Fast stable sorting algorithm implementation in javascript
Asked Answered
Y

16

110

I'm looking to sort an array of about 200-300 objects, sorting on a specific key and a given order (asc/desc). The order of results must be consistent and stable.

What would be the best algorithm to use, and could you provide an example of it's implementation in javascript?

Thanks!

Younker answered 15/9, 2009 at 14:40 Comment(8)
Since at least Chrome's array sort doesn't seem to be stable, relying on the built-in array sort is not an option for you.Greathearted
To summarize: I went with a hand rolled merge sort due to Array.sort stability inconsistencies between modern browsers (mainly chrome not implementing a stable sort at the time of this comment). Thanks everyone for your help!Younker
What do we mean by "stable" sort?Brookbrooke
@Brookbrooke Stable sort is a sort in which all of the items with the same sorting value are left in the same order as in the original collection. en.wikipedia.org/wiki/Sorting_algorithm#StabilityStalag
To answer "what is the best algorithm to use" we need to know if there is any underlying structure to your data. A lot of the answers below just talk about using merge sort, or quick sort, in reality it depends on the data. It's not a simple problem to just answer i wouldn't say. Google a few sorting algorithms and read about them to see what i mean. TimSort and Radix Sort are two good examples i'd reccomend reading about.Demeter
Please refer the following link. khan4019.github.io/front-end-Interview-Questions/sort.htmlInguinal
You can use lodash's _.sortBy() which is stable.Roughneck
With the exception of Internet Explorer, Array.sort is stable in all browsers as of November 2018 as mandated by the standard. Browser compatibilityTrypanosomiasis
P
115

It is possible to get a stable sorting from a non-stable sort function.

Before sorting you get the position of all the elements. In your sort condition, if both elements are equal, then you sort by the position.

Tada! You've got a stable sort.

I've written an article about it on my blog if you want to know more about this technique and how to implement it: http://blog.vjeux.com/2010/javascript/javascript-sorting-table.html

Paulus answered 18/1, 2010 at 10:13 Comment(4)
Could you please provide the answer here instead of posting and linking to your blog! ThanksMegillah
A link to a solution is welcome, but please ensure your answer is useful without it: add context around the link so your fellow users will have some idea what it is and why it’s there, then quote the most relevant part of the page you're linking to in case the target page is unavailable. Answers that are little more than a link may be deleted.Sumach
@Paulus since this is getting to be a popular, do you mind pasting the relevant code into this answer? It would help a lot! thanks!Younker
Array.sort is stable in all browsers as of November 2018 as mandated by the standard. Browser compatibilityTrypanosomiasis
M
34

Since you are looking for something stable, the merge sort should do.

http://www.stoimen.com/blog/2010/07/02/friday-algorithms-javascript-merge-sort/

The code can be found at the above website:

function mergeSort(arr)
{
    if (arr.length < 2)
        return arr;

    var middle = parseInt(arr.length / 2);
    var left   = arr.slice(0, middle);
    var right  = arr.slice(middle, arr.length);

    return merge(mergeSort(left), mergeSort(right));
}

function merge(left, right)
{
    var result = [];

    while (left.length && right.length) {
        if (left[0] <= right[0]) {
            result.push(left.shift());
        } else {
            result.push(right.shift());
        }
    }

    while (left.length)
        result.push(left.shift());

    while (right.length)
        result.push(right.shift());

    return result;
}

EDIT:

According to this post, it looks like Array.Sort in some implementations uses a merge sort.

Maliamalice answered 15/9, 2009 at 14:42 Comment(4)
++ Merge sort is my favorite. It's simple and stable with no bad worst cases.Counterspy
The link to the website is down :(Drinkwater
found a new website for the example.Maliamalice
Note: Array#shift may workn in O(n) time and so your merge in O(n*n).Hubie
E
27

Somewhat shorter version of the same thing using ES2017 features like arrow functions and destructuring:

Function

var stableSort = (arr, compare) => arr
  .map((item, index) => ({item, index}))
  .sort((a, b) => compare(a.item, b.item) || a.index - b.index)
  .map(({item}) => item)

It accepts input array and compare function:

stableSort([5,6,3,2,1], (a, b) => a - b)

It also returns new array instead of making in-place sort like the built-in Array.sort() function.

Test

If we take the following input array, initially sorted by weight:

// sorted by weight
var input = [
  { height: 100, weight: 80 },
  { height: 90, weight: 90 },
  { height: 70, weight: 95 },
  { height: 100, weight: 100 },
  { height: 80, weight: 110 },
  { height: 110, weight: 115 },
  { height: 100, weight: 120 },
  { height: 70, weight: 125 },
  { height: 70, weight: 130 },
  { height: 100, weight: 135 },
  { height: 75, weight: 140 },
  { height: 70, weight: 140 }
]

Then sort it by height using stableSort:

stableSort(input, (a, b) => a.height - b.height)

Results in:

// Items with the same height are still sorted by weight 
// which means they preserved their relative order.
var stable = [
  { height: 70, weight: 95 },
  { height: 70, weight: 125 },
  { height: 70, weight: 130 },
  { height: 70, weight: 140 },
  { height: 75, weight: 140 },
  { height: 80, weight: 110 },
  { height: 90, weight: 90 },
  { height: 100, weight: 80 },
  { height: 100, weight: 100 },
  { height: 100, weight: 120 },
  { height: 100, weight: 135 },
  { height: 110, weight: 115 }
]

However sorting the same input array using the built-in Array.sort() (in Chrome/NodeJS):

input.sort((a, b) => a.height - b.height)

Returns:

var unstable = [
  { height: 70, weight: 140 },
  { height: 70, weight: 95 },
  { height: 70, weight: 125 },
  { height: 70, weight: 130 },
  { height: 75, weight: 140 },
  { height: 80, weight: 110 },
  { height: 90, weight: 90 },
  { height: 100, weight: 100 },
  { height: 100, weight: 80 },
  { height: 100, weight: 135 },
  { height: 100, weight: 120 },
  { height: 110, weight: 115 }
]

Resources

Update

Array.prototype.sort is now stable in V8 v7.0 / Chrome 70!

Previously, V8 used an unstable QuickSort for arrays with more than 10 elements. Now, we use the stable TimSort algorithm.

source

Enabling answered 7/2, 2018 at 9:46 Comment(1)
The stableSort function is a really great solution!Raposa
D
17

I know that this question has been answered for some time, but I happen to have a good stable merge sort implementation for Array and jQuery in my clipboard, so I'll share it in the hopes that some future searchers might find it useful.

It allows you to specify your own comparison function just like the normal Array.sort implementation.

Implementation

// Add stable merge sort to Array and jQuery prototypes
// Note: We wrap it in a closure so it doesn't pollute the global
//       namespace, but we don't put it in $(document).ready, since it's
//       not dependent on the DOM
(function() {

  // expose to Array and jQuery
  Array.prototype.mergeSort = jQuery.fn.mergeSort = mergeSort;

  function mergeSort(compare) {

    var length = this.length,
        middle = Math.floor(length / 2);

    if (!compare) {
      compare = function(left, right) {
        if (left < right)
          return -1;
        if (left == right)
          return 0;
        else
          return 1;
      };
    }

    if (length < 2)
      return this;

    return merge(
      this.slice(0, middle).mergeSort(compare),
      this.slice(middle, length).mergeSort(compare),
      compare
    );
  }

  function merge(left, right, compare) {

    var result = [];

    while (left.length > 0 || right.length > 0) {
      if (left.length > 0 && right.length > 0) {
        if (compare(left[0], right[0]) <= 0) {
          result.push(left[0]);
          left = left.slice(1);
        }
        else {
          result.push(right[0]);
          right = right.slice(1);
        }
      }
      else if (left.length > 0) {
        result.push(left[0]);
        left = left.slice(1);
      }
      else if (right.length > 0) {
        result.push(right[0]);
        right = right.slice(1);
      }
    }
    return result;
  }
})();

Example Usage

var sorted = [
  'Finger',
  'Sandwich',
  'sandwich',
  '5 pork rinds',
  'a guy named Steve',
  'some noodles',
  'mops and brooms',
  'Potato Chip Brand® chips'
].mergeSort(function(left, right) {
  lval = left.toLowerCase();
  rval = right.toLowerCase();

  console.log(lval, rval);
  if (lval < rval)
    return -1;
  else if (lval == rval)
    return 0;
  else
    return 1;
});

sorted == ["5 pork rinds", "a guy named Steve", "Finger", "mops and brooms", "Potato Chip Brand® chips", "Sandwich", "sandwich", "some noodles"];
Deangelo answered 11/10, 2011 at 18:11 Comment(7)
Note this is at odds with the native sort, which works in place, meaning this cannot just be dropped in.Row
Maybe stable, but this method is about 20 times slower than native array.sort, see test here for both strings and integers -> jsfiddle.net/QC64jApparatus
Of course it's slower than native sort—it's not native. That's impossible. It's also true that it doesn't do an in place sort. That's also impossible (best case you create a copy then overwrite the original). Besides, returning a sorted copy is more JavaScript-y despite JavaScript's own native sort behavior. The function is also called mergeSort and not sort, so it's not intended as a drop in replacement. Sometimes you just need a stable merge sort, e.g. when sorting tables by column.Deangelo
Wrong, Node's Native sort is written in javascript. Its entirely possible for an algorithm programmed in javascript to out-speed the native sort. I built a sorting algorithm entirely in javascript(a type of adaptive merge sort) that Kremes/creams/Kreams The native quicksort in node. The point of this comment is to show that native does not matter in the case of javascript because the sorting algorithm is written in javascript and not in some higher language such as c++. Proof is here: github.com/nodejs/node/blob/master/deps/v8/src/js/array.jsDrinkwater
@Drinkwater So we know that's true for V8 and Node. The implementation of .sort() is not guaranteed to be in JavaScript. The whole point here is that ECMAScript's sort is not guaranteed to be stable. Any given implementation of JavaScript may have a stable sort, but since it's not guaranteed by the spec you shouldn't rely on it. Similarly, the spec doesn't stipulate that .sort() be implemented in JavaScript. And some minor nitpicks: C++ is a lower level language than JavaScript; and your tone sucks.Deangelo
@JustinForce if Array#slice works in O(n) time your merge will work in O(n*n).Hubie
For anyone who wants an in-place, drop-in solution that's much faster than this implementation, check out my answer.Honorary
H
11

You can use the following function to perform a stable sort regardless of the native implementation, based on the assertion made in this answer.

Do note that as of ECMAScript 2019, the specification requires that the builtin sort() method perform a stable sort. With that in mind, an explicit stable sort function like the one below is still relevant if you are required to support older browsers that are not specification compliant.

// ECMAScript 5 implementation
function stableSort(array, compareFunction) {
  'use strict';

  var length = array.length;
  var indices = new Uint32Array(length);
  var i;
  var slice;

  // reference values by indices
  for (i = 0; i < length; ++i) {
    indices[i] = i;
  }

  // sort with fallback based on indices
  indices.sort(function stableCompareFunction(compareFunction, a, b) {
    var order = Number(compareFunction(this[a], this[b]));
    return order || a - b;
  }.bind(array, compareFunction));

  slice = array.slice();

  // re-order original array to stable sorted values
  for (i = 0; i < length; ++i) {
    array[i] = slice[indices[i]];
  }

  return array;
}

// usage
const array = Array(500000).fill().map(() => Number(Math.random().toFixed(4)));

const alwaysEqual = () => 0;
const isUnmoved = (value, index) => value === array[index];

// not guaranteed to be stable before ES2019
console.log(
  'sort() stable?',
  array.slice().sort(alwaysEqual).every(isUnmoved)
);
// guaranteed to be stable
console.log(
  'stableSort() stable?',
  stableSort(array.slice(), alwaysEqual).every(isUnmoved)
);

// performance using realistic scenario with unsorted big data
function time(arraySlice, algorithm, compare) {
  var start;
  var stop;

  start = performance.now();
  algorithm(arraySlice, compare);
  stop = performance.now();

  return stop - start;
}

const ascending = (a, b) => a - b;

const msSort = time(array.slice(), (array, compare) => array.sort(compare), ascending);
const msStableSort = time(array.slice(), (array, compare) => stableSort(array, compare), ascending);

console.log('sort()', msSort.toFixed(3), 'ms');
console.log('stableSort()', msStableSort.toFixed(3), 'ms');
console.log('sort() / stableSort()', (100 * msSort / msStableSort).toFixed(3) + '%');

Running the performance tests implemented above, stableSort() appears to run at about 72% of the speed of sort() on version 88 of Google Chrome and Microsoft Edge.

Using .bind() on the inline function within stableSort() used to boost relative performance significantly by avoiding unneeded scoped references on each call.

In practice, this no longer makes a difference since modern engines automatically perform this optimization now, but it is left in the implementation anyway in order to continue improving performance in older browsers which don't ship with this optimization.

Honorary answered 31/7, 2017 at 18:13 Comment(0)
L
5

The following sorts the supplied array, by applying the supplied compare function, returning the original index comparison when the compare function returns 0:

function stableSort(arr, compare) {
    var original = arr.slice(0);

    arr.sort(function(a, b){
        var result = compare(a, b);
        return result === 0 ? original.indexOf(a) - original.indexOf(b) : result;
    });

    return arr;
}

The example below sorts an array of names by surname, retaining the order of equal surnames:

var names = [
	{ surname: "Williams", firstname: "Mary" },
	{ surname: "Doe", firstname: "Mary" }, 
	{ surname: "Johnson", firstname: "Alan" }, 
	{ surname: "Doe", firstname: "John" }, 
	{ surname: "White", firstname: "John" }, 
	{ surname: "Doe", firstname: "Sam" }
]

function stableSort(arr, compare) {
    var original = arr.slice(0);

    arr.sort(function(a, b){
        var result = compare(a, b);
        return result === 0 ? original.indexOf(a) - original.indexOf(b) : result;
    });
	
    return arr;
}

stableSort(names, function(a, b) { 
	return a.surname > b.surname ? 1 : a.surname < b.surname ? -1 : 0;
})

names.forEach(function(name) {
	console.log(name.surname + ', ' + name.firstname);
});
Laughlin answered 8/12, 2017 at 12:14 Comment(1)
Not stable for primitive types or duplicate elements in array. jQuery.expando.split( "" ).sort( ( a, b ) => 0 ).join( "" ) === jQuery.expando Ramiform
S
3

Here's a stable implementation. It works by using the native sort, but in cases where elements compare as equal, you break ties using the original index position.

function stableSort(arr, cmpFunc) {
    //wrap the arr elements in wrapper objects, so we can associate them with their origional starting index position
    var arrOfWrapper = arr.map(function(elem, idx){
        return {elem: elem, idx: idx};
    });

    //sort the wrappers, breaking sorting ties by using their elements orig index position
    arrOfWrapper.sort(function(wrapperA, wrapperB){
        var cmpDiff = cmpFunc(wrapperA.elem, wrapperB.elem);
        return cmpDiff === 0 
             ? wrapperA.idx - wrapperB.idx
             : cmpDiff;
    });

    //unwrap and return the elements
    return arrOfWrapper.map(function(wrapper){
        return wrapper.elem;
    });
}

a non-thorough test

var res = stableSort([{a:1, b:4}, {a:1, b:5}], function(a, b){
    return a.a - b.a;
});
console.log(res);

another answer alluded to this, but didn't post teh codez.

but, its not fast according to my benchmark. I modified a merge sort impl to accept a custom comparator function, and it was much faster.

Spotted answered 24/12, 2013 at 23:41 Comment(3)
Are your benchmark correct? Seems, your "stableSort" does not modify input array, other sorts - do, and as you did not recreate "arr" during "setup", other sorts sort already sorted arrays....Hubie
@Hubie did you read wrong? I said my stableSort func was NOT fast.Spotted
@gota, ah, pardon meHubie
F
3

You can also use Timsort. This is a really complicated algorithm (400+ lines, hence no source code here), so see Wikipedia's description or use one of the existing JavaScript implementations:

GPL 3 implementation. Packaged as Array.prototype.timsort. Appears to be an exact rewrite of Java code.

Public domain implementation Meant as a tutorial, the sample code only shows its use with integers.

Timsort is a highly optimized hybrid of mergesort and shuffle sort and is the default sorting algorithm in Python and in Java (1.7+). It is a complicated algorithm, since it uses different algorithms for many special cases. But as a result it's extremely fast under a wide variety of circumstances.

Foretop answered 27/7, 2015 at 15:22 Comment(0)
I
1

A simple one mergeSort from http://www.stoimen.com/blog/2010/07/02/friday-algorithms-javascript-merge-sort/

var a = [34, 203, 3, 746, 200, 984, 198, 764, 9];

function mergeSort(arr)
{
    if (arr.length < 2)
         return arr;

    var middle = parseInt(arr.length / 2);
    var left   = arr.slice(0, middle);
    var right  = arr.slice(middle, arr.length);

    return merge(mergeSort(left), mergeSort(right));
}

function merge(left, right)
{
     var result = [];

    while (left.length && right.length) {
         if (left[0] <= right[0]) {
             result.push(left.shift());
         } else {
            result.push(right.shift());
         }
    }

    while (left.length)
        result.push(left.shift());

    while (right.length)
        result.push(right.shift());

    return result;
}

console.log(mergeSort(a));
Ingest answered 29/7, 2012 at 17:35 Comment(0)
K
0

I have to sort multidimensional arrays by an arbitrary column, and then by another. I use this function to sort:

function sortMDArrayByColumn(ary, sortColumn){

    //Adds a sequential number to each row of the array
    //This is the part that adds stability to the sort
    for(var x=0; x<ary.length; x++){ary[x].index = x;}

    ary.sort(function(a,b){
        if(a[sortColumn]>b[sortColumn]){return 1;}
        if(a[sortColumn]<b[sortColumn]){return -1;}
        if(a.index>b.index){
            return 1;
        }
        return -1;
    });
}

Notice that ary.sort never returns zero, which is where some implementations of the "sort" function make decisions that might not be right.

This is pretty darn fast, too.

Koweit answered 24/6, 2014 at 23:13 Comment(0)
D
0

Here's how you could extend JS default Array object with a prototype method utilizing MERGE SORT. This method allows for sorting on a specific key (first parameter) and a given order ('asc'/'desc' as second parameter)

Array.prototype.mergeSort = function(sortKey, direction){
  var unsortedArray = this;
  if(unsortedArray.length < 2) return unsortedArray;

  var middle = Math.floor(unsortedArray.length/2);
  var leftSubArray = unsortedArray.slice(0,middle).mergeSort(sortKey, direction);
  var rightSubArray = unsortedArray.slice(middle).mergeSort(sortKey, direction);

  var sortedArray = merge(leftSubArray, rightSubArray);
  return sortedArray;

  function merge(left, right) {
    var combined = [];
    while(left.length>0 && right.length>0){
      var leftValue = (sortKey ? left[0][sortKey] : left[0]);
      var rightValue = (sortKey ? right[0][sortKey] : right[0]);
      combined.push((direction === 'desc' ? leftValue > rightValue : leftValue < rightValue) ? left.shift() : right.shift())
    }
    return combined.concat(left.length ? left : right)
  }
}

You can test this out yourself by dropping the above snippet into your browser console, then trying:

var x = [2,76,23,545,67,-9,12];
x.mergeSort(); //[-9, 2, 12, 23, 67, 76, 545]
x.mergeSort(undefined, 'desc'); //[545, 76, 67, 23, 12, 2, -9]

Or order based on a specific field in an array of objects:

var y = [
  {startTime: 100, value: 'cat'},
  {startTime: 5, value: 'dog'},
  {startTime: 23, value: 'fish'},
  {startTime: 288, value: 'pikachu'}
]
y.mergeSort('startTime');
y.mergeSort('startTime', 'desc');
Dyann answered 13/7, 2016 at 18:37 Comment(0)
J
0

So I needed a stable sort for my React+Redux app, and Vjeux's answer here helped me. However, my (generic) solution seems different than the others I see here so far, so I'm sharing it in case anyone else has a matching use-case:

  • I really just want to have something similar to the sort() API, where I can pass a comparator function.
  • Sometimes I can sort in-place, and sometimes my data is immutable (because Redux) and I need a sorted copy instead. So I need a stable sorting function for each use-case.
  • ES2015.

My solution is to create a typed array of indices, then use a comparison function to sort these indices based on the to-be-sorted array. Then we can use the sorted indices to either sort the original array or create a sorted copy in a single pass. If that's confusing, think of it this way: where you would normally pass a comparison function like:

(a, b) => { 
  /* some way to compare a and b, returning -1, 0, or 1 */ 
};

You now instead use:

(i, j) => { 
  let a = arrayToBeSorted[i], b = arrayToBeSorted[j]; 
  /* some way to compare a and b, returning -1 or 1 */
  return i - j; // fallback when a == b
}

Speed is good; it is basically the built-in sorting algorithm is, plus two linear passes in the end, and one extra layer of pointer indirection overhead.

Happy to receive feedback on this approach. Here is my full implementation of it it:

/**
 * - `array`: array to be sorted
 * - `comparator`: closure that expects indices `i` and `j`, and then
 *   compares `array[i]` to `array[j]` in some way. To force stability,
 *   end with `i - j` as the last "comparison".
 * 
 * Example:
 * ```
 *  let array = [{n: 1, s: "b"}, {n: 1, s: "a"}, {n:0, s: "a"}];
 *  const comparator = (i, j) => {
 *    const ni = array[i].n, nj = array[j].n;
 *    return ni < nj ? -1 :
 *      ni > nj ? 1 :
 *        i - j;
 *  };
 *  stableSortInPlace(array, comparator);
 *  // ==> [{n:0, s: "a"}, {n:1, s: "b"}, {n:1, s: "a"}]
 * ```
 */
function stableSortInPlace(array, comparator) {
  return sortFromIndices(array, findIndices(array, comparator));
}

function stableSortedCopy(array, comparator){
  let indices = findIndices(array, comparator);
  let sortedArray = [];
  for (let i = 0; i < array.length; i++){
    sortedArray.push(array[indices[i]]);
  }
  return sortedArray;
}

function findIndices(array, comparator){
  // Assumes we don't have to worry about sorting more than 
  // 4 billion elements; if you know the upper bounds of your
  // input you could replace it with a smaller typed array
  let indices = new Uint32Array(array.length);
  for (let i = 0; i < indices.length; i++) {
    indices[i] = i;
  }
  // after sorting, `indices[i]` gives the index from where
  // `array[i]` should take the value from, so to sort
  // move the value at at `array[indices[i]]` to `array[i]`
  return indices.sort(comparator);
}

// If I'm not mistaken this is O(2n) - each value is moved
// only once (not counting the vacancy temporaries), and 
// we also walk through the whole array once more to check
// for each cycle.
function sortFromIndices(array, indices) {
  // there might be multiple cycles, so we must
  // walk through the whole array to check.
  for (let k = 0; k < array.length; k++) {
    // advance until we find a value in
    // the "wrong" position
    if (k !== indices[k]) {
      // create vacancy to use "half-swaps" trick,
      // props to Andrei Alexandrescu
      let v0 = array[k];
      let i = k;
      let j = indices[k];
      while (j !== k) {
        // half-swap next value
        array[i] = array[j];
        // array[i] now contains the value it should have,
        // so we update indices[i] to reflect this
        indices[i] = i;
        // go to next index
        i = j;
        j = indices[j];
      }
      // put original array[k] back in
      // and update indices
      array[i] = v0;
      indices[i] = i;
    }
  }
  return array;
}
Jestinejesting answered 15/12, 2016 at 15:8 Comment(0)
R
0

I know this has been plenty answered. I just wanted to go ahead an post a quick TS implementation for anyone that landed here looking for that.

export function stableSort<T>( array: T[], compareFn: ( a: T, b: T ) => number ): T[] {
    const indices = array.map( ( x: T, i: number ) => ( { element: x, index: i } ) );

    return indices.sort( ( a, b ) => {
        const order = compareFn( a.element, b.element );
        return order === 0 ? a.index - b.index : order;
    } ).map( x => x.element );
}

The method does no longer run in-place, as the native sort does. I also want to point out that it is not the most efficient. It adds two loops of the order O(n). though sort itself is most likely O(n log(n)) so it's less than that.

Some of the solutions mentioned are more performant, thought this might be less code, also using internal Array.prototype.sort.

(For a Javascript solution, just remove all the types)

Recess answered 10/7, 2019 at 19:24 Comment(0)
S
0

According to the v8 dev blog and caniuse.com Array.sort is already stable as required by the spec in modern browsers, so you don't need to roll your own solution. The only exception I can see is Edge, which should soon move over to chromium and support it as well.

Stine answered 13/2, 2020 at 0:40 Comment(0)
J
0

function sort(data){
    var result=[];
    var array = data;
    const array2=data;
    const len=array2.length;
    for(var i=0;i<=len-1;i++){
    var min = Math.min.apply(Math,array)
    result.push(min);
    var index=array.indexOf(min)
    array.splice(index,1);
    }
    return result;
}   
sort([9,8,5,7,9,3,9,243,4,5,6,3,4,2,4,7,4,9,55,66,33,66]);
Jayson answered 15/6, 2020 at 7:25 Comment(0)
O
-1

Counting Sort is faster than merge sort (it performs in O(n) time) and it is intended for use on integers.

Math.counting_sort = function (m) {
    var i
    var j
    var k
    var step
    var start
    var Output
    var hash
    k = m.length
    Output = new Array ()
    hash = new Array ()
    // start at lowest possible value of m
    start = 0
    step = 1
    // hash all values
    i = 0
    while ( i < k ) {
        var _m = m[i]
        hash [_m] = _m
        i = i + 1
    }
    i = 0
    j = start
    // find all elements within x
    while ( i < k ) {
        while ( j != hash[j] ) {
            j = j + step
        }
        Output [i] = j
        i = i + 1
        j = j + step
    }
    return Output
}

Example:

var uArray = new Array ()<br/>
var sArray = new Array ()<br/><br/>
uArray = [ 10,1,9,2,8,3,7,4,6,5 ]<br/>
sArray = Math.counting_sort ( uArray ) // returns a sorted array
Ocreate answered 25/7, 2011 at 17:59 Comment(2)
A few things to be said: 1. Counting sort only works well in a dense number space. (Try sorting the array [1, 2e9, 1e9]) 2. Don't write for loops as while loops. 3. Don't randomly add things to the Math namespace. 4. You might want to consider making friends with semicolons.Conclave
Also, in case where array has duplicate values, it will run forever. For example, array [3, 1, 3] hashes to [undefined, 1, undefined, 3]. We get two non-undefined values, while algorithm expects there to be three of them.Claptrap

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