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); } } }