Java源码阅读-DualPivotQuicksort

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  之前一直对JDK的排序算法很好奇，抽了点时间看了下源码，收获良多。其核心思想在于，根据数据的不同特性，动态的采用不同的排序算法，充分利用各种排序算法的优势，以期达到更好的综合效果。这篇文章只是Sort相关的源码一部分内容，之后会继续深入探究其他部分的内容。

基本流程：

如果长度小于

QUICKSORT_THRESHOLD(286)

，则采用非归并排序

如果长度小于

INSERTION_SORT_THRESHOLD(47)

，则采用插入排序

最左区间（以初始left开始的区间）

leftmost

：普通插入排序

否则：

pair insertion sort

否则，快速排序

将数组划分为7段（大约），然后找出第2、3、4、5、6段的右端点对应的位置

对这5个位置上的数字进行插入排序，作为枢轴的候选

如果5个数都不相等

选取排序后的2、4作为枢轴，进行双枢轴排序

Dual-Pivot Quicksort

效果：

left part         center part                  right part
+----------------------------------------------------------+
| == pivot1 |  pivot1 < && < pivot2  |    ?    | == pivot2 |
+----------------------------------------------------------+

排序后如果，中间部分元素过多，可能原因是等于pivort1和等于pivort2的元素过多，则将其调整为：

left part         center part                  right part
+----------------------------------------------------------+
| == pivot1 |  pivot1 < && < pivot2  |    ?    | == pivot2 |
+----------------------------------------------------------+

否则，进行普通快速排序，枢轴排序后为第3个元素

否则，考虑Timsort(归并排序的优化版本，对一会升序、一会降序的混合情况处理比较好)

创建Timsort run数组，大小为

MAX_RUN_COUNT(67) + 1

a[run[i]] ~ a[run[i + 1]]之间为升序数组

检查当前待排序数组是否适合使用Timsort,即run数组中升序数组个数，如果个数不小于

MAX_RUN_COUNT

则认为数组内元素排序比较混乱，适合非归并排序

注：对于连续下降的元素会将其调整为连续上升

如果通过上述检测，则进行归并排序

代码详解：

普通的插入排序

for (int i = left, j = i; i < right; j = ++i) {
int ai = a[i + 1];      // 带插入元素
while (ai < a[j]) {     // 寻找插入位置
a[j + 1] = a[j];
if (j-- == left) {
break;
}
}
a[j + 1] = ai;          // 插入新元素
}

改进的插入排序：pair insertion sort,每次插入两个元素

pair insertion之所以不使用在做区间的原因，如果将其应用在左区间，需要增加额外的边界控制。但为什么没有左边界没有使用这种改良的插入排序呢？这一点还需要探究

/**
* 注：left左侧的内容均已排好序，默认的前提条件
*/
/*
* Skip the longest ascending sequence.
*/
do {
if (left >= right) {
return;
}
} while (a[++left] >= a[left - 1]);

/*
* Every element from adjoining part plays the role
* of sentinel, therefore this allows us to avoid the
* left range check on each iteration. Moreover, we use
* the more optimized algorithm, so called pair insertion
* sort, which is faster (in the context of Quicksort)
* than traditional implementation of insertion sort.
* 一次遍历插入两个元素
*/
for (int k = left; ++left <= right; k = ++left) {
int a1 = a[k], a2 = a[left];

// 使得 a1 >= a2
if (a1 < a2) {
a2 = a1; a1 = a[left];
}

// 寻找a1的插入位置，相隔距离为2
while (a1 < a[--k]) {
a[k + 2] = a[k];
}
a[++k + 1] = a1;

// 寻找a2的插入位置，相隔距离为1
while (a2 < a[--k]) {
a[k + 1] = a[k];
}
a[k + 1] = a2;
}

// 将最后一个位置插入到合适位置
int last = a[right];

while (last < a[--right]) {
a[right + 1] = a[right];
}
a[right + 1] = last;

双枢轴排序

Dual-Pivot Quicksort

// Inexpensive approximation of length / 7
int seventh = (length >> 3) + (length >> 6) + 1;

/*
* Sort five evenly spaced elements around (and including) the
* center element in the range. These elements will be used for
* pivot selection as described below. The choice for spacing
* these elements was empirically determined to work well on
* a wide variety of inputs.
*/
int e3 = (left + right) >>> 1; // The midpoint
int e2 = e3 - seventh;
int e1 = e2 - seventh;
int e4 = e3 + seventh;
int e5 = e4 + seventh;

// Sort these elements using insertion sort
if (a[e2] < a[e1]) { long t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

if (a[e3] < a[e2]) { long t = a[e3]; a[e3] = a[e2]; a[e2] = t;
if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
}
if (a[e4] < a[e3]) { long t = a[e4]; a[e4] = a[e3]; a[e3] = t;
if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
}
}
if (a[e5] < a[e4]) { long t = a[e5]; a[e5] = a[e4]; a[e4] = t;
if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
}
}
}

// Pointers
int less  = left;  // The index of the first element of center part
int great = right; // The index before the first element of right part

if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
/*
* Use the second and fourth of the five sorted elements as pivots.
* These values are inexpensive approximations of the first and
* second terciles of the array. Note that pivot1 <= pivot2.
*/
long pivot1 = a[e2];
long pivot2 = a[e4];

/*
* The first and the last elements to be sorted are moved to the
* locations formerly occupied by the pivots. When partitioning
* is complete, the pivots are swapped back into their final
* positions, and excluded from subsequent sorting.
*/
a[e2] = a[left];
a[e4] = a[right];

/*
* Skip elements, which are less or greater than pivot values.
*/
while (a[++less] < pivot1);
while (a[--great] > pivot2);

/*
* Partitioning:
*
*   left part           center part                   right part
* +--------------------------------------------------------------+
* |  < pivot1  |  pivot1 <= && <= pivot2  |    ?    |  > pivot2  |
* +--------------------------------------------------------------+
*               ^                          ^       ^
*               |                          |       |
*              less                        k     great
*
* Invariants:
*
*              all in (left, less)   < pivot1
*    pivot1 <= all in [less, k)     <= pivot2
*              all in (great, right) > pivot2
*
* Pointer k is the first index of ?-part.
*/
outer:
for (int k = less - 1; ++k <= great; ) {
long ak = a[k];
if (ak < pivot1) { // Move a[k] to left part
a[k] = a[less];
/*
- Here and below we use "a[i] = b; i++;" instead
- of "a[i++] = b;" due to performance issue.
*/
a[less] = ak;
++less;
} else if (ak > pivot2) { // Move a[k] to right part
while (a[great] > pivot2) {
if (great-- == k) {
break outer;
}
}
if (a[great] < pivot1) { // a[great] <= pivot2
a[k] = a[less];
a[less] = a[great];
++less;
} else { // pivot1 <= a[great] <= pivot2
a[k] = a[great];
}
/*
- Here and below we use "a[i] = b; i--;" instead
- of "a[i--] = b;" due to performance issue.
*/
a[great] = ak;
--great;
}
}

// Swap pivots into their final positions
a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
a[right] = a[great + 1]; a[great + 1] = pivot2;

// Sort left and right parts recursively, excluding known pivots
sort(a, left, less - 2, leftmost);
sort(a, great + 2, right, false);

/*
* If center part is too large (comprises > 4/7 of the array),
* swap internal pivot values to ends.
*/
if (less < e1 && e5 < great) {
/*
* Skip elements, which are equal to pivot values.
*/
while (a[less] == pivot1) {
++less;
}

while (a[great] == pivot2) {
--great;
}

/*
* Partitioning:
*
*   left part         center part                  right part
* +----------------------------------------------------------+
* | == pivot1 |  pivot1 < && < pivot2  |    ?    | == pivot2 |
* +----------------------------------------------------------+
*              ^                        ^       ^
*              |                        |       |
*             less                      k     great
*
* Invariants:
*
*              all in (*,  less) == pivot1
*     pivot1 < all in [less,  k)  < pivot2
*              all in (great, *) == pivot2
*
* Pointer k is the first index of ?-part.
*/
outer:
for (int k = less - 1; ++k <= great; ) {
long ak = a[k];
if (ak == pivot1) { // Move a[k] to left part
a[k] = a[less];
a[less] = ak;
++less;
} else if (ak == pivot2) { // Move a[k] to right part
while (a[great] == pivot2) {
if (great-- == k) {
break outer;
}
}
if (a[great] == pivot1) { // a[great] < pivot2
a[k] = a[less];
/*
* Even though a[great] equals to pivot1, the
* assignment a[less] = pivot1 may be incorrect,
* if a[great] and pivot1 are floating-point zeros
* of different signs. Therefore in float and
* double sorting methods we have to use more
* accurate assignment a[less] = a[great].
*/
a[less] = pivot1;
++less;
} else { // pivot1 < a[great] < pivot2
a[k] = a[great];
}
a[great] = ak;
--great;
}
}
}

// Sort center part recursively
sort(a, less, great, false);

} else { // Partitioning with one pivot
/*
* Use the third of the five sorted elements as pivot.
* This value is inexpensive approximation of the median.
*/
long pivot = a[e3];

/*
* Partitioning degenerates to the traditional 3-way
* (or "Dutch National Flag") schema:
*
*   left part    center part              right part
* +-------------------------------------------------+
* |  < pivot  |   == pivot   |     ?    |  > pivot  |
* +-------------------------------------------------+
*              ^              ^        ^
*              |              |        |
*             less            k      great
*
* Invariants:
*
*   all in (left, less)   < pivot
*   all in [less, k)     == pivot
*   all in (great, right) > pivot
*
* Pointer k is the first index of ?-part.
*/
for (int k = less; k <= great; ++k) {
if (a[k] == pivot) {
continue;
}
long ak = a[k];
if (ak < pivot) { // Move a[k] to left part
a[k] = a[less];
a[less] = ak;
++less;
} else { // a[k] > pivot - Move a[k] to right part
while (a[great] > pivot) {
--great;
}
if (a[great] < pivot) { // a[great] <= pivot
a[k] = a[less];
a[less] = a[great];
++less;
} else { // a[great] == pivot
/*
* Even though a[great] equals to pivot, the
* assignment a[k] = pivot may be incorrect,
* if a[great] and pivot are floating-point
* zeros of different signs. Therefore in float
* and double sorting methods we have to use
* more accurate assignment a[k] = a[great].
*/
a[k] = pivot;
}
a[great] = ak;
--great;
}
}

/*
* Sort left and right parts recursively.
* All elements from center part are equal
* and, therefore, already sorted.
*/
sort(a, left, less - 1, leftmost);
sort(a, great + 1, right, false);

Timsort归并排序

/*
* Index run[i] is the start of i-th run
* (ascending or descending sequence).
*/
int[] run = new int[MAX_RUN_COUNT + 1];
int count = 0; run[0] = left;

// Check if the array is nearly sorted
for (int k = left; k < right; run[count] = k) {
if (a[k] < a[k + 1]) { // ascending
while (++k <= right && a[k - 1] <= a[k]);
} else if (a[k] > a[k + 1]) { // descending
// 将降序数组变为升序
while (++k <= right && a[k - 1] >= a[k]);
for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
int t = a[lo]; a[lo] = a[hi]; a[hi] = t;
}
} else { // equal
for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
if (--m == 0) {
sort(a, left, right, true);
return;
}
}
}

/*
* The array is not highly structured,
* use Quicksort instead of merge sort.
*/
if (++count == MAX_RUN_COUNT) {
sort(a, left, right, true);
return;
}
}

// Check special cases
// Implementation note: variable "right" is increased by 1.
if (run[count] == right++) { // The last run contains one element
run[++count] = right;
} else if (count == 1) { // The array is already sorted
return;
}

// Determine alternation base for merge
// 确定归并排序的迭代次数（每迭代一次，将相邻升序子序列合并，即run内元素数目减半）
// 简单示例：a[1, 5, 2, 6, 3, 7, 4, 8] ==> a[1, 2, 5, 6, 3, 4, 7, 8]
byte odd = 0;
for (int n = 1; (n <<= 1) < count; odd ^= 1);

// Use or create temporary array b for merging
int[] b;                 // temp array; alternates with a
int ao, bo;              // array offsets from 'left'
int blen = right - left; // space needed for b
if (work == null || workLen < blen || workBase + blen > work.length) {
work = new int[blen];
workBase = 0;
}
// 根据归并的迭代次数，更改a,b
// a,b中必有一个数组为原始数组，另一个为临时数组
// 在归并的过程中，每迭代一次，run内元素数目减半，同时a,b会交换一次
// 为了保证最后一次迭代后，原始数组内存有归并好的数据，需要进行如下考虑
if (odd == 0) {
System.arraycopy(a, left, work, workBase, blen);
b = a;
bo = 0;
a = work;
ao = workBase - left;
} else {
b = work;
ao = 0;
bo = workBase - left;
}

// Merging
// a是原始数组，b是目标数组
for (int last; count > 1; count = last) {
// 合并两个相邻升序序列
for (int k = (last = 0) + 2; k <= count; k += 2) {
// 确定边界
int hi = run[k], mi = run[k - 1];
for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
b[i + bo] = a[p++ + ao];
} else {
b[i + bo] = a[q++ + ao];
}
}
// 更新子序列标示
run[++last] = hi;
}
// 如果升序子序列个数为奇数，之前两两合并时，最后会剩余一个，将剩余的直接拷贝到b, 等待下一次合并
if ((count & 1) != 0) {
for (int i = right, lo = run[count - 1]; --i >= lo;
b[i + bo] = a[i + ao]
);
run[++last] = right;
}

// 交换a b
int[] t = a; a = b; b = t;
int o = ao; ao = bo; bo = o;
}

Reference:

形式化方法的逆袭——如何找出Timsort算法和玉兔月球车中的Bug？

排序算法—快速排序（JDK1.7 DualPivotQuicksort 源码解析）

JDK源码解析(1)——数据数组排序：Arrays.sort()