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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.
I'm a new javascript developer from lua and I have some confusion about arrays. I'm trying to build a simple 2d array but after the initialization I keep getting an error that the array is "undefined"
here's the code :
var board = [];
function initBoard(){
for (var i = 0; i < 8; i++){
board.push([]);
for (var j = 0 ;i < 8; i++){
board[j].push([]);
}
}
}
function checkSquare(x, y){
if (typeof(board[x][y]) === ""){
return false;
} else {
return true;
}
}
initBoard();
console.log(checkSquare(3, 3));
Here's the error : Cannot read property '3' of undefined`
You need not only take a look to the loops, but also to the check of the value of an item of the array. The comparison with the result of typeof with an empty string is always false, because there is no data type in Javascript Which is an empty string.
For comparing the value, you could check with the value directly with a Identity/strict equality operator ===. This checks the type of the left and right side and the value as well. For objects, it check if the object has the same reference.
function initBoard() {
var board = [];
for (var i = 0; i < 8; i++) {
board.push([]);
for (var j = 0; j < 8; j++) {
board[i].push('');
}
}
return board;
}
function checkSquare(x, y) {
if (board[x][y] === '') { // check if the item is an empty string
return false;
} else {
return true;
}
}
var board = initBoard();
console.log(checkSquare(3, 3));
There's a little mistale you've made while initialising. You've misplaced j with i. Try this:
var board = [];
function initBoard(){
for (var i = 0; i < 8; i++){
board.push([]);
for (var j = 0 ;j< i; j++){ //There was a mistake here
board[j].push([]);
}
}
}
function checkSquare(x, y){
if (typeof(board[x][y]) === ""){
return false;
} else {
return true;
}
}
initBoard();
console.log(checkSquare(3, 3));
var board = [];
function initBoard(){
for (var i = 0; i < 8; i++){
board.push([]);
for (var j = 0 ;j< i; j++){
board[j].push([]);
}
}
}
function checkSquare(x, y){
if (typeof(board[x][y]) === ""){
return false;
} else {
return true;
}
}
initBoard();
console.log(checkSquare(3, 3));
Two main things, you have a typo in your second loop, it should be based on j not i
and when you are trying to initialize the second array you can't use push because board[j] is undefined and push is a method of an array
var board = [];
function initBoard(){
for (var i = 0; i < 8; i++){
board.push([]);
for (var j = 0 ;j < 8; j++){
board[j] = [];
}
}
}
function checkSquare(x, y){
if (typeof(board[x][y]) === ""){
return false;
} else {
return true;
}
}
initBoard();
console.log(checkSquare(3, 3));
Very new, trying to make a function that takes out and separates all negatives/positives/zeros in an array. So far Ive been able to make an acceptable for loop but only with hard coded numbers. Dont currently know how to convert it into a function. please help.
var arr=[1,3,5,-9,-3,0];
var new_arr = [];
var new_arr2 = [];
var new_arr3=[];
for(i =0; i < arr.length; i++){
if(arr[i]>0){
new_arr.push(arr[i]);
}
else if(arr[i]<0){
new_arr2.push(arr[i]);
}
else if(arr[i]===0){
new_arr3.push(arr[i]);
}
}
console.log(new_arr3.length/arr.length);
console.log(new_arr2.length/arr.length);
console.log(new_arr.length/arr.length);
How about something like this?
function division(arr) {
var new_arr = [];
var new_arr2 = [];
var new_arr3 = [];
for (i = 0; i < arr.length; i++) {
if (arr[i] > 0) {
new_arr.push(arr[i]);
} else if (arr[i] < 0) {
new_arr2.push(arr[i]);
} else if (arr[i] === 0) {
new_arr3.push(arr[i]);
}
}
console.log(new_arr3.length / arr.length);
console.log(new_arr2.length / arr.length);
console.log(new_arr.length / arr.length);
}
division([1, 3, 5, -9, -3, 0]);
This new function takes the array as a parameter, so all you need to do is call it and pass the array.
You can try this way also, here I am considering 0 as a positive number. If you want, you can tweak the condition as per requirement.
function positive_negative(array ){
positive = array.filter(function (a) { return a >= 0; });
negative = array.filter(function (a) { return a < 0; });
return [positive,negative];
}
var array = [1,3,5,-9,-3,0];
console.log(positive_negative(array));
I was wondering if there is a way to check for repeated characters in a string without using double loop. Can this be done with recursion?
An example of the code using double loop (return true or false based on if there are repeated characters in a string):
var charRepeats = function(str) {
for(var i = 0; i <= str.length; i++) {
for(var j = i+1; j <= str.length; j++) {
if(str[j] == str[i]) {
return false;
}
}
}
return true;
}
Many thanks in advance!
This will do:
function hasRepeats (str) {
return /(.).*\1/.test(str);
}
(A recursive solution can be found at the end of this answer)
You could simply use the builtin javascript Array functions some MDN some reference
var text = "test".split("");
text.some(function(v,i,a){
return a.lastIndexOf(v)!=i;
});
callback parameters:
v ... current value of the iteration
i ... current index of the iteration
a ... array being iterated
.split("") create an array from a string
.some(function(v,i,a){ ... }) goes through an array until the function returns true, and ends than right away. (it doesn't loop through the whole array, which is good for performance)
Details to the some function here in the documentation
Here some tests, with several different strings:
var texts = ["test", "rest", "why", "puss"];
for(var idx in texts){
var text = texts[idx].split("");
document.write(text + " -> " + text.some(function(v,i,a){return a.lastIndexOf(v)!=i;}) +"<br/>");
}
//tested on win7 in chrome 46+
If you will want recursion.
Update for recursion:
//recursive function
function checkString(text,index){
if((text.length - index)==0 ){ //stop condition
return false;
}else{
return checkString(text,index + 1)
|| text.substr(0, index).indexOf(text[index])!=-1;
}
}
// example Data to test
var texts = ["test", "rest", "why", "puss"];
for(var idx in texts){
var txt = texts[idx];
document.write( txt + " ->" + checkString(txt,0) + "<br/>");
}
//tested on win7 in chrome 46+
you can use .indexOf() and .lastIndexOf() to determine if an index is repeated. Meaning, if the first occurrence of the character is also the last occurrence, then you know it doesn't repeat. If not true, then it does repeat.
var example = 'hello';
var charRepeats = function(str) {
for (var i=0; i<str.length; i++) {
if ( str.indexOf(str[i]) !== str.lastIndexOf(str[i]) ) {
return false; // repeats
}
}
return true;
}
console.log( charRepeats(example) ); // 'false', because when it hits 'l', the indexOf and lastIndexOf are not the same.
function chkRepeat(word) {
var wordLower = word.toLowerCase();
var wordSet = new Set(wordLower);
var lenWord = wordLower.length;
var lenWordSet =wordSet.size;
if (lenWord === lenWordSet) {
return "false"
} else {
return'true'
}
}
Using regex to solve=>
function isIsogram(str){
return !/(\w).*\1/i.test(str);
}
console.log(isIsogram("isogram"), true );
console.log(isIsogram("aba"), false, "same chars may not be adjacent" );
console.log(isIsogram("moOse"), false, "same chars may not be same case" );
console.log(isIsogram("isIsogram"), false );
console.log(isIsogram(""), true, "an empty string is a valid isogram" );
The algorithm presented has a complexity of (1 + n - (1)) + (1 + n - (2)) + (1 + n - (3)) + ... + (1 + n - (n-1)) = (n-1)*(1 + n) - (n)(n-1)/2 = (n^2 + n - 2)/2 which is O(n2).
So it would be better to use an object to map and remember the characters to check for uniqueness or duplicates. Assuming a maximum data size for each character, this process will be an O(n) algorithm.
function charUnique(s) {
var r = {}, i, x;
for (i=0; i<s.length; i++) {
x = s[i];
if (r[x])
return false;
r[x] = true;
}
return true;
}
On a tiny test case, the function indeed runs a few times faster.
Note that JavaScript strings are defined as sequences of 16-bit unsigned integer values. http://bclary.com/2004/11/07/#a-4.3.16
Hence, we can still implement the same basic algorithm but using a much quicker array lookup rather than an object lookup. The result is approximately 100 times faster now.
var charRepeats = function(str) {
for (var i = 0; i <= str.length; i++) {
for (var j = i + 1; j <= str.length; j++) {
if (str[j] == str[i]) {
return false;
}
}
}
return true;
}
function charUnique(s) {
var r = {},
i, x;
for (i = 0; i < s.length; i++) {
x = s[i];
if (r[x])
return false;
r[x] = true;
}
return true;
}
function charUnique2(s) {
var r = {},
i, x;
for (i = s.length - 1; i > -1; i--) {
x = s[i];
if (r[x])
return false;
r[x] = true;
}
return true;
}
function charCodeUnique(s) {
var r = [],
i, x;
for (i = s.length - 1; i > -1; i--) {
x = s.charCodeAt(i);
if (r[x])
return false;
r[x] = true;
}
return true;
}
function regExpWay(s) {
return /(.).*\1/.test(s);
}
function timer(f) {
var i;
var t0;
var string = [];
for (i = 32; i < 127; i++)
string[string.length] = String.fromCharCode(i);
string = string.join('');
t0 = new Date();
for (i = 0; i < 10000; i++)
f(string);
return (new Date()) - t0;
}
document.write('O(n^2) = ',
timer(charRepeats), ';<br>O(n) = ',
timer(charUnique), ';<br>optimized O(n) = ',
timer(charUnique2), ';<br>more optimized O(n) = ',
timer(charCodeUnique), ';<br>regular expression way = ',
timer(regExpWay));
let myString = "Haammmzzzaaa";
myString = myString
.split("")
.filter((item, index, array) => array.indexOf(item) === index)
.join("");
console.log(myString); // "Hamza"
Another way of doing it using lodash
var _ = require("lodash");
var inputString = "HelLoo world!"
var checkRepeatition = function(inputString) {
let unique = _.uniq(inputString).join('');
if(inputString.length !== unique.length) {
return true; //duplicate characters present!
}
return false;
};
console.log(checkRepeatition(inputString.toLowerCase()));
const str = "afewreociwddwjej";
const repeatedChar=(str)=>{
const result = [];
const strArr = str.toLowerCase().split("").sort().join("").match(/(.)\1+/g);
if (strArr != null) {
strArr.forEach((elem) => {
result.push(elem[0]);
});
}
return result;
}
console.log(...repeatedChar(str));
You can also use the following code to find the repeated character in a string
//Finds character which are repeating in a string
var sample = "success";
function repeatFinder(str) {
let repeat="";
for (let i = 0; i < str.length; i++) {
for (let j = i + 1; j < str.length; j++) {
if (str.charAt(i) == str.charAt(j) && repeat.indexOf(str.charAt(j)) == -1) {
repeat += str.charAt(i);
}
}
}
return repeat;
}
console.log(repeatFinder(sample)); //output: sc
const checkRepeats = (str: string) => {
const arr = str.split('')
const obj: any = {}
for (let i = 0; i < arr.length; i++) {
if (obj[arr[i]]) {
return true
}
obj[arr[i]] = true
}
return false
}
console.log(checkRepeats('abcdea'))
function repeat(str){
let h =new Set()
for(let i=0;i<str.length-1;i++){
let a=str[i]
if(h.has(a)){
console.log(a)
}else{
h.add(a)
}
}
return 0
}
let str = '
function repeat(str){
let h =new Set()
for(let i=0;i<str.length-1;i++){
let a=str[i]
if(h.has(a)){
console.log(a)
}else{
h.add(a)
}
}
return 0
}
let str = 'haiiiiiiiiii'
console.log(repeat(str))
'
console.log(repeat(str))
Cleanest way for me:
Convert the string to an array
Make a set from the array
Compare the length of the set and the array
Example function:
function checkDuplicates(str) {
const strArray = str.split('');
if (strArray.length !== new Set(strArray).size) {
return true;
}
return false;
}
You can use "Set object"!
The Set object lets you store unique values of any type, whether
primitive values or object references. It has some methods to add or to check if a property exist in the object.
Read more about Sets at MDN
Here how i use it:
function isIsogram(str){
let obj = new Set();
for(let i = 0; i < str.length; i++){
if(obj.has(str[i])){
return false
}else{
obj.add(str[i])
}
}
return true
}
isIsogram("Dermatoglyphics") // true
isIsogram("aba")// false
Here is one to get your brain going! I've not had any luck with it.
[1,2,1,1,2,1,1,1,2,2]
[1,2,1,1,2,1]
I would like to use the second array to find the values in the first, but they must be in the same order.
Once for I would like it to return the next key up from the last key in the second array.
So in this example it would use the first six digits in the first array and then return 6 as the key after the final one in the second array.
var a2 = [1,2,1,1,2,1,1,1,2,2]
var a1 = [1,2,1,1,0,1]
function find(arr1, arr2) {
var len = 1
var result = 0;
var s2 = arr2.toString();
for (len=1;len <= a1.length; len++)
{
var aa1 = arr1.slice(0, len)
var s1 = aa1.toString();
if(s2.indexOf(s1)>=0){
result = aa1.length;
}
else {
break;
}
}
return result;
}
alert(find(a1, a2));
var find = function(haystack, needle) {
var doesMatch = function(offset) {
for (var i = 0; i < needle.length; i++) {
if (haystack[i+offset] !== needle[i]) {
return false;
}
}
return true;
};
for (var j=0; j < haystack.length - needle.length; j++) {
if (doesMatch(j)) {
return j;
}
}
return -1;
};
This is quick, this is dirty, and this is correct only if your data doesn't include any comma.
var needle = [1,2,1,1,2,1];
var haystack = [1,2,1,1,2,1,1,1,2,2];
if ( needle.length <= 0 ) return 0;
var fromStr = ','+haystack.toString()+','
var findStr = ','+needle.toString()+','
// Find ',1,2,1,1,2,1,' in ',1,2,1,1,2,1,1,1,2,2,'
var pos = fromStr.indexOf(findStr);
// Count the end position requested
return pos >= 0 ? fromStr.slice(0,pos+1).match(/,/g).length + needle.length - 1 : -1;
Note: The comma at head and tail is to make sure [22,12] doesn't match [2,1].