/**
     * The maximum number of runs in merge sort.
     */
    private static final int MAX_RUN_COUNT = 67;

    /**
     * The maximum length of run in merge sort.
     */
    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.
     */
    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.
     */
    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.
     */
    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.
     */
    private static final int COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR = 3200;

/*
 * Copyright (c) 2009, 2016, Oracle and/or its affiliates. All rights reserved.
 * ORACLE PROPRIETARY/CONFIDENTIAL. Use is subject to license terms.
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 *
 */

package java.utill;

/**
 * This class implements the Dual-Pivot Quicksort algorithm by
 * Vladimir Yaroslavskiy, Jon Bentley, and Josh Bloch. The 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.
 * 此处介绍一些双轴快排的优势，对很多两次循环后引起低效率的数据排列方式，
 * 都保持O(nlogn)复杂度
 *
 * All exposed methods are package-private, designed to be invoked
 * from public methods (in class Arrays) after performing any
 * necessary array bounds checks and expanding parameters into the
 * required forms.
 * 关于重用这个私有类的介绍
 *
 * @author Vladimir Yaroslavskiy
 * @author Jon Bentley
 * @author Josh Bloch
 *
 * @version 2011.02.11 m765.827.12i:5\7pm
 * @since 1.7
 */
final class DualPivotQuicksort {

    /**
     * Prevents instantiation.防止这个类被实例化
     */
    private DualPivotQuicksort() {
    }

    /*
     * Tuning parameters.一些设置好的阈值数据，这些数据经过实验证明是最优的
     */

    /**
     * The maximum number of runs in merge sort.
     * 待合并的序列的最大数量
     */
    private static final int MAX_RUN_COUNT = 67;

    /**
     * If the length of an array to be sorted is less than this
     * constant, Quicksort is used in preference to merge sort.
     * 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
     */
    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.
     * 如果参与排序的数组长度小于这个值，考虑插入排序，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.
     * 这个是byte数组的
     */
    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.
     * 这个是short或char数组的启用计数排序的阈值
     */
    private static final int COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR = 3200;

    /*
     * Sorting methods for seven primitive types.
     * 针对7种基本类型的排序方法，这篇博客仅讨论int类型和short类型
     */

    /**
     * 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
        //小序列直接使用快排。
        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] 意味着第i个有序数列开始的位置，（升序或者降序）
         */
        int[] run = new int[MAX_RUN_COUNT + 1];
        int count = 0;
        run[0] = left;

        // Check if the array is nearly sorted
        // 检查数组是不是已经接近有序状态。这个循环用于分割有序序列
        for (int k = left; k < right; run[count] = k) {
            // Equal items in the beginning of the sequence
            //跳过序列开头一些相等的元素
            while (k < right && a[k] == a[k + 1])
                k++;
            if (k == right) break;  // Sequence finishes with equal items直接结束，run[count]~k全部元素相等
            if (a[k] < a[k + 1]) { // ascending升序
                while (++k <= right && a[k - 1] <= a[k]) ;//run[account]~k的都是升序
            } else if (a[k] > a[k + 1]) { // descending降序
                while (++k <= right && a[k - 1] >= a[k]) ;
                // Transform into an ascending sequence转为升序
                for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
                    int t = a[lo];
                    a[lo] = a[hi];
                    a[hi] = t;
                }
                //处理到这里，run[account]~k的都是升序（包含相等）
            }

            // Merge a transformed descending sequence followed by an
            // ascending sequence
            //合并两个升序列，其中前一个是由降序转化而来，详情见图一
            if (run[count] > left && a[run[count]] >= a[run[count] - 1]) {
                count--;
            }

            /*
             * The array is not highly structured,
             * use Quicksort instead of merge sort.
             * 若待归并序列超过阈值，表明该序列不是高度接近有序，那么改为使用快速排序
             */
            if (++count == MAX_RUN_COUNT) {
                sort(a, left, right, true);
                return;
            }
        }

        // These invariants should hold true:
        //    run[0] = 0
        //    run[<last>] = right + 1; (terminator)

        //这三个判断用来处理最后一个子序列的终止问题
        if (count == 0) {
            // A single equal run
            //这个序列所有元素相等，则直接返回。显然是由上面的break打破循环产生。
            return;
        } else if (count == 1 && run[count] > right) {
            // Either a single ascending or a transformed descending run.
            // Always check that a final run is a proper terminator, otherwise
            // we have an unterminated trailing run, to handle downstream.
            //这里表示序列已经是升序列，直接返回
            return;
        }
        /*
         * 这个地方的触发有两种情况：
         * 1、run[count]==right（未加1前）
         * 2、由上面的break触发，也即run[count]后面是一串有相等元素的数列
         * 这样就必须为不在（left~right）范围内的下一个子序列设定起始，
         * 这样才能终止范围内的最后一个子序列
         * 详见图二
         * 同时，对照上面的elseif,也是一个有关终止条件的行为
         */
        right++;
        if (run[count] < right) {
            // Corner case: the final run is not a terminator. This may happen
            // if a final run is an equals run, or there is a single-element run
            // at the end. Fix up by adding a proper terminator at the end.
            // Note that we terminate with (right + 1), incremented earlier.
            run[++count] = right;
        }

        // Determine alternation base for merge
        //决定产生变化的基数组，即在work[]上变化还是在a[]上变化（参见下面第一个if-else）
        //这个优化确实看不太懂
        byte odd = 0;
        //由下面语句知n和odd的关系为：
        //odd   0   1   0   1   ……
        //n     1   2   4   8   ……
        for (int n = 1; (n <<= 1) < count; odd ^= 1) ;

        // Use or create temporary array b for merging
        //为归并过程创建一个临时数组b
        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;
        }
        //决定a和b的指向，指向原a[]还是work[]
        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
        // 归并
        // 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，标志归并成功
        // a 做原始数组，b 做临时数组
        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;
            }
            //这里为什么要交换a、b呢？详见图三
            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;

        // Use insertion sort on tiny arrays
        //小数组使用插入排序
        if (length < INSERTION_SORT_THRESHOLD) {
            if (leftmost) {//代表要比较的序列位于数组最左边
                /*
                 * Traditional (without sentinel) insertion sort,
                 * optimized for server VM, is used in case of
                 * the leftmost part.
                 *  经典的插入排序算法，不带哨兵。做了优化，在leftmost情况下使用
                 */
                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.
                 * 首先跨过开头的升序的部分
                 */
                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.
                 * 成对插入排序，常见图四
                 * 具体执行过程：上面的do-while循环已经排好的最前面的数据
    ​    ​    ​  * (1)将要插入的数据，第一个值赋值a1,第二个值赋值a2,
    ​    ​    ​  * (2)然后判断a1与a2的大小，使a1>a2(关键点)
    ​    ​    ​  * (3)接下来，首先是插入大的数值a1，将a1与k之前的数字一一比较，
                 *      直到数值小于a1为止，把a1插入到合适的位置，
                 *      注意：这里的相隔距离为2，目的是为了给a2留下插入的空隙
    ​            * (4)接下来，插入小的数值a2,将a2与此时k之前的数字一一比较，
                 *      （这个k已经变化到a1左边，即此时，只需在a1左边插入a2即可，
                 *      因此减少了a2的比较次数，优化就发生在这里）
                 *      直到数值小于a2为止，将a2插入到合适的位置，
                 *      注意：这里的相隔距离为1
                 * (5)最后把最后一个没有遍历到的数据插入到合适位置
                 *
                 * 还有一个问题：为什么一定不能是leftmost呢？
                 * 如果是leftmost,可能会发生左越界
                 *
                 */
                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;
        }

        //开始双轴快排的选轴工作
        // Inexpensive approximation of length / 7
        // length / 7 的一种低复杂度的实现
        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.
         * 取五个靠近中间位置的元素，这五个位置的间隔为length/7，
         * 对这五个元素进行排序，这些元素最终会被用来做轴(见图五)
         */
        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
        //使用插入排序将这五个位置的元素排序起来
        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
        int less = left;  // 中间区域的首个元素的位置
        int great = right; // 右边区域的首个元素的位置

        //若满足下面这个条件，则以e2和e4进行双轴快排，否则以e3进行单轴快排
        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.
             * 使用5个元素中的e2，e4两个位置的元素作为轴，他们两个大致处在四分位的位置上。
             * 需要注意的是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.
             *  这里就是用第一个和最后一个元素覆盖e2和e4上的元素，
             *  这样e2和e4上的元素就被排除出下面的排序，因为他们已经是枢轴
             */
            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.
             * 这里的思想是：k不断右移，将元素放进相对应的区域，直到碰上great。
             */
            outer:
            for (int k = less - 1; ++k <= great; ) {
                int ak = a[k];
                if (ak < pivot1) {
                    // Move a[k] to left part
                    //挪到left part中去
                    a[k] = a[less];
                    /*
                     * Here and below we use "a[i] = b; i++;" instead
                     * of "a[i++] = b;" due to performance issue.
                     * "a[i] = b; i--;"的效率比"a[i--] = b;"的效率高
                     */
                    a[less] = ak;
                    ++less;
                } else if (ak > pivot2) {
                    // Move a[k] to right part
                    //挪到right part中去
                    while (a[great] > pivot2) {//这里先将一波大于privot2的放在right part
                        if (great-- == k) { // k遇到great，本次分割终止
                            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
            /*
             * 由于排序的部分是a[left+1...right-1]，而原范围却是a[left...right]，
             * 而那两个被扣掉的元素就是pivot1和pivot2，
             * 因此，经过交换，将pivot1和pivot2插入到a[less-1]和a[great+1]中去，
             * 作为真正的枢轴，并维持left-part、center-part、right-part三个部分
             */
            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
            //递归快排左右两边的序列
            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
                        //挪到left part中去
                        a[k] = a[less];
                        a[less] = ak;
                        ++less;
                    } else if (ak == pivot2) {
                        // Move a[k] to right part
                        //挪到left part中去
                        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
            //用单轴3-way进行分区，因为e1-e5至少存在一对相等的元素，
            // 因此判定这个数组中重复的元素居多
            /*
             * 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.
             * 单轴快排和上面的双轴差不多，一个动图介绍图五
             */
            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.
             * 递归快排
             */
            sort(a, left, less - 1, leftmost);
            sort(a, great + 1, right, false);
        }
    }

    /**
     * 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
     * 统计排序适用于元素个数远大于元素种数的情况，
     * 适用于Short、Byte、Char等元素种数较少的类型。
     */
    static void sort(short[] a, int left, int right,
                     short[] work, int workBase, int workLen) {
        // Use counting sort on large arrays
        if (right - left > COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR) {
            int[] count = new int[NUM_SHORT_VALUES];

            //从左往右，存在的数的count加1
            for (int i = left - 1; ++i <= right;
                 count[a[i] - Short.MIN_VALUE]++
                    );
            //从右往左，将count不是0的数提取出来，这样的序列就变成有序了
            for (int i = NUM_SHORT_VALUES, k = right + 1; k > left; ) {
                while (count[--i] == 0);
                short value = (short) (i + Short.MIN_VALUE);
                int s = count[i];

                do {
                    a[--k] = value;
                } while (--s > 0);
            }
        } else { // Use Dual-Pivot Quicksort on small arrays
            //doSort(a, left, right, work, workBase, workLen);
            //上面的doSort与之前的快排sort差不多，这里就不赘述了。
        }
    }
    /** The number of distinct short values. */
    private static final int NUM_SHORT_VALUES = 1 << 16;
}