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本文转载自于晓飞93，原文链接 DualPivotQuickSort 双轴快速排序 源码 笔记
DualPivotQuicksort是Arrays类中提供的给基本类型的数据排序的算法。它针对每种基本数据类型都有对应的实现，实现方式有细微差异，但思路都是相同的，所以这里只挑选int类型的排序。
整个实现中的思路是：首先检查数组的长度，比一个阈值小的时候直接使用双轴快排。其它情况下，先检查数组中数据的顺序连续性。把数组中连续升序或者连续降序的信息记录下来，顺便把连续降序的部分倒置。这样数据就被切割成一段段连续升序的数列。
如果顺序连续性好，直接使用TimSort算法。TimSort算法的核心在于利用数列中的原始顺序，所以可以提高很多效率。
顺序连续性不好的数组直接使用了 双轴快排 + 成对插入排序。成对插入排序是插入排序的改进版，它采用了同时插入两个元素的方式调高效率。双轴快排是从传统的单轴快排到3-way快排演化过来的。参考：QUICKSORTING – 3-WAY AND DUAL PIVOT
final class DualPivotQuicksort {
/**
* Prevents instantiation.
*/
private DualPivotQuicksort() {}
/**
* 待合并的序列的最大数量
* The maximum number of runs in merge sort.
*/
private static final int MAX_RUN_COUNT = 67;
/**
* 待合并的序列的最大长度
* The maximum length of run in merge sort.
*/
private static final int MAX_RUN_LENGTH = 33;
/**
* 如果参与排序的数组长度小于这个值，优先使用快速排序而不是归并排序
* If the length of an array to be sorted is less than this
* constant, Quicksort is used in preference to merge sort.
*/
private static final int QUICKSORT_THRESHOLD = 286;
/**
* 如果参与排序的数组长度小于这个值，优先考虑插入排序，而不是快速排序
* If the length of an array to be sorted is less than this
* constant, insertion sort is used in preference to Quicksort.
*/
private static final int INSERTION_SORT_THRESHOLD = 47;
/**
* Sorts the specified range of the array using the given
* workspace array slice if possible for merging
*
* @param a the array to be sorted
* @param left the index of the first element, inclusive, to be sorted
* @param right the index of the last element, inclusive, to be sorted
* @param work a workspace array (slice)
* @param workBase origin of usable space in work array
* @param workLen usable size of work array
*/
static void sort(int[] a, int left, int right,
int[] work, int workBase, int workLen) {
// Use Quicksort on small arrays
if (right - left < QUICKSORT_THRESHOLD) {
sort(a, left, right, true);
return;
}
/*
* run[i] 意味着第i个有序数列开始的位置，（升序或者降序）
* 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;
// 检查数组是不是已经接近有序状态
// Check if the array is nearly sorted
for (int k = left; k < right; run[count] = k) {
if (a[k] < a[k + 1]) { // ascending 升序
while (++k <= right && a[k - 1] <= a[k]);
} else if (a[k] > a[k + 1]) { // descending 降序
while (++k <= right && a[k - 1] >= a[k]);
// 如果是降序的，找出k之后，把数列倒置
for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
int t = a[lo]; a[lo] = a[hi]; a[hi] = t;
}
} else { // equal 相等
for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
// 数列中有至少MAX_RUN_LENGTH的数据相等的时候，直接使用快排
if (--m == 0) {
sort(a, left, right, true);
return;
}
}
}
/*
* 数组并非高度有序，使用快速排序，因为数组中有序数列的个数超过了MAX_RUN_COUNT
* The array is not highly structured,
* use Quicksort instead of merge sort.
*/
if (++count == MAX_RUN_COUNT) {
sort(a, left, right, true);
return;
}
}
// 检查特殊情况
// Check special cases
// Implementation note: variable "right" is increased by 1.
if (run[count] == right++) { // The last run contains one element // 最后一个有序数列只有最后一个元素
run[++count] = right; // 那给最后一个元素的后面加一个哨兵
} else if (count == 1) { // The array is already sorted // 整个数组中只有一个有序数列，说明数组已经有序啦，不需要排序了
return;
}
// Determine alternation base for merge
byte odd = 0;
for (int n = 1; (n <<= 1) < count; odd ^= 1);
// 创建合并用的临时数组
// 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;
}
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;
}
// 合并
// 最外层循环，直到count为1，也就是栈中待合并的序列只有一个的时候，标志合并成功
// a 做原始数组，b 做目标数组
// Merging
for (int last; count > 1; count = last) {
// 遍历数组，合并相邻的两个升序序列
for (int k = (last = 0) + 2; k <= count; k += 2) {
// 合并run[k-2] 与 run[k-1]两个序列
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];
}
}
// 这里把合并之后的数列往前移动
run[++last] = hi;
}
// 如果栈的长度为奇数，那么把最后落单的有序数列copy过对面
if ((count & 1) != 0) {
for (int i = right, lo = run[count - 1]; --i >= lo;
b[i + bo] = a[i + ao]
);
run[++last] = right;
}
// 临时数组，与原始数组对调，保持a做原始数组，b 做目标数组
int[] t = a; a = b; b = t;
int o = ao; ao = bo; bo = o;
}
}
/**
* Sorts the specified range of the array by Dual-Pivot Quicksort.
*
* @param a the array to be sorted
* @param left the index of the first element, inclusive, to be sorted
* @param right the index of the last element, inclusive, to be sorted
* @param leftmost indicates if this part is the leftmost in the range
*/
private static void sort(int[] a, int left, int right, boolean leftmost) {
int length = right - left + 1;
// 小数组使用插入排序
// Use insertion sort on tiny arrays
if (length < INSERTION_SORT_THRESHOLD) {
if (leftmost) {
/*
* 经典的插入排序算法，不带哨兵。做了优化，在leftmost情况下使用
* Traditional (without sentinel) insertion sort,
* optimized for server VM, is used in case of
* the leftmost part.
*/
for (int i = left, j = i; i < right; j = ++i) {
int ai = a[i + 1];
while (ai < a[j]) {
a[j + 1] = a[j];
if (j-- == left) {
break;
}
}
a[j + 1] = ai;
}
} else {
/*
* 首先跨过开头的升序的部分
* Skip the longest ascending sequence.
*/
do {
if (left >= right) {
return;
}
} while (a[++left] >= a[left - 1]);
/*
* 这里用到了成对插入排序方法，它比简单的插入排序算法效率要高一些
* 因为这个分支执行的条件是左边是有元素的
* 所以可以直接从left开始往前查找
*
* 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.
*/
for (int k = left; ++left <= right; k = ++left) {
int a1 = a[k], a2 = a[left];
// 保证a1>=a2
if (a1 < a2) {
a2 = a1; a1 = a[left];
}
// 先把两个数字中较大的那个移动到合适的位置
while (a1 < a[--k]) {
a[k + 2] = a[k]; // 这里每次需要向左移动两个元素
}
a[++k + 1] = a1;
// 再把两个数字中较小的那个移动到合适的位置
while (a2 < a[--k]) {
a[k + 1] = a[k]; // 这里每次需要向左移动一个元素
}
a[k + 1] = a2;
}
int last = a[right];
while (last < a[--right]) {
a[right + 1] = a[right];
}
a[right + 1] = last;
}
return;
}
// length / 7 的一种低复杂度的实现, 近似值(length * 9 / 64 + 1)
// Inexpensive approximation of length / 7
int seventh = (length >> 3) + (length >> 6) + 1;
/*
* 对5段靠近中间位置的数列排序，这些元素最终会被用来做轴(下面会讲)
* 他们的选定是根据大量数据积累经验确定的
*
* 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 // 中间值
int e2 = e3 - seventh;
int e1 = e2 - seventh;
int e4 = e3 + seventh;
int e5 = e4 + seventh;
// 插入排序
// Sort these elements using insertion sort
if (a[e2] < a[e1]) { int t = a[e2]; a[e2] = a[e1]; a[e1] = t; }
if (a[e3] < a[e2]) { int 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]) { int 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]) { int 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; }
}
}
}
// 指针
// Pointers
int less = left; // The index of the first element of center part // 中间区域的首个元素的位置
int great = right; // The index before the first element of right part //右边区域的首个元素的位置
if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
/*
* 使用5个元素中的2，4两个位置，他们两个大致处在四分位的位置上
* 需要注意的是pivot1 <= pivot2
*
* 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.
*/
int pivot1 = a[e2];
int 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.
* 第一个和最后一个元素被放到两个轴所在的位置。当阶段性的分段结束后
* 他们会被分配到最终的位置并从子排序阶段排除
*/
a[e2] = a[left];
a[e4] = a[right];
/*
* Skip elements, which are less or greater than pivot values.
* 跳过一些队首的小于pivot1的值，跳过队尾的大于pivot2的值
*/
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; ) {
int 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.
* 这里考虑的好细致，"a[i] = b; i++"的效率要好过
* 'a[i++] = b'
*/
a[less] = ak;
++less;
} else if (ak > pivot2) { // Move a[k] to right part
while (a[great] > pivot2) {
if (great-- == k) { // k遇到great本次分割
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.
* 同上，用"a[i]=b;i--"代替"a[i--] = b"
*/
a[great] = ak;
--great;
}
} // 分割阶段结束出来的位置,上一个outer结束的位置
// 把两个放在外面的轴放回他们应该在的位置上
// Swap pivots into their final positions
a[left] = a[less - 1]; a[less - 1] = pivot1;
a[right] = a[great + 1]; a[great + 1] = pivot2;
// 把左边和右边递归排序，跟普通的快速排序差不多
// 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.
* 如果中心区域太大，超过数组长度的 4/7。就先进行预处理，再参与递归排序
* 预处理的方法是把等于pivot1的元素统一放到左边，等于pivot2的元素统一
* 放到右边，最终产生一个不包含pivot1和pivot2的数列，再拿去参与快排中的递归
*/
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; ) {
int 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;
}
} // outer结束的位置
}
// Sort center part recursively
sort(a, less, great, false);
} else { // Partitioning with one pivot // 这里选取的5个元素刚好相等，使用传统的3-way快排
/*
* Use the third of the five sorted elements as pivot.
* This value is inexpensive approximation of the median.
* 在5个元素中取中值
*/
int 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;
}
int ak = a[k];
if (ak < pivot) { // Move a[k] to left part // 把a[k]移动到左边去，把center区向右滚动一个单位
a[k] = a[less];
a[less] = ak;
++less;
} else { // a[k] > pivot - Move a[k] to right part // 把a[k]移动到右边
while (a[great] > pivot) { // 先找到右边最后一个比pivot小的值
--great;
}
if (a[great] < pivot) { // a[great] <= pivot 把他移到左边
a[k] = a[less];
a[less] = a[great];
++less;
} else { // a[great] == pivot //如果相等，中心区直接扩展
/*
* 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].
* 这里因为是整型值，所以a[k] == a[less] == pivot
*/
a[k] = pivot;
}
a[great] = ak;
--great;
}
}
/*
* Sort left and right parts recursively.
* All elements from center part are equal
* and, therefore, already sorted.
* 左右两边还没有完全排序，所以递归解决
* 中心区只有一个值，不再需要排序
*/
sort(a, left, less - 1, leftmost);
sort(a, great + 1, right, false);
}
}
}
