Java1.8源码阅读-DualPivotQuicksort

在刷leetcode时，涉及到一个排序算法时，自己采用的冒泡，和快速排序，出现超时问题，于是看了解决方案，解决方案采用的是Arrays.sort,于是对其内部实现排序进行了查看。总的来说，Arrays.sort在对int[] 数组进行排序时，根据其长度进行动态的排序算法选择。

基本流程：

• 1：如果长度小于 QUICKSORT_THRESHOLD(286),则采用非归并并排序

        // Use Quicksort on small arrays        if (right - left < QUICKSORT_THRESHOLD) {            sort(a, left, right, true);            return;        }

• 1.1 如果长度小于INSERTION_SORT_THRESHOLD（47），则采用插入排序

 int length = right - left + 1;        // Use insertion sort on tiny arrays        if (length < INSERTION_SORT_THRESHOLD) {

• 1.1.1 如果最左区间（以初始left开始的区间） leftmost：普通插入排序

             if (leftmost) {                /*                 * Traditional (without sentinel) insertion sort,                 * optimized for server VM, is used in case of                 * the leftmost part.                 */

• 1.1.2 否则 pair insertion sort,

/*                 * 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.                 */

• 1.2 否则采用快速排序算法

  • 将数组划分为7段（大约），然后找出第2、3、4、5、6段的右端点对应的位置
  • 对这5个位置上的数字进行插入排序，作为枢轴的候选

  • 如果5个数都不相等

    • 选取排序后的2、4作为枢轴，进行双枢轴排序 Dual-Pivot Quicksort

    • 排序后如果，中间部分元素过多，可能原因是等于pivort1和等于pivort2的元素过多，则将其调整

    • 否则，进行普通快速排序，枢轴排序后为第3个元素

• 2 否则，考虑timesort(归并排序的优化版本，对一会升序、一会降序的混合情况处理比较好)

  • 2.1 创建Timesort run数组，大小为 MAX_RUN_COUNT(67)+1

/*         * 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;

• 2.1.1 a[run[i]] ~ a[run[i + 1]]之间为升序数组

• 2.1.2检查当前待排序数组是否适合使用Timsort,即run数组中升序数组个数，如果个数不小于MAX_RUN_COUNT则认为数组内元素排序比较混乱，适合非归并排序

  注：对于连续下降的元素会将其调整为连续上升

• 2.2 如果通过上述检测，则进行归并排序

代码

插入排序

for (int i = left, j = i; i < right; j = ++i) {    int ai = a[i + 1];      // 带插入元素    while (ai < a[j]) {     // 寻找插入位置        a[j + 1] = a[j];        if (j-- == left) {            break;        }    }    a[j + 1] = ai;          // 插入新元素}

改进版插入排序

pair insertion sort,每次插入两个元素

/** * 注：left左侧的内容均已排好序，默认的前提条件 *//* * 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. * 一次遍历插入两个元素 */for (int k = left; ++left <= right; k = ++left) {    int a1 = a[k], a2 = a[left];    // 使得 a1 >= a2    if (a1 < a2) {        a2 = a1; a1 = a[left];    }    // 寻找a1的插入位置，相隔距离为2    while (a1 < a[--k]) {        a[k + 2] = a[k];    }    a[++k + 1] = a1;    // 寻找a2的插入位置，相隔距离为1    while (a2 < a[--k]) {        a[k + 1] = a[k];    }    a[k + 1] = a2;}// 将最后一个位置插入到合适位置int last = a[right];while (last < a[--right]) {    a[right + 1] = a[right];}a[right + 1] = last;

双枢轴排序 Dual-Pivot Quicksort

// 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]) { long t = a[e2]; a[e2] = a[e1]; a[e1] = t; }if (a[e3] < a[e2]) { long 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]) { long 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]) { long 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.     */    long pivot1 = a[e2];    long 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; ) {        long 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; ) {            long 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.     */    long 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;        }        long 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);

timesort 归并排序

/* * 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;// Check if the array is nearly sortedfor (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        // 将降序数组变为升序        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 { // 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;    }}// Check special cases// Implementation note: variable "right" is increased by 1.if (run[count] == right++) { // The last run contains one element    run[++count] = right;} else if (count == 1) { // The array is already sorted    return;}// Determine alternation base for merge// 确定归并排序的迭代次数（每迭代一次，将相邻升序子序列合并，即run内元素数目减半）// 简单示例：a[1, 5, 2, 6, 3, 7, 4, 8] ==> a[1, 2, 5, 6, 3, 4, 7, 8]byte 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;}// 根据归并的迭代次数，更改a,b// a,b中必有一个数组为原始数组，另一个为临时数组// 在归并的过程中，每迭代一次，run内元素数目减半，同时a,b会交换一次// 为了保证最后一次迭代后，原始数组内存有归并好的数据，需要进行如下考虑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;}// Merging// a是原始数组，b是目标数组for (int last; count > 1; count = last) {    // 合并两个相邻升序序列    for (int k = (last = 0) + 2; k <= count; k += 2) {        // 确定边界        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];            }        }        // 更新子序列标示        run[++last] = hi;    }    // 如果升序子序列个数为奇数，之前两两合并时，最后会剩余一个，将剩余的直接拷贝到b, 等待下一次合并    if ((count & 1) != 0) {        for (int i = right, lo = run[count - 1]; --i >= lo;            b[i + bo] = a[i + ao]        );        run[++last] = right;    }    // 交换a b    int[] t = a; a = b; b = t;    int o = ao; ao = bo; bo = o;