Arrays.sort()底层源码学习（1）
通过查看Arrays.sort()源码发现，从JDK7开始，底层实现在DualPivotQuicksort类，这个类是ladimir Yaroslavskiy，Jon Bentley和Josh Bloch等人实现的Dual-Pivot Quicksort算法。该算法在许多数据集上提供了O(nlogn)的时间复杂度，比传统的one-pivot 快排实现更快。
该类提供了些阈值，如下所示：
/**
* The maximum number of runs in merge sort.
* 合并排序的最大值
*/
private static final int MAX_RUN_COUNT = 67;
/**
* The maximum length of run in merge sort.
* 合并排序的最大长度
*/
private static final int MAX_RUN_LENGTH = 33;
/**
* If the length of an array to be sorted is less than this
* constant, Quicksort is used in preference to merge sort.
* 如果待排序数组的长度小于此常数，则优先选择采用快速排序而不是归并排序
*/
private static final int QUICKSORT_THRESHOLD = 286;
/**
* If the length of an array to be sorted is less than this
* constant, insertion sort is used in preference to Quicksort.
* 如果待排序数组的长度小于此常数，那么优先选择插入排序而不是快速排序
*/
private static final int INSERTION_SORT_THRESHOLD = 47;
/**
* If the length of a byte array to be sorted is greater than this
* constant, counting sort is used in preference to insertion sort.
* 如果byte数组的长度大于此常数，那么优先使用计数排序而不是插入排序。
*/
private static final int COUNTING_SORT_THRESHOLD_FOR_BYTE = 29;
/**
* If the length of a short or char array to be sorted is greater
* than this constant, counting sort is used in preference to Quicksort.
* 如果待排序数组未short数组或字符数组，且长度大于此常数，那么优先使用计数排序而不是快
* 速排序。
*/
private static final int COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR = 3200;该类中分别实现了不同数据类型数组的排序。
如sort(int[] a),sort(long[] a),sort(short[] a),sort(char[] a),sort(byte[] a),sort(float[] a),sort(double[] a)
下面以sort(int[] a)为例介绍一下（下面为sort(int[] a)的源码：
/**
* Sorts the specified array.
*
* @param a the array to be sorted
*/
public static void sort(int[] a) {
sort(a, 0, a.length - 1);
}
/**
* Sorts the specified range of the array.
*
* @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
*/
public static void sort(int[] a, int left, int right) {
// Use Quicksort on small arrays 小数组则采用快速排序
if (right - left < QUICKSORT_THRESHOLD) {
sort(a, left, right, true);
return;
}
/*
* Index run[i] is the start of i-th run
* (ascending or descending sequence).
* run[i]是第i个有序数列开始的位置（升序或降序）
*/
int[] run = new int[MAX_RUN_COUNT + 1];
int count = 0; run[0] = left;
// Check if the array is nearly sorted
//判断数组是否接近有序
for (int k = left; k < right; run[count] = k) {
if (a[k] < a[k + 1]) { // ascending 升序
//如果是升序，一直循环找到k,直到不再升序或到末尾
while (++k <= right && a[k - 1] <= a[k]);
} else if (a[k] > a[k + 1]) { // descending 降序
//如果是降序，一直循环找到k值，直到不再升序或者末尾，然后倒置
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 相等
//如果有至少MAX_RUN_LENGTH个连续相等数据，则直接使用快速排序
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.
* 如果这个数组不是高度结构化（这里表示的应该是是否高度排序）的，
* 则直接使用快速排序而不是归并排序
*/
if (++count == MAX_RUN_COUNT) {
sort(a, left, right, true);
return;
}
}
// Check special cases
//针对特殊情况
//1.如果一个有序数列只有最后一个元素，那么就给最后一个元素的后面加一个哨兵
if (run[count] == right++) { // The last run contains one element
run[++count] = right;
} else if (count == 1) { // The array is already sorted
//2.整个数组中只有一个有序数列，说明该数组已经有序，不需要再排序了，直接返回
return;
}
/*
* Create temporary array, which is used for merging.
* Implementation note: variable "right" is increased by 1.
* 创建合并时使用的临时数组 right = right+1
*/
int[] b; byte odd = 0;
for (int n = 1; (n <<= 1) < count; odd ^= 1);
if (odd == 0) {//偶数
b = a; a = new int[b.length];
for (int i = left - 1; ++i < right; a[i] = b[i]);
} else {//奇数
b = new int[a.length];
}
// Merging 合并操作 a为原始数组，b为目标数组
for (int last; count > 1; count = last) {
//遍历两个数组，合并相邻的两个升序数列
for (int k = (last = 0) + 2; k <= count; k += 2) {
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] <= a[q]) {
b[i] = a[p++];
} else {
b[i] = a[q++];
}
}
run[++last] = hi;
}
if ((count & 1) != 0) {
for (int i = right, lo = run[count - 1]; --i >= lo;
b[i] = a[i]
);
run[++last] = right;
}
//临时数组，保证a作为原始数组，b为目标数组
int[] t = a; a = b; b = t;
}
}
/**
* Sorts the specified range of the array by Dual-Pivot Quicksort.
* 通过双轴快排实现排序
* @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
* leftmost 参数：指定的范围是否在数组的最左边
*/
private static void sort(int[] a, int left, int right, boolean leftmost) {
int length = right - left + 1;
// Use insertion sort on tiny arrays
//小数组采用插入排序
if (length < INSERTION_SORT_THRESHOLD) {
if (leftmost) {
/*
* Traditional (without sentinel) insertion sort,
* optimized for server VM, is used in case of
* the leftmost part.
* 经典插入排序算法（不使用哨兵），优化服务器VM，用于leftmost情况
*/
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 {
/*
* Skip the longest ascending sequence.
* 跳过开头的升序部分
*/
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.
* 相邻部分的每个元素都起到哨兵的作用，
* 因此可以避免每次迭代时检查left的范围
* 这里采用了更加优化的算法，叫做“成对插入算法”，
* 这比传统的插入排序要快（在快速排序的情况下）
*/
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];
}
//将a1插入到合适的位置
while (a1 < a[--k]) {
a[k + 2] = a[k];
}
a[++k + 1] = a1;
//将a2插入到合适的位置
while (a2 < a[--k]) {
a[k + 1] = a[k];
}
a[k + 1] = a2;
}
int last = a[right];
while (last < a[--right]) {
a[right + 1] = a[right];
}
a[right + 1] = last;
}
return;
}
// Inexpensive approximation of length / 7
//低复杂度的实现
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.
* 对5段靠近中间位置的数列排序，这些元素将被作为轴
*/
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
//插入排序
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; }
}
}
}
// Pointers 指针
//less为中间部分首个元素的位置，great为右边部分首个元素的位置
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.
*/
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.
*/
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; ) {
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.
*/
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; ) {
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;
}
}
}
// 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.
*/
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
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.
* 将会左右两部分递归排序
*/
sort(a, left, less - 1, leftmost);
sort(a, great + 1, right, false);
}
}