Timsort 最快的排序算法
背景
Timsort是一种混合、稳定高效的排序算法，源自合并排序和插入排序，旨在很好地处理多种真实数据。它由Tim Peters于2002年实施使用在Python编程语言中。该算法查找已经排序的数据的子序列，并使用该知识更有效地对其余部分进行排序。这是通过将已识别的子序列（称为运行）与现有运行合并直到满足某些条件来完成的。从版本2.3开始，Timsort一直是Python的标准排序算法。如今，Timsort 已是是 Python、 Java、 Android平台 和 GNU Octave 的默认排序算法。
上文讲到v8 7.0版本后排序会使用Timsort，本文就接着讲解Timsort的原理。深入解析V8中sort的工作原理
性能分析
图片来源：https://www.bigocheatsheet.com/
不难看出，Timsort跟Mergesort的时间复杂度相近，快于其他排序算法。
Timsort实际上借鉴了插入排序和归并排序的方法
原理
run(分割槽)
其实在现实中，给定的大部分数组中通常是有部分已经排好序的（无论是升序还是降序），Timsort利用了这一点，按照升序(后一元素大于等于前一元素，a[i] <= a[i+1])和严格降序(前一元素大于后一元素a[i]>a[i+1])的规则，将原来的数组分解为若干个run，升序的run就保持不变，严格降序的run就翻转，最终得到若干个升序的run。
比如：
- 现在有一个目标的数组
[1, 5, 9, 8, 6, 4, 5, 6, 7] - 我们可以看到
[1, 5, 9]符合升序，[8, 6, 4]符合严格降序，[4, 5, 6, 7]符合升序 - 将严格降序的
run翻转，然后拼凑得到新的数组[1, 5, 9, 6, 8, 4, 5, 6, 7]
合并run
假设得到的新数组为[... 6, 7, 8, 9, 10, 1, 2, 3, 4, 5 ...]
如果我们单纯使用Mergesort将两个run进行合并
run1: [6, 7, 8, 9, 10]
run2: [1, 2, 3, 4, 5]Mergesort会将两个run的第一个元素进行比较
1 < 6, 得到 1
2 < 6, 得到 2
3 < 6, 得到 3
4 < 6, 得到 4
5 < 6, 得到 5我们发现，直接遍历了整个run，那如果run的长度很长，那将非常消耗性能。
那么既然每个run都是升序，我们没必要逐个逐个比较
这里我们引入一个新的概念Galloping（倍增搜寻法）
Galloping（倍增搜寻法）
比如我们将要合并以下两个run
run1: [101, 102, 103, ... 200]
run2: [1, 2, 3, ..., 100]这里我们不再逐个元素比较，而是以2^n（2的n次方）递增作比较
run1[0] > run2[0]
run1[0] > run2[1]
run1[0] > run2[3]
run1[0] > run2[7]
run1[0] > run2[15]
...
run1[0] > run2[2^n - 1]
run1[0] <= run2[2^(n+1) - 1]所以我们得到一个结果run2[2^n - 1] < run1[0] <= run2[2^(n+1) - 1]，从而得知run1[0]在run2中的大概位置
由于run2是升序的，我们限定左边界left = run2[2^n - 1]，右边界run2[2^(n+1) - 1]，然后在该区域用Binary Search（二分查找法）便能很高效地定位到run1[0]在run2中的具体位置。
这样，我们合并两个run的比较次数就由O(N)变为O(logN)
但是，我们能不能直接跳过Galloping，直接做Binary Search呢？
我们来举个例子：
run1: [1, 3, 5, 7, 9 ... 2n+1]
run2: [0, 2, 4, 6, 8 ... 2n]这样的话，run1[0]只大于run2[0]，每次做Binary Search只能得到一个数的结果，再加上要做n次Binary Search，所以时间复杂度由O(N)变为O(NlogN)
合并run的顺序
当原始数组变为若干个升序run的集合时，我们需要两两合并run，但是如果我们将一个长度很大的run和一个长度很短的run合并，需要很长时间而且只能得到一个较短的run的效益。
所以Timsort维护了一个stack,将所有的run入栈，同时满足后入栈的run长度大于前两个入栈的run的长度之和，让run的长度递减
[... runA, runB, runC]
runA > runB + runC
runB > runC避免长度差太多的run合并。
插入排序
当两个run的长度较短（小于32）时，相对来说，Binary Insertion Search(二分插入排序)会较快。
具体run的长度多少算短，这个量会在计算中动态变化。
总结
该排序充分利用了原始数组中已经有序的部分，从而提高了排序所消耗的时间。
如果数据非常随机或者很多重复的数组，是不适用Timsort的。
本文结合该视频观看会更好https://www.youtube.com/watch...（需要翻墙）
JavaScript实现版本
代码来源https://stackoverflow.com/que...
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];
}
}
python
java
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