
1.0001 > 1.00000.1.0.0.0.01
/*use strict*/
const compareVer = ((prep, repl) =>
{
prep = t => ("" + t)
//treat non-numerical characters as lower version
//replacing them with a negative number based on charcode of first character
.replace(/[^0-9\.]+/g, c => "." + (c.replace(/[\W_]+/, "").toLowerCase().charCodeAt(0) - 65536) + ".")
//remove trailing "." and "0" if followed by non-numerical characters (1.0.0b);
.replace(/(?:\.0+)*(\.-[0-9]+)(\.[0-9]+)?\.*$/g, "$1$2")
.split('.');
return (a, b, c, i, r) =>
{
a = prep(a);
b = prep(b);
for (i = 0, r = 0, c = Math.max(a.length, b.length); !r && i++ < c;)
{
r = -1 * ((a[i] = ~~a[i]) < (b[i] = ~~b[i])) + (a[i] > b[i]);
}
return r;
}
})();
//examples
let list = [
["1.0", "1.0.0.0.0.0"],
["1.0", "1.0.1"],
["1.0b1", "1.0"],
["1.0b", "1.0b"],
["1.1", "1.0.1b"],
["1.1alpha", "1.1beta"],
["1.1rc1", "1.1beta"],
["1.1rc1", "1.1rc2"],
["1.1.0a1", "1.1a2"],
["1.1.0a10", "1.1.0a1"],
["1.1.0alpha", "1.1a"],
["1.1.0alpha2", "1.1b1"],
["1.0001", "1.00000.1.0.0.0.01"]
]
for(let i = 0; i < list.length; i++)
{
console.log( list[i][0] + " " + "<=>"[compareVer(list[i][0], list[i][1]) + 1] + " " + list[i][1] );
}
https://jsfiddle.net/vanowm/p7uvtbor/

Couldn't find a function doing what I wanted here. So I wrote my own. This is my contribution. I hope someone find it useful.
Pros:
Handles version strings of arbitrary length. '1' or '1.1.1.1.1'.
Defaults each value to 0 if not specified. Just because a string is longer doesn't mean it's a bigger version. ('1' should be the same as '1.0' and '1.0.0.0'.)
Compare numbers not strings. ('3'<'21' should be true. Not false.)
Don't waste time on useless compares in the loop. (Comparing for ==)
You can choose your own comparator.
Cons:
It does not handle letters in the version string. (I don't know how that would even work?)
My code, similar to the accepted answer by Jon:
function compareVersions(v1, comparator, v2) {
"use strict";
var comparator = comparator == '=' ? '==' : comparator;
if(['==','===','<','<=','>','>=','!=','!=='].indexOf(comparator) == -1) {
throw new Error('Invalid comparator. ' + comparator);
}
var v1parts = v1.split('.'), v2parts = v2.split('.');
var maxLen = Math.max(v1parts.length, v2parts.length);
var part1, part2;
var cmp = 0;
for(var i = 0; i < maxLen && !cmp; i++) {
part1 = parseInt(v1parts[i], 10) || 0;
part2 = parseInt(v2parts[i], 10) || 0;
if(part1 < part2)
cmp = 1;
if(part1 > part2)
cmp = -1;
}
return eval('0' + comparator + cmp);
}
Examples:
compareVersions('1.2.0', '==', '1.2'); // true
compareVersions('00001', '==', '1.0.0'); // true
compareVersions('1.2.0', '<=', '1.2'); // true
compareVersions('2.2.0', '<=', '1.2'); // false

2017 answer:
v1 = '20.0.12';
v2 = '3.123.12';
compareVersions(v1,v2)
// return positive: v1 > v2, zero:v1 == v2, negative: v1 < v2
function compareVersions(v1, v2) {
v1= v1.split('.')
v2= v2.split('.')
var len = Math.max(v1.length,v2.length)
/*default is true*/
for( let i=0; i < len; i++)
v1 = Number(v1[i] || 0);
v2 = Number(v2[i] || 0);
if (v1 !== v2) return v1 - v2 ;
i++;
}
return 0;
}
Simplest code for modern browsers:
function compareVersion2(ver1, ver2) {
ver1 = ver1.split('.').map( s => s.padStart(10) ).join('.');
ver2 = ver2.split('.').map( s => s.padStart(10) ).join('.');
return ver1 <= ver2;
}
The idea here is to compare numbers but in the form of string. to make the comparison work the two strings must be at the same length. so:
"123" > "99" become "123" > "099"
padding the short number "fix" the comparison
Here I padding each part with zeros to lengths of 10. then just use simple string compare for the answer
Example :
var ver1 = '0.2.10', ver2=`0.10.2`
//become
ver1 = '0000000000.0000000002.0000000010'
ver2 = '0000000000.0000000010.0000000002'
// then it easy to see that
ver1 <= ver2 // true

I faced the similar issue, and I had already created a solution for it. Feel free to give it a try.
It returns 0 for equal, 1 if the version is greater and -1 if it is less
function compareVersion(currentVersion, minVersion) {
let current = currentVersion.replace(/\./g," .").split(' ').map(x=>parseFloat(x,10))
let min = minVersion.replace(/\./g," .").split(' ').map(x=>parseFloat(x,10))
for(let i = 0; i < Math.max(current.length, min.length); i++) {
if((current[i] || 0) < (min[i] || 0)) {
return -1
} else if ((current[i] || 0) > (min[i] || 0)) {
return 1
}
}
return 0
}
console.log(compareVersion("81.0.1212.121","80.4.1121.121"));
console.log(compareVersion("81.0.1212.121","80.4.9921.121"));
console.log(compareVersion("80.0.1212.121","80.4.9921.121"));
console.log(compareVersion("4.4.0","4.4.1"));
console.log(compareVersion("5.24","5.2"));
console.log(compareVersion("4.1","4.1.2"));
console.log(compareVersion("4.1.2","4.1"));
console.log(compareVersion("4.4.4.4","4.4.4.4.4"));
console.log(compareVersion("4.4.4.4.4.4","4.4.4.4.4"));
console.log(compareVersion("0","1"));
console.log(compareVersion("1","1"));
console.log(compareVersion("1","1.0.00000.0000"));
console.log(compareVersion("","1"));
console.log(compareVersion("10.0.1","10.1"));

Although this question already has a lot of answers, each one promotes their own backyard-brewn solution, whilst we have a whole ecosystem of (battle-)tested libraries for this.
A quick search on NPM, GitHub, X will give us some lovely libs, and I'd want to run through some:
semver-compare is a great lightweight (~230 bytes) lib that's especially useful if you want to sort by version numbers, as the library's exposed method returns -1, 0 or 1 appropriately.
The core of the library:
module.exports = function cmp (a, b) {
var pa = a.split('.');
var pb = b.split('.');
for (var i = 0; i < 3; i++) {
var na = Number(pa[i]);
var nb = Number(pb[i]);
if (na > nb) return 1;
if (nb > na) return -1;
if (!isNaN(na) && isNaN(nb)) return 1;
if (isNaN(na) && !isNaN(nb)) return -1;
}
return 0;
};
compare-semver is rather hefty in size (~4.4 kB gzipped), but allows for some nice unique comparisons like to find the minimum/maximum of a stack of versions or to find out if the provided version is unique or less than anything else in a collection of versions.
compare-versions is another small library (~630 bytes gzipped) and follows the spec nicely, meaning you can compare versions with alpha/beta flags and even wildcards (like for minor/patch versions: 1.0.x or 1.0.*)
The point being: there's not always a need to copy-paste code from Stack Overflow, if you can find decent, (unit-)tested versions via your package manager of choice.

Forgive me if this idea already been visited in a link I have not seen.
I have had some success with conversion of the parts into a weighted sum like so:
partSum = this.major * Math.Pow(10,9);
partSum += this.minor * Math.Pow(10, 6);
partSum += this.revision * Math.Pow(10, 3);
partSum += this.build * Math.Pow(10, 0);
Which made comparisons very easy (comparing a double).
Our version fields are never more than 4 digits.
7.10.2.184 -> 7010002184.0
7.11.0.1385 -> 7011001385.0
I hope this helps someone, as the multiple conditionals seem a bit overkill.

We can now use Intl.Collator API now to create numeric comparators. Browser support is pretty decent, but not supported in Node.js at the time of writing.
const semverCompare = new Intl.Collator("en", { numeric: true }).compare;
const versions = ['1.0.1', '1.10.2', '1.1.1', '1.10.1', '1.5.10', '2.10.0', '2.0.1'];
console.log(versions.sort(semverCompare))
const example2 = ["1.0", "1.0.1", "2.0", "2.0.0.1", "2.0.1"];
console.log(example2.sort(semverCompare))

A dead simple way:
function compareVer(previousVersion, currentVersion) {
try {
const [prevMajor, prevMinor = 0, prevPatch = 0] = previousVersion.split('.').map(Number);
const [curMajor, curMinor = 0, curPatch = 0] = currentVersion.split('.').map(Number);
if (curMajor > prevMajor) {
return 'major update';
}
if (curMajor < prevMajor) {
return 'major downgrade';
}
if (curMinor > prevMinor) {
return 'minor update';
}
if (curMinor < prevMinor) {
return 'minor downgrade';
}
if (curPatch > prevPatch) {
return 'patch update';
}
if (curPatch < prevPatch) {
return 'patch downgrade';
}
return 'same version';
} catch (e) {
return 'invalid format';
}
}
Output:
compareVer("3.1", "3.1.1") // patch update
compareVer("3.1.1", "3.2") // minor update
compareVer("2.1.1", "1.1.1") // major downgrade
compareVer("1.1.1", "1.1.1") // same version

Check the function version_compare() from the php.js project. It's is similar to PHP's version_compare().
You can simply use it like this:
version_compare('2.0', '2.0.0.1', '<');
// returns true

My less verbose answer than most of the answers here
/**
* Compare two semver versions. Returns true if version A is greater than
* version B
* #param {string} versionA
* #param {string} versionB
* #returns {boolean}
*/
export const semverGreaterThan = function(versionA, versionB){
var versionsA = versionA.split(/\./g),
versionsB = versionB.split(/\./g)
while (versionsA.length || versionsB.length) {
var a = Number(versionsA.shift()), b = Number(versionsB.shift())
if (a == b)
continue
return (a > b || isNaN(b))
}
return false
}

You could use String#localeCompare with options
sensitivity
Which differences in the strings should lead to non-zero result values. Possible values are:
"base": Only strings that differ in base letters compare as unequal. Examples: a ≠ b, a = á, a = A.
"accent": Only strings that differ in base letters or accents and other diacritic marks compare as unequal. Examples: a ≠ b, a ≠ á, a = A.
"case": Only strings that differ in base letters or case compare as unequal. Examples: a ≠ b, a = á, a ≠ A.
"variant": Strings that differ in base letters, accents and other diacritic marks, or case compare as unequal. Other differences may also be taken into consideration. Examples: a ≠ b, a ≠ á, a ≠ A.
The default is "variant" for usage "sort"; it's locale dependent for usage "search".
numeric
Whether numeric collation should be used, such that "1" < "2" < "10". Possible values are true and false; the default is false. This option can be set through an options property or through a Unicode extension key; if both are provided, the options property takes precedence. Implementations are not required to support this property.
var versions = ["2.0.1", "2.0", "1.0", "1.0.1", "2.0.0.1"];
versions.sort((a, b) => a.localeCompare(b, undefined, { numeric: true, sensitivity: 'base' }));
console.log(versions);

The (most of the time) correct JavaScript answer in 2020
Both Nina Scholz in March 2020 and Sid Vishnoi in April 2020 post the modern answer:
var versions = ["2.0.1", "2.0", "1.0", "1.0.1", "2.0.0.1"];
versions.sort((a, b) =>
a.localeCompare(b, undefined, { numeric: true, sensitivity: 'base' })
);
console.log(versions);
localCompare has been around for some time
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Intl/Collator/Collator
But what about 1.0a and 1.0.1
localCompare doesn't solve that, still returns 1.0.1 , 1.0a
Michael Deal in his (longish &complex) solution already cracked that in 2013
He converts Numbers to another Base, so they can be sorted better
His answer got me thinking...
666 - Don't think in numbers - 999
Sorting is alphanumeric, based on the ASCII values, so let's (ab)use ASCII as the "base"
My solution is to convert 1.0.2.1 to b.a.c.b to bacb , and then sort
This solves 1.1 vs. 1.0.0.0.1 with: bb vs. baaab
And immediately solves the 1.0a and 1.0.1 sorting problem with notation: baa and bab
Conversion is done with:
const str = s => s.match(/(\d+)|[a-z]/g)
.map(c => c == ~~c ? String.fromCharCode(97 + c) : c);
= Calculate ASCII value for 0...999 Numbers, otherwise concat letter
1.0a >>> [ "1" , "0" , "a" ] >>> [ "b" , "a" , "a" ]
For comparison sake there is no need to concatenate it to one string with .join("")
Oneliner
const sortVersions=(x,v=s=>s.match(/(\d+)|[a-z]/g)
.map(c=>c==~~c?String.fromCharCode(97+c):c))
=>x.sort((a,b)=>v(b)<v(a)?1:-1)
Test snippet:
function log(label,val){
document.body.append(label,String(val).replace(/,/g," - "),document.createElement("BR"));
}
let v = ["1.90.1", "1.9.1", "1.89", "1.090", "1.2", "1.0a", "1.0.1", "1.10", "1.0.0a"];
log('not sorted input :',v);
v.sort((a, b) => a.localeCompare(b,undefined,{numeric:true,sensitivity:'base' }));
log(' locale Compare :', v); // 1.0a AFTER 1.0.1
const str = s => s.match(/(\d+)|[a-z]/g)
.map(c => c == ~~c ? String.fromCharCode(97 + c) : c);
const versionCompare = (a, b) => {
a = str(a);
b = str(b);
return b < a ? 1 : a == b ? 0 : -1;
}
v.sort(versionCompare);
log('versionCompare:', v);
Note how 1.090 is sorted in both results.
My code will not solve the 001.012.001 notation mentioned in one answer, but the localeCompare gets that part of the challenge right.
You could combine the two methods:
sort with .localCompare OR versionCompare when there is a letter involved
Final JavaScript solution
const sortVersions = (
x,
v = s => s.match(/[a-z]|\d+/g).map(c => c==~~c ? String.fromCharCode(97 + c) : c)
) => x.sort((a, b) => (a + b).match(/[a-z]/)
? v(b) < v(a) ? 1 : -1
: a.localeCompare(b, 0, {numeric: true}))
let v=["1.90.1","1.090","1.0a","1.0.1","1.0.0a","1.0.0b","1.0.0.1"];
console.log(sortVersions(v));

Few lines of code and good if you don't want to allow letters or symbols. This works if you control the versioning scheme and it's not something a 3rd party provides.
// we presume all versions are of this format "1.4" or "1.10.2.3", without letters
// returns: 1 (bigger), 0 (same), -1 (smaller)
function versionCompare (v1, v2) {
const v1Parts = v1.split('.')
const v2Parts = v2.split('.')
const length = Math.max(v1Parts.length, v2Parts.length)
for (let i = 0; i < length; i++) {
const value = (parseInt(v1Parts[i]) || 0) - (parseInt(v2Parts[i]) || 0)
if (value < 0) return -1
if (value > 0) return 1
}
return 0
}
console.log(versionCompare('1.2.0', '1.2.4') === -1)
console.log(versionCompare('1.2', '1.2.0') === 0)
console.log(versionCompare('1.2', '1') === 1)
console.log(versionCompare('1.2.10', '1.2.1') === 1)
console.log(versionCompare('1.2.134230', '1.2.2') === 1)
console.log(versionCompare('1.2.134230', '1.3.0.1.2.3.1') === -1)

You can use a JavaScript localeCompare method:
a.localeCompare(b, undefined, { numeric: true })
Here is an example:
"1.1".localeCompare("2.1.1", undefined, { numeric: true }) => -1
"1.0.0".localeCompare("1.0", undefined, { numeric: true }) => 1
"1.0.0".localeCompare("1.0.0", undefined, { numeric: true }) => 0

// Returns true if v1 is bigger than v2, and false if otherwise.
function isNewerThan(v1, v2) {
v1=v1.split('.');
v2=v2.split('.');
for(var i = 0; i<Math.max(v1.length,v2.length); i++){
if(v1[i] == undefined) return false; // If there is no digit, v2 is automatically bigger
if(v2[i] == undefined) return true; // if there is no digit, v1 is automatically bigger
if(v1[i] > v2[i]) return true;
if(v1[i] < v2[i]) return false;
}
return false; // Returns false if they are equal
}

The idea is to compare two versions and know which is the biggest. We delete "." and we compare each position of the vector with the other.
// Return 1 if a > b
// Return -1 if a < b
// Return 0 if a == b
function compareVersions(a_components, b_components) {
if (a_components === b_components) {
return 0;
}
var partsNumberA = a_components.split(".");
var partsNumberB = b_components.split(".");
for (var i = 0; i < partsNumberA.length; i++) {
var valueA = parseInt(partsNumberA[i]);
var valueB = parseInt(partsNumberB[i]);
// A bigger than B
if (valueA > valueB || isNaN(valueB)) {
return 1;
}
// B bigger than A
if (valueA < valueB) {
return -1;
}
}
}

The replace() function only replaces the first occurence in the string. So, lets replace the . with ,. Afterwards delete all . and make the , to . again and parse it to float.
for(i=0; i<versions.length; i++) {
v = versions[i].replace('.', ',');
v = v.replace(/\./g, '');
versions[i] = parseFloat(v.replace(',', '.'));
}
finally, sort it:
versions.sort();

Check out this blog post. This function works for numeric version numbers.
function compVersions(strV1, strV2) {
var nRes = 0
, parts1 = strV1.split('.')
, parts2 = strV2.split('.')
, nLen = Math.max(parts1.length, parts2.length);
for (var i = 0; i < nLen; i++) {
var nP1 = (i < parts1.length) ? parseInt(parts1[i], 10) : 0
, nP2 = (i < parts2.length) ? parseInt(parts2[i], 10) : 0;
if (isNaN(nP1)) { nP1 = 0; }
if (isNaN(nP2)) { nP2 = 0; }
if (nP1 != nP2) {
nRes = (nP1 > nP2) ? 1 : -1;
break;
}
}
return nRes;
};
compVersions('10', '10.0'); // 0
compVersions('10.1', '10.01.0'); // 0
compVersions('10.0.1', '10.0'); // 1
compVersions('10.0.1', '10.1'); // -1

If, for example, we want to check if the current jQuery version is less than 1.8, parseFloat($.ui.version) < 1.8 ) would give a wrong result if version is "1.10.1", since parseFloat("1.10.1") returns 1.1.
A string compare would also go wrong, since "1.8" < "1.10" evaluates to false.
So we need a test like this
if(versionCompare($.ui.version, "1.8") < 0){
alert("please update jQuery");
}
The following function handles this correctly:
/** Compare two dotted version strings (like '10.2.3').
* #returns {Integer} 0: v1 == v2, -1: v1 < v2, 1: v1 > v2
*/
function versionCompare(v1, v2) {
var v1parts = ("" + v1).split("."),
v2parts = ("" + v2).split("."),
minLength = Math.min(v1parts.length, v2parts.length),
p1, p2, i;
// Compare tuple pair-by-pair.
for(i = 0; i < minLength; i++) {
// Convert to integer if possible, because "8" > "10".
p1 = parseInt(v1parts[i], 10);
p2 = parseInt(v2parts[i], 10);
if (isNaN(p1)){ p1 = v1parts[i]; }
if (isNaN(p2)){ p2 = v2parts[i]; }
if (p1 == p2) {
continue;
}else if (p1 > p2) {
return 1;
}else if (p1 < p2) {
return -1;
}
// one operand is NaN
return NaN;
}
// The longer tuple is always considered 'greater'
if (v1parts.length === v2parts.length) {
return 0;
}
return (v1parts.length < v2parts.length) ? -1 : 1;
}
Here are some examples:
// compare dotted version strings
console.assert(versionCompare("1.8", "1.8.1") < 0);
console.assert(versionCompare("1.8.3", "1.8.1") > 0);
console.assert(versionCompare("1.8", "1.10") < 0);
console.assert(versionCompare("1.10.1", "1.10.1") === 0);
// Longer is considered 'greater'
console.assert(versionCompare("1.10.1.0", "1.10.1") > 0);
console.assert(versionCompare("1.10.1", "1.10.1.0") < 0);
// Strings pairs are accepted
console.assert(versionCompare("1.x", "1.x") === 0);
// Mixed int/string pairs return NaN
console.assert(isNaN(versionCompare("1.8", "1.x")));
//works with plain numbers
console.assert(versionCompare("4", 3) > 0);
See here for a live sample and test suite:
http://jsfiddle.net/mar10/8KjvP/

This is a neat trick. If you are dealing with numeric values, between a specific range of values, you can assign a value to each level of the version object. For instance "largestValue" is set to 0xFF here, which creates a very "IP" sort of look to your versioning.
This also handles alpha-numeric versioning (i.e. 1.2a < 1.2b)
// The version compare function
function compareVersion(data0, data1, levels) {
function getVersionHash(version) {
var value = 0;
version = version.split(".").map(function (a) {
var n = parseInt(a);
var letter = a.replace(n, "");
if (letter) {
return n + letter[0].charCodeAt() / 0xFF;
} else {
return n;
}
});
for (var i = 0; i < version.length; ++i) {
if (levels === i) break;
value += version[i] / 0xFF * Math.pow(0xFF, levels - i + 1);
}
return value;
};
var v1 = getVersionHash(data0);
var v2 = getVersionHash(data1);
return v1 === v2 ? -1 : v1 > v2 ? 0 : 1;
};
// Returns 0 or 1, correlating to input A and input B
// Direct match returns -1
var version = compareVersion("1.254.253", "1.254.253a", 3);

I made this based on Kons idea, and optimized it for Java version "1.7.0_45". It's just a function meant to convert a version string to a float. This is the function:
function parseVersionFloat(versionString) {
var versionArray = ("" + versionString)
.replace("_", ".")
.replace(/[^0-9.]/g, "")
.split("."),
sum = 0;
for (var i = 0; i < versionArray.length; ++i) {
sum += Number(versionArray[i]) / Math.pow(10, i * 3);
}
console.log(versionString + " -> " + sum);
return sum;
}
String "1.7.0_45" is converted to 1.0070000450000001 and this is good enough for a normal comparison. Error explained here: How to deal with floating point number precision in JavaScript?. If need more then 3 digits on any part you can change the divider Math.pow(10, i * 3);.
Output will look like this:
1.7.0_45 > 1.007000045
ver 1.7.build_45 > 1.007000045
1.234.567.890 > 1.23456789

Here's a coffeescript implementation suitable for use with Array.sort inspired by other answers here:
# Returns > 0 if v1 > v2 and < 0 if v1 < v2 and 0 if v1 == v2
compareVersions = (v1, v2) ->
v1Parts = v1.split('.')
v2Parts = v2.split('.')
minLength = Math.min(v1Parts.length, v2Parts.length)
if minLength > 0
for idx in [0..minLength - 1]
diff = Number(v1Parts[idx]) - Number(v2Parts[idx])
return diff unless diff is 0
return v1Parts.length - v2Parts.length

I wrote a node module for sorting versions, you can find it here: version-sort
Features:
no limit of sequences '1.0.1.5.53.54654.114.1.154.45' works
no limit of sequence length: '1.1546515465451654654654654138754431574364321353734' works
can sort objects by version (see README)
stages (like alpha, beta, rc1, rc2)
Do not hesitate to open an issue if you need an other feature.

How to use Javascript math on a version number

I use jQuery to get the browser version like this:
var x = $.browser.version;
I get a string like this: 1.9.1.1
Now, I want to do an evaluation so if x is >= 1.9.1 then do some stuff. Unfortunately, with multiple decimal points, I cannot do a parseFloat() because it converts 1.9.1.1 to simply 1.9, and the if evaluation would match a 1.9.0 version (which I do not want).
Has someone figured out a way to accomplish turning a version number (with multiple decimals) into something that can be used as a number for evaluation (or some other way to accomplish what I am trying to do here)?
Thanks -

You could do something with string.split and then do a digit by digit comparison
// arr[0] = 1
// arr[1] = 9
// arr[2] = 1
// arr[3] = 1
var arr = ($.browser.version).split('.');
The following is taken from this post
This is a function that will parse your version string and give you back a JSON object
function parseVersionString (str) {
if (typeof(str) != 'string') { return false; }
var x = str.split('.');
// parse from string or default to 0 if can't parse
var maj = parseInt(x[0]) || 0;
var min = parseInt(x[1]) || 0;
var bld = parseInt(x[2]) || 0;
var rev = parseInt(x[3]) || 0;
return {
major: maj,
minor: min,
build: bld,
revision: rev
}
}
Then you could use the following syntax
var version = parseVersionString($.browser.version);
// version.major == 1
// version.minor == 9
// version.build == 1
// version.revision == 1

Here's another version of versionCmp():
function versionCmp(v1, v2) {
v1 = String(v1).split('.');
v2 = String(v2).split('.');
var diff = 0;
while((v1.length || v2.length) && !diff)
diff = (+v1.shift() || 0) - (+v2.shift() || 0);
return (diff > 0) - (diff < 0);
}
Another possibility would be to assign a numeric value to each version number:
function valueOfVersion(ver) {
ver = String(ver).split('.');
var value = 0;
for(var i = ver.length; i--;)
value += ver[i] / Math.pow(2, i * 8) || 0;
return value;
}
This only works if each digit is less than 256 (because of the hard-coded divisor) and has a limited precision (ie the version strings can't get arbitrarily long).

You need to treat each portion of the string as a seperate integer, so split and iterate, and cmp:
// perform cmp(a, b)
// -1 = a is smaller
// 0 = equal
// 1 = a is bigger
function versionCmp(a, b) {
a = a.split(".");
b = b.split(".");
for(var i=0; i < a.length; i++) {
av = parseInt(a[i]);
bv = parseInt(b[i]);
if (av < bv) {
return -1;
} else if (av > bv) {
return 1;
}
}
return 0;
}
console.log(versionCmp("1.1.2.3", "1.2.1.0")); // should be -1
console.log(versionCmp("1.19.0.1", "1.2.0.4")); // should be 1
console.log(versionCmp("1.2.3.4", "1.2.3.4")); // should be 0

You could remove all dots and then parse it as an integer.
Take note tho, this solution doesn't work in the long term.