Edit: I'm sorry, but I forgot to mention that I'll need the values of the counter variables. So making one loop isn't a solution I'm afraid.
I'm not sure if this is possible at all, but I would like to do the following.
To a function, an array of numbers is passed. Each number is the upper limit of a for loop, for example, if the array is [2, 3, 5], the following code should be executed:
for(var a = 0; a < 2; a++) {
for(var b = 0; b < 3; b++) {
for(var c = 0; c < 5; c++) {
doSomething([a, b, c]);
}
}
}
So the amount of nested for loops is equal to the length of the array. Would there be any way to make this work? I was thinking of creating a piece of code which adds each for loop to a string, and then evaluates it through eval. I've read however that eval should not be one's first choice as it can have dangerous results too.
What technique might be appropriate here?
Recursion can solve this problem neatly:
function callManyTimes(maxIndices, func) {
doCallManyTimes(maxIndices, func, [], 0);
}
function doCallManyTimes(maxIndices, func, args, index) {
if (maxIndices.length == 0) {
func(args);
} else {
var rest = maxIndices.slice(1);
for (args[index] = 0; args[index] < maxIndices[0]; ++args[index]) {
doCallManyTimes(rest, func, args, index + 1);
}
}
}
Call it like this:
callManyTimes([2,3,5], doSomething);
Recursion is overkill here. You can use generators:
function* allPossibleCombinations(lengths) {
const n = lengths.length;
let indices = [];
for (let i = n; --i >= 0;) {
if (lengths[i] === 0) { return; }
if (lengths[i] !== (lengths[i] & 0x7fffffff)) { throw new Error(); }
indices[i] = 0;
}
while (true) {
yield indices;
// Increment indices.
++indices[n - 1];
for (let j = n; --j >= 0 && indices[j] === lengths[j];) {
if (j === 0) { return; }
indices[j] = 0;
++indices[j - 1];
}
}
}
for ([a, b, c] of allPossibleCombinations([3, 2, 2])) {
console.log(`${a}, ${b}, ${c}`);
}
The intuition here is that we keep a list of indices that are always less than the corresponding length.
The second loop handles carry. As when incrementing a decimal number 199, we go to (1, 9, 10), and then carry to get (1, 10, 0) and finally (2, 0, 0). If we don't have enough digits to carry into, we're done.
Set up an array of counters with the same length as the limit array. Use a single loop, and increment the last item in each iteration. When it reaches it's limit you restart it and increment the next item.
function loop(limits) {
var cnt = new Array(limits.length);
for (var i = 0; i < cnt.length; i++) cnt[i] = 0;
var pos;
do {
doSomething(cnt);
pos = cnt.length - 1;
cnt[pos]++;
while (pos >= 0 && cnt[pos] >= limits[pos]) {
cnt[pos] = 0;
pos--;
if (pos >= 0) cnt[pos]++;
}
} while (pos >= 0);
}
One solution that works without getting complicated programatically would be to take the integers and multiply them all. Since you're only nesting the ifs, and only the innermost one has functionality, this should work:
var product = 0;
for(var i = 0; i < array.length; i++){
product *= array[i];
}
for(var i = 0; i < product; i++){
doSomething();
}
Alternatively:
for(var i = 0; i < array.length; i++){
for(var j = 0; j < array[i]; j++){
doSomething();
}
}
Instead of thinking in terms of nested for loops, think about recursive function invocations. To do your iteration, you'd make the following decision (pseudo code):
if the list of counters is empty
then "doSomething()"
else
for (counter = 0 to first counter limit in the list)
recurse with the tail of the list
That might look something like this:
function forEachCounter(counters, fn) {
function impl(counters, curCount) {
if (counters.length === 0)
fn(curCount);
else {
var limit = counters[0];
curCount.push(0);
for (var i = 0; i < limit; ++i) {
curCount[curCount.length - 1] = i;
impl(counters.slice(1), curCount);
}
curCount.length--;
}
}
impl(counters, []);
}
You'd call the function with an argument that's your list of count limits, and an argument that's your function to execute for each effective count array (the "doSomething" part). The main function above does all the real work in an inner function. In that inner function, the first argument is the counter limit list, which will be "whittled down" as the function is called recursively. The second argument is used to hold the current set of counter values, so that "doSomething" can know that it's on an iteration corresponding to a particular list of actual counts.
Calling the function would look like this:
forEachCounter([4, 2, 5], function(c) { /* something */ });
This is my attempt at simplifying the non-recursive solution by Mike Samuel. I also add the ability to set a range (not just maximum) for every integer argument.
function everyPermutation(args, fn) {
var indices = args.map(a => a.min);
for (var j = args.length; j >= 0;) {
fn.apply(null, indices);
// go through indices from right to left setting them to 0
for (j = args.length; j--;) {
// until we find the last index not at max which we increment
if (indices[j] < args[j].max) {
++indices[j];
break;
}
indices[j] = args[j].min;
}
}
}
everyPermutation([
{min:4, max:6},
{min:2, max:3},
{min:0, max:1}
], function(a, b, c) {
console.log(a + ',' + b + ',' + c);
});
There's no difference between doing three loops of 2, 3, 5, and one loop of 30 (2*3*5).
function doLots (howMany, what) {
var amount = 0;
// Aggregate amount
for (var i=0; i<howMany.length;i++) {
amount *= howMany[i];
};
// Execute that many times.
while(i--) {
what();
};
}
Use:
doLots([2,3,5], doSomething);
You can use the greedy algorithm to enumerate all elements of the cartesian product 0:2 x 0:3 x 0:5. This algorithm is performed by my function greedy_backward below. I am not an expert in Javascript and maybe this function could be improved.
function greedy_backward(sizes, n) {
for (var G = [1], i = 0; i<sizes.length; i++) G[i+1] = G[i] * sizes[i];
if (n>=_.last(G)) throw new Error("n must be <" + _.last(G));
for (i = 0; i<sizes.length; i++) if (sizes[i]!=parseInt(sizes[i]) || sizes[i]<1){ throw new Error("sizes must be a vector of integers be >1"); };
for (var epsilon=[], i=0; i < sizes.length; i++) epsilon[i]=0;
while(n > 0){
var k = _.findIndex(G, function(x){ return n < x; }) - 1;
var e = (n/G[k])>>0;
epsilon[k] = e;
n = n-e*G[k];
}
return epsilon;
}
It enumerates the elements of the Cartesian product in the anti-lexicographic order (you will see the full enumeration in the doSomething example):
~ var sizes = [2, 3, 5];
~ greedy_backward(sizes,0);
0,0,0
~ greedy_backward(sizes,1);
1,0,0
~ greedy_backward(sizes,2);
0,1,0
~ greedy_backward(sizes,3);
1,1,0
~ greedy_backward(sizes,4);
0,2,0
~ greedy_backward(sizes,5);
1,2,0
This is a generalization of the binary representation (the case when sizes=[2,2,2,...]).
Example:
~ function doSomething(v){
for (var message = v[0], i = 1; i<v.length; i++) message = message + '-' + v[i].toString();
console.log(message);
}
~ doSomething(["a","b","c"])
a-b-c
~ for (var max = [1], i = 0; i<sizes.length; i++) max = max * sizes[i];
30
~ for(i=0; i<max; i++){
doSomething(greedy_backward(sizes,i));
}
0-0-0
1-0-0
0-1-0
1-1-0
0-2-0
1-2-0
0-0-1
1-0-1
0-1-1
1-1-1
0-2-1
1-2-1
0-0-2
1-0-2
0-1-2
1-1-2
0-2-2
1-2-2
0-0-3
1-0-3
0-1-3
1-1-3
0-2-3
1-2-3
0-0-4
1-0-4
0-1-4
1-1-4
0-2-4
1-2-4
If needed, the reverse operation is simple:
function greedy_forward(sizes, epsilon) {
if (sizes.length!=epsilon.length) throw new Error("sizes and epsilon must have the same length");
for (i = 0; i<sizes.length; i++) if (epsilon[i] <0 || epsilon[i] >= sizes[i]){ throw new Error("condition `0 <= epsilon[i] < sizes[i]` not fulfilled for all i"); };
for (var G = [1], i = 0; i<sizes.length-1; i++) G[i+1] = G[i] * sizes[i];
for (var n = 0, i = 0; i<sizes.length; i++) n += G[i] * epsilon[i];
return n;
}
Example :
~ epsilon = greedy_backward(sizes, 29)
1,2,4
~ greedy_forward(sizes, epsilon)
29
One could also use a generator for that:
function loop(...times) {
function* looper(times, prev = []) {
if(!times.length) {
yield prev;
return;
}
const [max, ...rest] = times;
for(let current = 0; current < max; current++) {
yield* looper(rest, [...prev, current]);
}
}
return looper(times);
}
That can then be used as:
for(const [j, k, l, m] of loop(1, 2, 3, 4)) {
//...
}
Related
I've been trying to solve this kata on codewars.
I've got an algorithm, but it's apparently too slow to pass the test. It can create sequence of 2450 numbers in just under 1.6 seconds. I don't need the solution but the hint or something to help me to make my algorithm faster.
function ulamSequence(u0, u1, n) {
// create an array with first two elements in it
const seq = [u0, u1];
// create a loop that checks if next number is valid and if it is, push it in seq
num: for (let i = u1 + 1; seq.length < n; i++) {
let sumCount = 0;
for (let k = 0; k < seq.length - 1; k++) {
if (seq.indexOf(i - seq[k]) > k && ++sumCount === 2) { continue num; }
}
sumCount === 1 ? seq.push(i) : "";
}
return seq;
}
Here's an idea: have an array sums so that sums[N] keeps the number of possible summations for N. For example, for U(1,2) sums[3] will be 1 and sums[5] will be 2 (1+4, 2+3). On each step, locate the minimal N so that N > last item and sums[N] == 1. Add N to the result, then sum it with all previous items and update sums accordingly.
function ulam(u0, u1, len) {
let seq = [u0, u1]
let sums = []
let last = u1
sums[u0 + u1] = 1
while (seq.length < len) {
last += 1
while (sums[last] !== 1) {
last += 1
}
for (let n of seq) {
let s = n + last
sums[s] = (sums[s] || 0) + 1
}
seq.push(last)
}
return seq
}
console.time()
ulam(1, 2, 2450)
console.timeEnd()
function ulamSequence(u0, u1, n) {
const seq = [u0, u1];
const set = new Set(seq);
for (let i = u1 + 2; seq.length < n; i++) {
let sumCount = 0;
for (let k = 0; k < seq.length - 1; k++) {
if (set.has(i - seq[k])) {
sumCount++;
if (sumCount === 2) {
continue;
}
}
}
if (sumCount === 1) {
seq.push(i);
set.add(i);
}
}
return seq;
}
I need help with writing some code that will create a random number from an array of 12 numbers and print it 9 times without dupes. This has been tough for me to accomplish. Any ideas?
var nums = [1,2,3,4,5,6,7,8,9,10,11,12];
var gen_nums = [];
function in_array(array, el) {
for(var i = 0 ; i < array.length; i++)
if(array[i] == el) return true;
return false;
}
function get_rand(array) {
var rand = array[Math.floor(Math.random()*array.length)];
if(!in_array(gen_nums, rand)) {
gen_nums.push(rand);
return rand;
}
return get_rand(array);
}
for(var i = 0; i < 9; i++) {
console.log(get_rand(nums));
}
The most effective and efficient way to do this is to shuffle your numbers then print the first nine of them. Use a good shuffle algorithm.What Thilo suggested will give you poor results. See here.
Edit
Here's a brief Knuth Shuffle algorithm example:
void shuffle(vector<int> nums)
{
for (int i = nums.size()-1; i >= 0; i--)
{
// this line is really shorthand, but gets the point across, I hope.
swap(nums[i],nums[rand()%i]);
}
}
Try this once:
//Here o is the array;
var testArr = [6, 7, 12, 15, 17, 20, 21];
shuffle = function(o){ //v1.0
for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
return o;
};
shuffle(testArr);
This is relatively simple to do, the theory behind it is creating another array which keeps track of which elements of the array you have used.
var tempArray = new Array(12),i,r;
for (i=0;i<9;i++)
{
r = Math.floor(Math.random()*12); // Get a random index
if (tempArray[r] === undefined) // If the index hasn't been used yet
{
document.write(numberArray[r]); // Display it
tempArray[r] = true; // Flag it as have been used
}
else // Otherwise
{
i--; // Try again
}
}
Other methods include shuffling the array, removing used elements from the array, or moving used elements to the end of the array.
If I understand you correctly, you want to shuffle your array.
Loop a couple of times (length of array should do), and in every iteration, get two random array indexes and swap the two elements there. (Update: if you are really serious about this, this may not be the best algorithm).
You can then print the first nine array elements, which will be in random order and not repeat.
Here is a generic way of getting random numbers between min and max without duplicates:
function inArray(arr, el) {
for(var i = 0 ; i < arr.length; i++)
if(arr[i] == el) return true;
return false;
}
function getRandomIntNoDuplicates(min, max, DuplicateArr) {
var RandomInt = Math.floor(Math.random() * (max - min + 1)) + min;
if (DuplicateArr.length > (max-min) ) return false; // break endless recursion
if(!inArray(DuplicateArr, RandomInt)) {
DuplicateArr.push(RandomInt);
return RandomInt;
}
return getRandomIntNoDuplicates(min, max, DuplicateArr); //recurse
}
call with:
var duplicates =[];
for (var i = 1; i <= 6 ; i++) {
console.log(getRandomIntNoDuplicates(1,10,duplicates));
}
const nums = [1,2,3,4,5,6,7,8,9,10,11,12];
for(var i = 1 ; i < 10; i++){
result = nums[Math.floor(Math.random()*nums.length)];
const index = nums.indexOf(result);
nums.splice(index, 1);
console.log(i+' - '+result);
}
I've been trying to find a O(n) solution to the following problem: Find the number of anagrams (permutations) of string s in string b, where s.length will always be smaller than b.length
I read that the optimal solution involves keeping track of the frequencies of the characters in the smaller string and doing the same for the sliding window as it moves across the larger string, but I'm not sure how that implementation actually works. Right now my solution doesn't work (see comments) but even if it did, it would take O(s + sn) time.
EDIT: Sample input: ('aba', 'abaab'). Output: 3, because 'aba' exists in b starting at index 0, and 'baa' at 1, and 'aab' at 2.
function anagramsInStr(s,b) {
//O(s)
let freq = s.split("").reduce((map, el) => {
map[el] = (map[el] + 1) || 1;
return map;
}, {});
let i = 0, j = s.length;
// O(n)
for (let char in b.split("")) {
// O(s)
if (b.length - char + 1 > s.length) {
let window = b.slice(i,j);
let windowFreq = window.split("").reduce((map, el) => {
map[el] = (map[el] + 1) || 1;
return map;
}, {});
// Somewhere about here compare the frequencies of chars found in the window to the frequencies hash defined in the outer scope.
i++;
j++;
}
}
}
Read through the comments and let me know if you have any questions:
function countAnagramOccurrences(s, b) {
var matchCount = 0;
var sCounts = {}; // counts for the letters in s
var bCounts = {}; // counts for the letters in b
// construct sCounts
for (var i = 0; i < s.length; i++) {
sCounts[s[i]] = (sCounts[s[i]] || 0) + 1;
}
// all letters that occur in sCounts
var letters = Object.keys(sCounts);
// for each letter in b
for (var i = 0; i < b.length; i++) {
// maintain a sliding window
// if we already have s.length items in the counts, remove the oldest one
if (i >= s.length) {
bCounts[b[i-s.length]] -= 1;
}
// increment the count for the letter we're currently looking at
bCounts[b[i]] = (bCounts[b[i]] || 0) + 1;
// test for a match (b counts == s counts)
var match = true;
for (var j = 0; j < letters.length; j++) {
if (sCounts[letters[j]] !== bCounts[letters[j]]) {
match = false;
break;
}
}
if (match) {
matchCount += 1;
}
}
return matchCount;
}
console.log(countAnagramOccurrences('aba', 'abaab')); // 3
EDIT
A note about the runtime: this is sort of O(nk + m), where n is the length of s, m is the length of b, and k is the number of unique characters in b. Since m is always less than n, we can reduce to O(nk), and since k is bounded by a fixed constant (the size of the alphabet), we can further reduce to O(n).
If I have n balls and k containers then this -> ( (n+k-1)! / n!(k-1)! ) will work out how many combinations there are.
I am having difficulty changing this to produce a list of all combinations in javascript.
In a function taking an array of balls and some amount of containers.
combinations([1,2,3,4,5,6], 3)
Each container can have any number of balls and containers can be empty.
Here is something i attempted but im only getting one ball in each container.
function generateCombinations(array, r, callback) {
function equal(a, b) {
for (var i = 0; i < a.length; i++) {
if (a[i] != b[i]) return false;
}
return true;
}
function values(i, a) {
var ret = [];
for (var j = 0; j < i.length; j++) ret.push(a[i[j]]);
return ret;
}
var n = array.length;
var indices = [];
for (var i = 0; i < r; i++) indices.push(i);
var final = [];
for (var i = n - r; i < n; i++) final.push(i);
while (!equal(indices, final)) {
callback(values(indices, array));
var i = r - 1;
while (indices[i] == n - r + i) i -= 1;
indices[i] += 1;
for (var j = i + 1; j < r; j++) indices[j] = indices[i] + j - i;
}
callback(values(indices, array));
}
count = 0
generateCombinations([1,2,3,4,5,6,7,8,9,1],3,function(first){
$("#hello").append(first+"<br />")
count = count +1
})
$("#hello").append(count)
You can do it in this way:
var containers = [];
// n - number of balls, k - number of containers
function dfs(n, k) {
// Ending point of recursion, all balls are placed
if(n == 0) {
var output = [];
for(var i = 0; i < k; i++) {
output.push('{' + containers[i].join(', ') + '}');
}
output = '[' + output.join(', ') + ']';
console.log(output);
return;
}
// Try to put ball #n
for(var i = 0; i < k; i++) {
containers[i].push(n);
// Now we have placed ball #n, so we have 1 .. n - 1 balls only
dfs(n - 1, k);
// Remove ball when back to use again
containers[i].pop();
}
}
var n = 4;
var k = 3;
for(var i = 0; i < k; i++) {
containers[i] = [];
}
dfs(n, k);
I initially thought you wanted all the combinations of k elements out of n, but your problem is different, it's partitioning n elements in k parts.
When going through the elements, at each steps, you may choose to put the current element in any container, that's k possibilities. In total, you will have kn possible solutions.
Therefore, it would be faster to iterate through all the solutions, rather than storing them in an array.
You can represent a solution as a unique number in base k, with n digits, and iterate through the solutions by incrementing that number.
In your example, the base is 3, and the number of digits is 6. The first solution is to put all the balls in container 0, ie.
000000
The next solution is to put all the balls in container 0, excepted the last which goes in container 1.
000001
...
000002
000010
000011
000020
Hopefully you should get the idea.
Closed. This question is opinion-based. It is not currently accepting answers.
Want to improve this question? Update the question so it can be answered with facts and citations by editing this post.
Closed 6 months ago.
Improve this question
What's the best way to break from nested loops in Javascript?
//Write the links to the page.
for (var x = 0; x < Args.length; x++)
{
for (var Heading in Navigation.Headings)
{
for (var Item in Navigation.Headings[Heading])
{
if (Args[x] == Navigation.Headings[Heading][Item].Name)
{
document.write("<a href=\""
+ Navigation.Headings[Heading][Item].URL + "\">"
+ Navigation.Headings[Heading][Item].Name + "</a> : ");
break; // <---HERE, I need to break out of two loops.
}
}
}
}
Just like Perl,
loop1:
for (var i in set1) {
loop2:
for (var j in set2) {
loop3:
for (var k in set3) {
break loop2; // breaks out of loop3 and loop2
}
}
}
as defined in EMCA-262 section 12.12. [MDN Docs]
Unlike C, these labels can only be used for continue and break, as Javascript does not have goto.
Wrap that up in a function and then just return.
I'm a little late to the party but the following is a language-agnostic approach which doesn't use GOTO/labels or function wrapping:
for (var x = Set1.length; x > 0; x--)
{
for (var y = Set2.length; y > 0; y--)
{
for (var z = Set3.length; z > 0; z--)
{
z = y = -1; // terminates second loop
// z = y = x = -1; // terminate first loop
}
}
}
On the upside it flows naturally which should please the non-GOTO crowd. On the downside, the inner loop needs to complete the current iteration before terminating so it might not be applicable in some scenarios.
I realize this is a really old topic, but since my standard approach is not here yet, I thought I post it for the future googlers.
var a, b, abort = false;
for (a = 0; a < 10 && !abort; a++) {
for (b = 0; b < 10 && !abort; b++) {
if (condition) {
doSomeThing();
abort = true;
}
}
}
Quite simple:
var a = [1, 2, 3];
var b = [4, 5, 6];
var breakCheck1 = false;
for (var i in a) {
for (var j in b) {
breakCheck1 = true;
break;
}
if (breakCheck1) break;
}
Here are five ways to break out of nested loops in JavaScript:
1) Set parent(s) loop to the end
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
if (j === 2)
{
i = 5;
break;
}
}
}
2) Use label
exit_loops:
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
if (j === 2)
break exit_loops;
}
}
3) Use variable
var exit_loops = false;
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
if (j === 2)
{
exit_loops = true;
break;
}
}
if (exit_loops)
break;
}
4) Use self executing function
(function()
{
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
if (j === 2)
return;
}
}
})();
5) Use regular function
function nested_loops()
{
for (i = 0; i < 5; i++)
{
for (j = 0; j < 5; j++)
{
if (j === 2)
return;
}
}
}
nested_loops();
var str = "";
for (var x = 0; x < 3; x++) {
(function() { // here's an anonymous function
for (var y = 0; y < 3; y++) {
for (var z = 0; z < 3; z++) {
// you have access to 'x' because of closures
str += "x=" + x + " y=" + y + " z=" + z + "<br />";
if (x == z && z == 2) {
return;
}
}
}
})(); // here, you execute your anonymous function
}
How's that? :)
How about using no breaks at all, no abort flags, and no extra condition checks. This version just blasts the loop variables (makes them Number.MAX_VALUE) when the condition is met and forces all the loops to terminate elegantly.
// No breaks needed
for (var i = 0; i < 10; i++) {
for (var j = 0; j < 10; j++) {
if (condition) {
console.log("condition met");
i = j = Number.MAX_VALUE; // Blast the loop variables
}
}
}
There was a similar-ish answer for decrementing-type nested loops, but this works for incrementing-type nested loops without needing to consider each loop's termination value for simple loops.
Another example:
// No breaks needed
for (var i = 0; i < 89; i++) {
for (var j = 0; j < 1002; j++) {
for (var k = 0; k < 16; k++) {
for (var l = 0; l < 2382; l++) {
if (condition) {
console.log("condition met");
i = j = k = l = Number.MAX_VALUE; // Blast the loop variables
}
}
}
}
}
If you use Coffeescript, there is a convenient "do" keyword that makes it easier to define and immediately execute an anonymous function:
do ->
for a in first_loop
for b in second_loop
if condition(...)
return
...so you can simply use "return" to get out of the loops.
I thought I'd show a functional-programming approach. You can break out of nested Array.prototype.some() and/or Array.prototype.every() functions, as in my solutions. An added benefit of this approach is that Object.keys() enumerates only an object's own enumerable properties, whereas "a for-in loop enumerates properties in the prototype chain as well".
Close to the OP's solution:
Args.forEach(function (arg) {
// This guard is not necessary,
// since writing an empty string to document would not change it.
if (!getAnchorTag(arg))
return;
document.write(getAnchorTag(arg));
});
function getAnchorTag (name) {
var res = '';
Object.keys(Navigation.Headings).some(function (Heading) {
return Object.keys(Navigation.Headings[Heading]).some(function (Item) {
if (name == Navigation.Headings[Heading][Item].Name) {
res = ("<a href=\""
+ Navigation.Headings[Heading][Item].URL + "\">"
+ Navigation.Headings[Heading][Item].Name + "</a> : ");
return true;
}
});
});
return res;
}
Solution that reduces iterating over the Headings/Items:
var remainingArgs = Args.slice(0);
Object.keys(Navigation.Headings).some(function (Heading) {
return Object.keys(Navigation.Headings[Heading]).some(function (Item) {
var i = remainingArgs.indexOf(Navigation.Headings[Heading][Item].Name);
if (i === -1)
return;
document.write("<a href=\""
+ Navigation.Headings[Heading][Item].URL + "\">"
+ Navigation.Headings[Heading][Item].Name + "</a> : ");
remainingArgs.splice(i, 1);
if (remainingArgs.length === 0)
return true;
}
});
});
How about pushing loops to their end limits
for(var a=0; a<data_a.length; a++){
for(var b=0; b<data_b.length; b++){
for(var c=0; c<data_c.length; c++){
for(var d=0; d<data_d.length; d++){
a = data_a.length;
b = data_b.length;
c = data_b.length;
d = data_d.length;
}
}
}
}
Already mentioned previously by swilliams, but with an example below (Javascript):
// Function wrapping inner for loop
function CriteriaMatch(record, criteria) {
for (var k in criteria) {
if (!(k in record))
return false;
if (record[k] != criteria[k])
return false;
}
return true;
}
// Outer for loop implementing continue if inner for loop returns false
var result = [];
for (var i = 0; i < _table.length; i++) {
var r = _table[i];
if (!CriteriaMatch(r[i], criteria))
continue;
result.add(r);
}
There are many excellent solutions above.
IMO, if your break conditions are exceptions,
you can use try-catch:
try{
for (var i in set1) {
for (var j in set2) {
for (var k in set3) {
throw error;
}
}
}
}catch (error) {
}
Hmmm hi to the 10 years old party ?
Why not put some condition in your for ?
var condition = true
for (var i = 0 ; i < Args.length && condition ; i++) {
for (var j = 0 ; j < Args[i].length && condition ; j++) {
if (Args[i].obj[j] == "[condition]") {
condition = false
}
}
}
Like this you stop when you want
In my case, using Typescript, we can use some() which go through the array and stop when condition is met
So my code become like this :
Args.some((listObj) => {
return listObj.some((obj) => {
return !(obj == "[condition]")
})
})
Like this, the loop stopped right after the condition is met
Reminder : This code run in TypeScript
Assign the values which are in comparison condition
function test(){
for(var i=0;i<10;i++)
{
for(var j=0;j<10;j++)
{
if(somecondition)
{
//code to Break out of both loops here
i=10;
j=10;
}
}
}
//Continue from here
}
An example with for .. of, close to the example further up which checks for the abort condition:
test()
function test() {
var arr = [1, 2, 3,]
var abort = false;
for (var elem of arr) {
console.log(1, elem)
for (var elem2 of arr) {
if (elem2 == 2) abort = true;
if (!abort) {
console.log(2, elem2)
}
}
}
}
Condition 1 - outer loop - will always run
The top voted and accepted answer also works for this kind of for loop.
Result: the inner loop will run once as expected
1 1
2 1
1 2
1 3
XXX.Validation = function() {
var ok = false;
loop:
do {
for (...) {
while (...) {
if (...) {
break loop; // Exist the outermost do-while loop
}
if (...) {
continue; // skips current iteration in the while loop
}
}
}
if (...) {
break loop;
}
if (...) {
break loop;
}
if (...) {
break loop;
}
if (...) {
break loop;
}
ok = true;
break;
} while(true);
CleanupAndCallbackBeforeReturning(ok);
return ok;
};
the best way is -
1) Sort the both array which are used in first and second loop.
2) if item matched then break the inner loop and hold the index value.
3) when start next iteration start inner loop with hold index value.