Sorting Algorithms (v31)

Revision 31 of this benchmark created on


Description

Sort 10000 [small number of] values, for one value go to

Preparation HTML

<script>
  function swap(ary, a, b) {
      var t = ary[a];
      ary[a] = ary[b];
      ary[b] = t;
  }
  
  // Built-in with comparison function (default sorting is "dictionary-style")
  function builtin_sort(ary) {
      return ary.sort(function(a, b) {
          return a - b;
      });
  }
  
  // Bubble Sort
  function bubble_sort(ary) {
      var a, b;
      for (a = 0; a < ary.length; a++) {
          for (b = a + 1; b < ary.length; b++) {
              if (ary[a] > ary[b]) {
                  swap(ary, a, b);
              }
          }
      }
  
      return ary;
  }
  //Insertion sort
  function insertion_sort(ary) {
                for(var i=1,l=ary.length;i<l;i++) {
                        var value = ary[i];
                        for(var j=i - 1;j>=0;j--) {
                                if(ary[j] <= value)
                                        break;
                                ary[j+1] = ary[j];
                        }
                        ary[j+1] = value;
                }
                return ary;
  }
  
  // Naive (but understandable) quicksort (memory hog)
  function naive_quicksort(ary) {
      if (ary.length <= 1) {
          return ary;
      }
  
      var less = [],
          greater = [],
          pivot = ary.pop();
  
      for (var i = 0; i < ary.length; i++) {
          if (ary[i] < pivot) {
              less.push(ary[i]);
          } else {
              greater.push(ary[i]);
          }
      }
  
      less = naive_quicksort(less);
      greater = naive_quicksort(greater);
  
      return less.concat(pivot, greater);
  }
  
  // In place quicksort
  function inplace_quicksort_partition(ary, start, end, pivotIndex) {
      var i = start, j = end;
      var pivot = ary[pivotIndex];
      
      while(true) {
          while(ary[i] < pivot) {i++};
          j--;
          while(pivot < ary[j]) {j--};
          if(!(i<j)) {
              return i;
          }
          swap(ary,i,j);
          i++;
     }
  }
  
  function inplace_quicksort(ary, start, end) {
      if (start < end - 1) {
          var pivotIndex = Math.round((start + end) / 2);
  
          pivotIndex = inplace_quicksort_partition(ary, start, end, pivotIndex);
  
          inplace_quicksort(ary, start, pivotIndex - 1);
          inplace_quicksort(ary, pivotIndex + 1, end);
      }
  
      return ary;
  }
  
  // Heap sort
  function heapSort(ary) {
      heapify(ary);
  
      for (var end = ary.length - 1; end > 0; end--) {
          swap(ary, end, 0);
          shiftDown(ary, 0, end - 1);
      }
  
      return ary;
  }
  
  function heapify(ary) {
      for (var start = (ary.length >> 1) - 1; start >= 0; start--) {
          shiftDown(ary, start, ary.length - 1);
      }
  }
  
  function shiftDown(ary, start, end) {
      var root = start,
          child, s;
  
      while (root * 2 + 1 <= end) {
          child = root * 2 + 1;
          s = root;
  
          if (ary[s] < ary[child]) {
              s = child;
          }
  
          if (child + 1 <= end && ary[s] < ary[child + 1]) {
              s = child + 1;
          }
  
          if (s !== root) {
              swap(ary, root, s);
              root = s;
          } else {
              return;
          }
      }
  }
  
  // Merge sort
  function merge_sort(ary) {
      if (ary.length <= 1) {
          return ary;
      }
  
      var m = ary.length >> 1;
  
      var left = ary.slice(0, m),
          right = ary.slice(m);
  
      return merge(merge_sort(left), merge_sort(right));
  }
  
  function merge(left, right) {
      var result = [],
          li = 0, ri = 0;
  
      while (li < left.length || ri < right.length) {
          if (li < left.length && ri < right.length) {
              if (left[li] <= right[ri]) {
                  result.push(left[li]);
                  li++;
              } else {
                  result.push(right[ri]);
                  ri++;
              }
          } else if (li < left.length) {
              result.push(left[li]);
              li++;
          } else if (ri < right.length) {
              result.push(right[ri]);
              ri++;
          }
      }
  
      return result;
  }
  
  // Shell sort
  function shell_sort(ary) {
      var inc = Math.round(ary.length / 2),
          i, j, t;
  
      while (inc > 0) {
          for (i = inc; i < ary.length; i++) {
              t = ary[i];
              j = i;
              while (j >= inc && ary[j - inc] > t) {
                  ary[j] = ary[j - inc];
                  j -= inc;
              }
              ary[j] = t;
          }
          inc = Math.round(inc / 2.2);
      }
  
      return ary;
  }
  //Comb Sort (Basically bubblesort with a small modification, but heaps faster)
function comb_sort(ary) {
    var gap = ary.length;
    while(true) {
        gap = newgap(gap);
        var swapped = false;
        for(var i=0,l=ary.length;i<l;i++) {
            var j = i + gap;
            if(ary[i] < ary[j]) {
                swap(ary,i,j);
                swapped = true;
            }
        }
        if(gap == 1 && !swapped)
             break;
    }
    return ary;   
}
function newgap(gap) {
    gap = ((gap * 10) / 13) | 0;
    if(gap == 9 || gap == 10)
        gap = 11;
    if(gap < 1)
         gap = 1;
    return gap;
}
//faster quicksort using a stack to eliminate recursion, sorting inplace to reduce memory usage, and using insertion sort for small partition sizes
//It should NOT be running at the EXACT same speed as the inplace version due to the number of changes, but it does on jsPerf for some reason....
function fast_quicksort(ary) {
    var stack = [];
    var entry = [0,ary.length,2 * Math.floor(Math.log(ary.length)/Math.log(2))];
    stack.push(entry);
    while(stack.length > 0) {
        entry = stack.pop();
        var start = entry[0];
        var end = entry[1];
        var depth = entry[2];
        if(depth == 0) {
         ary = shell_sort_bound(ary,start,end);
         continue;
        }
        depth--;
        var pivot = Math.round((start + end) / 2);
            
        var pivotNewIndex = inplace_quicksort_partition(ary,start,end, pivot);
        if(end - pivotNewIndex > 16) {
            entry = [pivotNewIndex,end,depth];
            stack.push(entry);
        }
        if(pivotNewIndex - start > 16) {
            entry = [start,pivotNewIndex,depth];
            stack.push(entry);
        }
    }
    ary = insertion_sort(ary);
    return ary;
}
function shell_sort_bound(ary,start,end) {
      var inc = Math.round((start + end)/2),
          i, j, t;
  
      while (inc >= start) {
          for (i = inc; i < end; i++) {
              t = ary[i];
              j = i;
              while (j >= inc && ary[j - inc] > t) {
                  ary[j] = ary[j - inc];
                  j -= inc;
              }
              ary[j] = t;
          }
          inc = Math.round(inc / 2.2);
      }
  
      return ary;
  }

function mysort(arr) {
  var len = arr.length;
  var lp = -1;

  for (var i=0; i<len-1; ) {
    var p1 = arr[i], p2 = arr[i+1];
    if (p1 > p2) {
      arr[i] = p2; arr[i+1] = p1; lp = i; i -= 1;
      if (i < 0) { 
        i = 0;
      }
    }
    else if (lp == -1) { i++; }
    else { i = lp + 1; lp++ }
  }
  return arr;
}
Array.prototype.timsort = function(comp){

    var global_a=this
    var MIN_MERGE = 32;
    var MIN_GALLOP = 7
    var runBase=[];
    var runLen=[];
    var stackSize = 0;
    var compare = comp;

    sort(this,0,this.length,compare);

    /*
         * The next two methods (which are package private and static) constitute the entire API of this class. Each of these methods
         * obeys the contract of the public method with the same signature in java.util.Arrays.
         */

    function sort (a, lo, hi, compare) {

        if (typeof compare != "function") {
            throw new Error("Compare is not a function.");
            return;
        }

        stackSize = 0;
        runBase=[];
        runLen=[];

        rangeCheck(a.length, lo, hi);
        var nRemaining = hi - lo;
        if (nRemaining < 2) return; // Arrays of size 0 and 1 are always sorted

        // If array is small, do a "mini-TimSort" with no merges
        if (nRemaining < MIN_MERGE) {
            var initRunLen = countRunAndMakeAscending(a, lo, hi, compare);
            binarySort(a, lo, hi, lo + initRunLen, compare);
            return;
        }

        /**
                 * March over the array once, left to right, finding natural runs, extending short natural runs to minRun elements, and
                 * merging runs to maintain stack invariant.
                 */
        var ts = [];
        var minRun = minRunLength(nRemaining);
        do {
            // Identify next run
            var runLenVar = countRunAndMakeAscending(a, lo, hi, compare);

            // If run is short, extend to min(minRun, nRemaining)
            if (runLenVar < minRun) {
                var force = nRemaining <= minRun ? nRemaining : minRun;
                binarySort(a, lo, lo + force, lo + runLenVar, compare);
                runLenVar = force;
            }

            // Push run onto pending-run stack, and maybe merge
            pushRun(lo, runLenVar);
            mergeCollapse();

            // Advance to find next run
            lo += runLenVar;
            nRemaining -= runLenVar;
        } while (nRemaining != 0);

        // Merge all remaining runs to complete sort
        mergeForceCollapse();
    }


    /**
         * Sorts the specified portion of the specified array using a binary insertion sort. This is the best method for sorting small
         * numbers of elements. It requires O(n log n) compares, but O(n^2) data movement (worst case).
         *
         * If the initial part of the specified range is already sorted, this method can take advantage of it: the method assumes that
         * the elements from index {@code lo}, inclusive, to {@code start}, exclusive are already sorted.
         *
         * @param a the array in which a range is to be sorted
         * @param lo the index of the first element in the range to be sorted
         * @param hi the index after the last element in the range to be sorted
         * @param start the index of the first element in the range that is not already known to be sorted (@code lo <= start <= hi}
         * @param c comparator to used for the sort
         */
    function binarySort (a, lo, hi, start, compare) {
        if (start == lo) start++;
        for (; start < hi; start++) {
            var pivot = a[start];

            // Set left (and right) to the index where a[start] (pivot) belongs
            var left = lo;
            var right = start;
            /*
            * Invariants: pivot >= all in [lo, left). pivot < all in [right, start).
            */
            while (left < right) {
                var mid = (left + right) >>> 1;
                if (compare(pivot, a[mid]) < 0)
                    right = mid;
                else
                    left = mid + 1;
            }
            /*
            * The invariants still hold: pivot >= all in [lo, left) and pivot < all in [left, start), so pivot belongs at left. Note
            * that if there are elements equal to pivot, left points to the first slot after them -- that's why this sort is stable.
            * Slide elements over to make room to make room for pivot.
            */
            var n = start - left; // The number of elements to move
            // Switch is just an optimization for arraycopy in default case
            switch (n) {
            case 2:
                a[left + 2] = a[left + 1];
            case 1:
                a[left + 1] = a[left];
                break;
            default:
            arraycopy(a, left, a, left + 1, n);
            }
            a[left] = pivot;
        }
    }


    /**
         * Returns the length of the run beginning at the specified position in the specified array and reverses the run if it is
         * descending (ensuring that the run will always be ascending when the method returns).
         *
         * A run is the longest ascending sequence with:
         *
         * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
         *
         * or the longest descending sequence with:
         *
         * a[lo] > a[lo + 1] > a[lo + 2] > ...
         *
         * For its intended use in a stable mergesort, the strictness of the definition of "descending" is needed so that the call can
         * safely reverse a descending sequence without violating stability.
         *
         * @param a the array in which a run is to be counted and possibly reversed
         * @param lo index of the first element in the run
         * @param hi index after the last element that may be contained in the run. It is required that @code{lo < hi}.
         * @param c the comparator to used for the sort
         * @return the length of the run beginning at the specified position in the specified array
         */
    function countRunAndMakeAscending (a, lo, hi, compare) {
        var runHi = lo + 1;

        // Find end of run, and reverse range if descending
        if (compare(a[runHi++], a[lo]) < 0) { // Descending
            while (runHi < hi && compare(a[runHi], a[runHi - 1]) < 0){
                runHi++;
            }
            reverseRange(a, lo, runHi);
        } else { // Ascending
            while (runHi < hi && compare(a[runHi], a[runHi - 1]) >= 0){
                runHi++;
            }
        }

        return runHi - lo;
    }

    /**
         * Reverse the specified range of the specified array.
         *
         * @param a the array in which a range is to be reversed
         * @param lo the index of the first element in the range to be reversed
         * @param hi the index after the last element in the range to be reversed
         */
    function /*private static void*/ reverseRange (/*Object[]*/ a, /*int*/ lo, /*int*/ hi) {
        hi--;
        while (lo < hi) {
            var t = a[lo];
            a[lo++] = a[hi];
            a[hi--] = t;
        }
    }


    /**
         * Returns the minimum acceptable run length for an array of the specified length. Natural runs shorter than this will be
         * extended with {@link #binarySort}.
         *
         * Roughly speaking, the computation is:
         *
         * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff). Else if n is an exact power of 2, return
         * MIN_MERGE/2. Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k is close to, but strictly less than, an
         * exact power of 2.
         *
         * For the rationale, see listsort.txt.
         *
         * @param n the length of the array to be sorted
         * @return the length of the minimum run to be merged
         */
    function /*private static int*/ minRunLength (/*int*/ n) {
        //var v=0;
        var r = 0; // Becomes 1 if any 1 bits are shifted off
        /*while (n >= MIN_MERGE) { v++;
            r |= (n & 1);
            n >>= 1;
        }*/
        //console.log("minRunLength("+n+") "+v+" vueltas, result="+(n+r));
        //return n + r;
        return n + 1;
    }

    /**
         * Pushes the specified run onto the pending-run stack.
         *
         * @param runBase index of the first element in the run
         * @param runLen the number of elements in the run
         */
    function pushRun (runBaseArg, runLenArg) {
        //console.log("pushRun("+runBaseArg+","+runLenArg+")");
        //this.runBase[stackSize] = runBase;
        //runBase.push(runBaseArg);
        runBase[stackSize] = runBaseArg;

        //this.runLen[stackSize] = runLen;
        //runLen.push(runLenArg);
        runLen[stackSize] = runLenArg;
        stackSize++;
    }

    /**
         * Examines the stack of runs waiting to be merged and merges adjacent runs until the stack invariants are reestablished:
         *
         * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] 2. runLen[i - 2] > runLen[i - 1]
         *
         * This method is called each time a new run is pushed onto the stack, so the invariants are guaranteed to hold for i <
         * stackSize upon entry to the method.
         */
    function mergeCollapse () {
        while (stackSize > 1) {
            var n = stackSize - 2;
            if (n > 0 && runLen[n - 1] <= runLen[n] + runLen[n + 1]) {
                if (runLen[n - 1] < runLen[n + 1]) n--;
                mergeAt(n);
            } else if (runLen[n] <= runLen[n + 1]) {
                mergeAt(n);
            } else {
                break; // Invariant is established
            }
        }
    }

    /**
         * Merges all runs on the stack until only one remains. This method is called once, to complete the sort.
         */
    function mergeForceCollapse () {
        while (stackSize > 1) {
            var n = stackSize - 2;
            if (n > 0 && runLen[n - 1] < runLen[n + 1]) n--;
            mergeAt(n);
        }
    }


    /**
         * Merges the two runs at stack indices i and i+1. Run i must be the penultimate or antepenultimate run on the stack. In other
         * words, i must be equal to stackSize-2 or stackSize-3.
         *
         * @param i stack index of the first of the two runs to merge
         */
    function mergeAt (i) {

        var base1 = runBase[i];
        var len1 = runLen[i];
        var base2 = runBase[i + 1];
        var len2 = runLen[i + 1];

        /*
        * Record the length of the combined runs; if i is the 3rd-last run now, also slide over the last run (which isn't involved
        * in this merge). The current run (i+1) goes away in any case.
        */
        //var stackSize = runLen.length;
        runLen[i] = len1 + len2;
        if (i == stackSize  - 3) {
            runBase[i + 1] = runBase[i + 2];
            runLen[i + 1] = runLen[i + 2];
        }
        stackSize--;

        /*
        * Find where the first element of run2 goes in run1. Prior elements in run1 can be ignored (because they're already in
        * place).
        */

        var k = gallopRight(global_a[base2], global_a, base1, len1, 0, compare);
        base1 += k;
        len1 -= k;
        if (len1 == 0) return;

        /*
        * Find where the last element of run1 goes in run2. Subsequent elements in run2 can be ignored (because they're already in
        * place).
        */
        len2 = gallopLeft(global_a[base1 + len1 - 1], global_a, base2, len2, len2 - 1, compare);

        if (len2 == 0) return;

        // Merge remaining runs, using tmp array with min(len1, len2) elements
        if (len1 <= len2)
            mergeLo(base1, len1, base2, len2);
        else
            mergeHi(base1, len1, base2, len2);
    }


    /**
         * Locates the position at which to insert the specified key into the specified sorted range; if the range contains an element
         * equal to key, returns the index of the leftmost equal element.
         *
         * @param key the key whose insertion point to search for
         * @param a the array in which to search
         * @param base the index of the first element in the range
         * @param len the length of the range; must be > 0
         * @param hint the index at which to begin the search, 0 <= hint < n. The closer hint is to the result, the faster this method
         *           will run.
         * @param c the comparator used to order the range, and to search
         * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k], pretending that a[b - 1] is minus infinity and a[b
         *         + n] is infinity. In other words, key belongs at index b + k; or in other words, the first k elements of a should
         *         precede key, and the last n - k should follow it.
         */
    function gallopLeft (key, a, base, len, hint, compare) {
        var lastOfs = 0;
        var ofs = 1;
        if (compare(key, a[base + hint]) > 0) {
            // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
            var maxOfs = len - hint;
            while (ofs < maxOfs && compare(key, a[base + hint + ofs]) > 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0) // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs) ofs = maxOfs;

            // Make offsets relative to base
            lastOfs += hint;
            ofs += hint;
        } else { // key <= a[base + hint]
            // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
            var maxOfs = hint + 1;
            while (ofs < maxOfs && compare(key, a[base + hint - ofs]) <= 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0) // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs) ofs = maxOfs;

            // Make offsets relative to base
            var tmp = lastOfs;
            lastOfs = hint - ofs;
            ofs = hint - tmp;
        }

        /*
        * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere to the right of lastOfs but no farther right than ofs.
        * Do a binary search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
        */
        lastOfs++;
        while (lastOfs < ofs) {
            var m = lastOfs + ((ofs - lastOfs) >>> 1);

            if (compare(key, a[base + m]) > 0)
                lastOfs = m + 1; // a[base + m] < key
            else
                ofs = m; // key <= a[base + m]
        }
        return ofs;
    }

    /**
         * Like gallopLeft, except that if the range contains an element equal to key, gallopRight returns the index after the
         * rightmost equal element.
         *
         * @param key the key whose insertion point to search for
         * @param a the array [] in which to search
         * @param base the index of the first element in the range
         * @param len the length of the range; must be > 0
         * @param hint the index at which to begin the search, 0 <= hint < n. The closer hint is to the result, the faster this method
         *           will run.
         * @param c the comparator used to order the range, and to search
         * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
         */
    function gallopRight (key, a, base, len, hint,  compare) {

        var ofs = 1;
        var lastOfs = 0;
        if (compare(key, a[base + hint]) < 0) {
            // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
            var maxOfs = hint + 1;
            while (ofs < maxOfs && compare(key, a[base + hint - ofs]) < 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0) // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs) ofs = maxOfs;

            // Make offsets relative to b
            var tmp = lastOfs;
            lastOfs = hint - ofs;
            ofs = hint - tmp;
        } else { // a[b + hint] <= key
            // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
            var maxOfs = len - hint;
            while (ofs < maxOfs && compare(key, a[base + hint + ofs]) >= 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0) // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs) ofs = maxOfs;

            // Make offsets relative to b
            lastOfs += hint;
            ofs += hint;
        }

        /*
        * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to the right of lastOfs but no farther right than ofs.
        * Do a binary search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
        */
        lastOfs++;
        while (lastOfs < ofs) {
            var m = lastOfs + ((ofs - lastOfs) >>> 1);

            if (compare(key, a[base + m]) < 0)
                ofs = m; // key < a[b + m]
            else
                lastOfs = m + 1; // a[b + m] <= key
        }
        return ofs;
    }

    /**
    * Merges two adjacent runs in place, in a stable fashion. The first element of the first run must be greater than the first
    * element of the second run (a[base1] > a[base2]), and the last element of the first run (a[base1 + len1-1]) must be greater
    * than all elements of the second run.
    *
    * For performance, this method should be called only when len1 <= len2; its twin, mergeHi should be called if len1 >= len2.
    * (Either method may be called if len1 == len2.)
    *
    * @param base1 index of first element in first run to be merged
    * @param len1 length of first run to be merged (must be > 0)
    * @param base2 index of first element in second run to be merged (must be aBase + aLen)
    * @param len2 length of second run to be merged (must be > 0)
    */
    function mergeLo (base1, len1, base2, len2) {

        // Copy first run into temp array
        var a = global_a;// For performance
        var tmp=a.slice(base1,base1+len1);

        var cursor1 = 0; // Indexes into tmp array
        var cursor2 = base2; // Indexes int a
        var dest = base1; // Indexes int a

        // Move first element of second run and deal with degenerate cases
        a[dest++] = a[cursor2++];
        if (--len2 == 0) {
            arraycopy(tmp, cursor1, a, dest, len1);
            return;
        }
        if (len1 == 1) {
            arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
            return;
        }

        var c = compare;// Use local variable for performance

        var minGallop = MIN_GALLOP; // "    " "     " "
        outer:
        while (true) {
            var count1 = 0; // Number of times in a row that first run won
            var count2 = 0; // Number of times in a row that second run won

            /*
            * Do the straightforward thing until (if ever) one run starts winning consistently.
            */
            do {
                if (compare(a[cursor2], tmp[cursor1]) < 0) {
                    a[dest++] = a[cursor2++];
                    count2++;
                    count1 = 0;
                    if (--len2 == 0) break outer;
                } else {
                    a[dest++] = tmp[cursor1++];
                    count1++;
                    count2 = 0;
                    if (--len1 == 1) break outer;
                }
            } while ((count1 | count2) < minGallop);

            /*
            * One run is winning so consistently that galloping may be a huge win. So try that, and continue galloping until (if
            * ever) neither run appears to be winning consistently anymore.
            */
            do {
                count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
                if (count1 != 0) {
                    arraycopy(tmp, cursor1, a, dest, count1);
                    dest += count1;
                    cursor1 += count1;
                    len1 -= count1;
                    if (len1 <= 1) // len1 == 1 || len1 == 0
                        break outer;
                }
                a[dest++] = a[cursor2++];
                if (--len2 == 0) break outer;

                count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
                if (count2 != 0) {
                    arraycopy(a, cursor2, a, dest, count2);
                    dest += count2;
                    cursor2 += count2;
                    len2 -= count2;
                    if (len2 == 0) break outer;
                }
                a[dest++] = tmp[cursor1++];
                if (--len1 == 1) break outer;
                minGallop--;
            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if (minGallop < 0) minGallop = 0;
            minGallop += 2; // Penalize for leaving gallop mode
        } // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field

        if (len1 == 1) {
            arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
        } else if (len1 == 0) {
            throw new Error("IllegalArgumentException. Comparison method violates its general contract!");
        } else {
            arraycopy(tmp, cursor1, a, dest, len1);
        }
    }


    /**
         * Like mergeLo, except that this method should be called only if len1 >= len2; mergeLo should be called if len1 <= len2.
         * (Either method may be called if len1 == len2.)
         *
         * @param base1 index of first element in first run to be merged
         * @param len1 length of first run to be merged (must be > 0)
         * @param base2 index of first element in second run to be merged (must be aBase + aLen)
         * @param len2 length of second run to be merged (must be > 0)
         */
    function mergeHi ( base1, len1, base2, len2) {

        // Copy second run into temp array
        var a = global_a;// For performance
        var tmp=a.slice(base2, base2+len2);

        var cursor1 = base1 + len1 - 1; // Indexes into a
        var cursor2 = len2 - 1; // Indexes into tmp array
        var dest = base2 + len2 - 1; // Indexes into a

        // Move last element of first run and deal with degenerate cases
        a[dest--] = a[cursor1--];
        if (--len1 == 0) {
            arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
            return;
        }
        if (len2 == 1) {
            dest -= len1;
            cursor1 -= len1;
            arraycopy(a, cursor1 + 1, a, dest + 1, len1);
            a[dest] = tmp[cursor2];
            return;
        }

        var c = compare;// Use local variable for performance

        var minGallop = MIN_GALLOP; // "    " "     " "
        outer:
        while (true) {
            var count1 = 0; // Number of times in a row that first run won
            var count2 = 0; // Number of times in a row that second run won

            /*
            * Do the straightforward thing until (if ever) one run appears to win consistently.
            */
            do {
                if (compare(tmp[cursor2], a[cursor1]) < 0) {
                    a[dest--] = a[cursor1--];
                    count1++;
                    count2 = 0;
                    if (--len1 == 0) break outer;
                    } else {
                        a[dest--] = tmp[cursor2--];
                        count2++;
                        count1 = 0;
                        if (--len2 == 1) break outer;
                    }
            } while ((count1 | count2) < minGallop);

            /*
            * One run is winning so consistently that galloping may be a huge win. So try that, and continue galloping until (if
            * ever) neither run appears to be winning consistently anymore.
            */
            do {
                count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
                if (count1 != 0) {
                    dest -= count1;
                    cursor1 -= count1;
                    len1 -= count1;
                    arraycopy(a, cursor1 + 1, a, dest + 1, count1);
                    if (len1 == 0) break outer;
                }
                a[dest--] = tmp[cursor2--];
                if (--len2 == 1) break outer;

                count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);
                if (count2 != 0) {
                    dest -= count2;
                    cursor2 -= count2;
                    len2 -= count2;
                    arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
                    if (len2 <= 1) // len2 == 1 || len2 == 0
                        break outer;
                }
                a[dest--] = a[cursor1--];
                if (--len1 == 0) break outer;
                    minGallop--;
            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if (minGallop < 0) minGallop = 0;
            minGallop += 2; // Penalize for leaving gallop mode
        } // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field

        if (len2 == 1) {
            dest -= len1;
            cursor1 -= len1;
            arraycopy(a, cursor1 + 1, a, dest + 1, len1);
            a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
        } else if (len2 == 0) {
            throw new Error("IllegalArgumentException. Comparison method violates its general contract!");
        } else {
            arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
        }
    }


    /**
    * Checks that fromIndex and toIndex are in range, and throws an appropriate exception if they aren't.
    *
    * @param arrayLen the length of the array
    * @param fromIndex the index of the first element of the range
    * @param toIndex the index after the last element of the range
    * @throws IllegalArgumentException if fromIndex > toIndex
    * @throws ArrayIndexOutOfBoundsException if fromIndex < 0 or toIndex > arrayLen
    */
    function rangeCheck (arrayLen, fromIndex, toIndex) {
        if (fromIndex > toIndex) throw new Error( "IllegalArgument fromIndex(" + fromIndex + ") > toIndex(" + toIndex + ")");
        if (fromIndex < 0) throw new Error( "ArrayIndexOutOfBounds "+fromIndex);
        if (toIndex > arrayLen) throw new Error( "ArrayIndexOutOfBounds "+toIndex);
    }
}

// java System.arraycopy(Object src, int srcPos, Object dest, int destPos, int length)
function arraycopy(s,spos,d,dpos,len){
    var a=s.slice(spos,spos+len);
    while(len--){
        d[dpos+len]=a[len];
    }
}
function compare(a ,b){return a-b;}
function bucketSort(a, key) {
    key = key || function(x) { return x };
    var len = a.length,
        buckets = [],
        i, j, b, d = 0;
    for (; d < 32; d += 4) {
        for (i = 16; i--;)
            buckets[i] = [];
        for (i = len; i--;)
            buckets[(key(a[i]) >> d) & 15].push(a[i]);
        for (b = 0; b < 16; b++)
            for (j = buckets[b].length; j--;)
                a[++i] = buckets[b][j];
    }
    return a;
}
function inplaceRadix(nums) {
        var k = Math.max.apply(null, nums.map(function(i) {
                return Math.ceil(Math.log(i)/Math.log(2));
        }));
         
        for (var d = 0; d < k; ++d) {
                for (var i = 0, p = 0, b = 1 << d, n = nums.length; i < n; ++i) {
                        var o = nums[i];
                        if ((o & b) == 0) {
                                nums.splice(p++, 0, nums.splice(i, 1)[0]);
                        }
                }
        }
  return nums;
}
Array.prototype.pancake_sort = function () {
    for (var i = this.length - 1; i >= 1; i--) {
        // find the index of the largest element not yet sorted
        var max_idx = 0;
        var max = this[0];
        for (var j = 1; j <= i; j++) {
            if (this[j] > max) {
                max = this[j];
                max_idx = j;
            }
        }
 
        if (max_idx == i) 
            continue; // element already in place
 
        var new_slice;
 
        // flip this max element to index 0
        if (max_idx > 0) {
            new_slice = this.slice(0, max_idx+1).reverse();
            for (var j = 0; j <= max_idx; j++) 
                this[j] = new_slice[j];
        }
 
        // then flip the max element to its place
        new_slice = this.slice(0, i+1).reverse();
        for (var j = 0; j <= i; j++) 
            this[j] = new_slice[j];
    }
    return this;
}
</script>

Setup

var input = [];
    for (var i = 0; i < 10000; i++) {
      input[i] = Math.round(Math.random() * 60000);
    }

Teardown


    for (var i = 0; i < 9999; i++) {
      if (input[i] > input[i] + 1) {
        throw Error('Array is not sorted');
      }
    }
  

Test runner

Ready to run.

Testing in
TestOps/sec
Inplace Quicksort
input = inplace_quicksort(input, 0, input.length);
ready
Heap Sort
input = heapSort(input);
ready
Fast QuickSort
input = fast_quicksort(input);
ready
TimSort
input.timsort(compare);
ready
BucketSort
input = bucketSort(input);
ready
Inplace Radix
input = inplaceRadix(input);
ready
Pancake Sort
input = input.pancake_sort();
ready

Revisions

You can edit these tests or add more tests to this page by appending /edit to the URL.