Dual-Pivot QuickSort
今天做题的时候想看看 Arrays.sort() 的快排是怎么实现的，然后惊讶地发现它居然不是简单的快排，是一个很长名字的快排——Dual-Pivot QuickSort。
论文 Dual-Pivot Quicksort algorithm
1. Dual-Pivot Quick Sort 简介
dual 的意思是"双的，双数的"。顾名思义，显然 Dual-Pivot 就是指在经典快排的基础上创造了两个主元。
一般而言，选择最左和最右两个主元分别为 p1 和 p2, 确保 p1 < p2。
将数组分成 (-∞, p1]; (p1, p2); [p2, +∞) 三个部分。
2. Dual-Pivot Quick Sort 源码
理论上使用双轴快速排序,如果数组小于QUICKSORT_THRESHOLD 286 就使用经典快速排序,如果小于 INSERTION_SORT_THRESHOLD 47 就使用插入排序,否则使用双轴快速排序。
(1) 以下是 Arrays 的 sort。
public static void sort(int[] a) {DualPivotQuicksort.sort(a, 0, a.length - 1, null, 0, 0);
}(2) 以下是 DualPivotQuicksort 的 sort 方法。
/*** Sorts the specified range of the array using the given* workspace array slice if possible for merging* 如果可能的话，使用给定的工作空间数组切片对数组的指定范围进行排序* * @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 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 小数组上使用快排if (right - left < QUICKSORT_THRESHOLD) {// 这个 sort 见下面的分析sort(a, left, right, true); return;}/** 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 sortedfor (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]);for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {int t = a[lo]; a[lo] = a[hi]; a[hi] = t;}} else { // equalfor (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 elementrun[++count] = right;} else if (count == 1) { // The array is already sortedreturn;}// Determine alternation base for mergebyte odd = 0;for (int n = 1; (n <<= 1) < count; odd ^= 1);// Use or create temporary array b for mergingint[] b; // temp array; alternates with aint ao, bo; // array offsets from 'left'int blen = right - left; // space needed for bif (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;}// Mergingfor (int last; count > 1; count = last) {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 + ao] <= a[q + ao]) {b[i + bo] = a[p++ + ao];} else {b[i + bo] = a[q++ + ao];}}run[++last] = hi;}if ((count & 1) != 0) {for (int i = right, lo = run[count - 1]; --i >= lo;b[i + bo] = a[i + ao]);run[++last] = right;}int[] t = a; a = b; b = t;int o = ao; ao = bo; bo = o;}
}(3) 本场重点——双主元快速排序
/*** Sorts the specified range of the array by Dual-Pivot Quicksort.* 使用双主元快速排序对数组的指定范围进行排序。** @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*/
private static void sort(int[] a, int left, int right, boolean leftmost) {int length = right - left + 1;// Use insertion sort on tiny arrays 对小数组使用插入排序if (length < INSERTION_SORT_THRESHOLD) {if (leftmost) {/** Traditional (without sentinel) insertion sort,* optimized for server VM, is used in case of* the leftmost part.* 传统的(没有标记的)插入排序，为服务器VM优化，用于最左边的部分。*/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.* 跳过最长的升序序列。*/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;}// Inexpensive approximation of length / 7int 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 midpointint e2 = e3 - seventh;int e1 = e2 - seventh;int e4 = e3 + seventh;int e5 = e4 + seventh;// Sort these elements using insertion sortif (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; }}}}// Pointersint less = left; // The index of the first element of center partint great = right; // The index before the first element of right partif (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);/** 第一次看到图形化的注释* 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 parta[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 partwhile (a[great] > pivot2) {if (great-- == k) {break outer;}}if (a[great] < pivot1) { // a[great] <= pivot2a[k] = a[less];a[less] = a[great];++less;} else { // pivot1 <= a[great] <= pivot2a[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 positionsa[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 pivotssort(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 parta[k] = a[less];a[less] = ak;++less;} else if (ak == pivot2) { // Move a[k] to right partwhile (a[great] == pivot2) {if (great-- == k) {break outer;}}if (a[great] == pivot1) { // a[great] < pivot2a[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] < pivot2a[k] = a[great];}a[great] = ak;--great;}}}// Sort center part recursivelysort(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 parta[k] = a[less];a[less] = ak;++less;} else { // a[k] > pivot - Move a[k] to right partwhile (a[great] > pivot) {--great;}if (a[great] < pivot) { // a[great] <= pivota[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);}
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