Arrays.sort()的底层实现原理
文章目录
- 前言
- 一、Arrays.sort()总览
-
- 数组长度小于286
- 数组长度小于47
- 数组长度大于等于47
- 数组长度大于等于286
- 二、总结
前言
最近在leetcode刷题，遇到挺多Arrays.sort()这个API，今天在牛客网看到有面试官问这个API对于快排做了什么优化，那么面向面试学习的我也要去学习一下啦。
本文基于JDK1.8。
一、Arrays.sort()总览
先上答案：jdk1.8之前，Arrays.sort()方法使用的是传统快排的方式进行排序。jdk1.8后，Arrays.sort()方法使用的是双轴快排。
双轴快排(DualPivotQuicksort)的基本思想是：
顾名思义有两个轴元素pivot1，pivot2，且pivot ≤pivot2 将序列分成三段：x < pivot1、pivot1 ≤ x ≤ pivot2、x >pivot2 然后分别对三段进行递归
一幅我觉得很好的流程图。图片以及本文主要内容来自深入理解Arrays.sort()底层实现
sort方法进来后，调用了DualPivotQuicksort.sort()。传统快排（单轴快排）的时间复杂度最差的情况为n²，DualPivotQuicksort（双轴快排）能保证大多数数组排序的时间复杂度保持在O(nlogn) 。
/**
* Sorts the specified array into ascending numerical order.
*
* Implementation note: The sorting algorithm is a Dual-Pivot Quicksort
* by Vladimir Yaroslavskiy, Jon Bentley, and Joshua Bloch. This algorithm
* offers O(n log(n)) performance on many data sets that cause other
* quicksorts to degrade to quadratic performance, and is typically
* faster than traditional (one-pivot) Quicksort implementations.
*
* @param a the array to be sorted
*/
public static void sort(int[] a) {
DualPivotQuicksort.sort(a, 0, a.length - 1, null, 0, 0);
}
数组长度小于286
DualPivotQuicksort.sort()进来后，会碰到第一个阀值：QUICKSORT_THRESHOLD（286），数组长度小于这个阀值的进入插入排序或者Quicksort （快速排序）
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;
}
//······
}
数组长度小于47
进入sort(a, left, right, true)，遇到们第二个阀值INSERTION_SORT_THRESHOLD（47）。如果元素少于47这个阀值，就用插入排序。参数leftmost的含义是给定的范围，是不是这个数组最左边的部分。
private static void sort(int[] a, int left, int right, boolean leftmost) {
int length = right - left + 1;
// Use insertion sort on tiny arrays
if (length < INSERTION_SORT_THRESHOLD) {
if (leftmost) {
/*
* 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 {
/*
* 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.
*/
for (int k = left; ++left <= right; k = ++left) {
int a1 = a[k], a2 = a[left];
if (a1 < a2) {
a2 = a1; a1 = a[left];
}
while (a1 < a[--k]) {
a[k + 2] = a[k];
}
a[++k + 1] = a1;
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;
}
数组长度大于等于47
如果大于47这个阀值，则选择一种快速排序的方法：选出e1,e2,e3,e4,e5五个点，将数组等分为6份，称为 “基准”（pivot）；针对这个5个元素，进行插入排序。
// 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]) { 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; }
}
}
}
选取a[e2]，a[e4]分别作为pivot1，pivot2。由于进行了插入排序，pivot1 <=pivot2 。定义两个指针less和great。less从最左边开始向右遍历，一直找到第一个不小于pivot1的元素；great从右边开始向左遍历，一直找到第一个不大于pivot2的元素。
/*
* 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);
指针k从less-1开始向右遍历至great，把小于pivot1的元素移动到less左边，大于pivot2的元素移动到great右边。
/*
* 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;
}
}
将less-1处的元素移动到队头，great+1处的元素移动到队尾，并把pivot1和pivot2分别放到less-1和great+1处。至此，less左边的元素都小于pivot1，great右边的元素都大于pivot2，分别对两部分进行同样的递归排序。
/*
* 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);
在这里其实有一个分支，如果e1，e2，e3，e4，e5有相等的情况，
则选取a[e3]作为pivot，即经典的单轴快排。
/*
* 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.
*/
sort(a, left, less - 1, leftmost);
sort(a, great + 1, right, false);
数组长度大于等于286
数组长度小于286的情况已经介绍完了，当大于等于286的时候先对数组进行一个Check if the array is nearly sorted判断，看看是否适合使用归并排序。这个判断主要作用是看数组具不具备有序结构：每降序为一个组，像1,9,8,7,6,8。9到6是降序，为一个组，然后把降序的一组排成升序：1,6,7,8,9,8。然后最后的8后面继续往后面找。每遇到这样一个降序组，++count，当count大于MAX_RUN_COUNT（67）或者有超过33个相同元素即MAX_RUN_LENGTH（33），被判断为这个数组不具备有序结构，送给之前的sort(int[] a, int left, int right, boolean leftmost)（The array is not highly structured,use Quicksort instead of merge sort）。反之进入归并排序。
// 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
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;
}
}
二、总结
Arrays.sort()并不是单一的排序，而是插入排序，快速排序，归并排序三种排序的组合。数组长度大于或等于47或小于286会进入快排，而在大于或等于286后，会// Check if the array is nearly sorted，即使大于286，但在降序组太多的时候（The array is not highly structured,use Quicksort instead of merge sort），要转回快排。