关于JDK中的Arrays.sort排序方法

public static void sort(int[] a) {
    DualPivotQuicksort.sort(a, 0, a.length - 1, null, 0, 0);
}

/**
    如果可能合并，使用给定的工作空间数组片对数组的指定范围进行排序
    参数:
    a -要排序的数组
    left - 要排序的第一个元素的索引(包括第一个元素)
    right - 最后一个要排序的元素的下标
    Work -工作空间数组(片)
    workBase -工作数组中可用空间的来源
    workLen -工作数组的可用大小
*/

//如果要排序的数组长度小于这个常量，则优先使用快速排序，而不是归并排序。
private static final int QUICKSORT_THRESHOLD = 286;
//归并排序的最大运行次数。
private static final int MAX_RUN_COUNT = 67;

static void sort(int[] a, int left, int right,int[] work, int workBase, int workLen) {
    // 小数组使用快排，阈值为286
    if (right - left < QUICKSORT_THRESHOLD) {
        //快排
        sort(a, left, right, true);
        return;
    }

    //索引运行[i]是第i次运行的开始(升序或降序)。
    int[] run = new int[MAX_RUN_COUNT + 1];
    int count = 0; run[0] = left;

    // 检查数组是否接近排序 见解释 1
    for (int k = left; k < right; run[count] = k) {
        // 序列开头相等的项
        while (k < right && a[k] == a[k + 1])
            k++;
        if (k == right) break;  // 序列以相等的项结束
        if (a[k] < a[k + 1]) { // 升序
            while (++k <= right && a[k - 1] <= a[k]);
        } else if (a[k] > a[k + 1]) { // 降序
            while (++k <= right && a[k - 1] >= a[k]);
            //转换成一个升序序列
            for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
                int t = a[lo]; 
                a[lo] = a[hi]; 
                a[hi] = t;
            }
        }

        // 合并一个转换后的降序序列，后面跟着升序序列
        if (run[count] > left && a[run[count]] >= a[run[count] - 1]) {
            count--;
        }

        /*
        * 如果count == 归并排序的最大次数说明数组不是高度结构化的，使用快速排序代替归并排序。
        */
        if (++count == MAX_RUN_COUNT) {
            //进入到快排
            sort(a, left, right, true);
            return;
        }
    }

    // 这些不变量应该是正确的:
    //    run[0] = 0
    //    run[<last>] = right + 1; (terminator)

    if (count == 0) {
        // 单次相等跑
        return;
    } else if (count == 1 && run[count] > right) {
        //单个升序或转换后的降序运行。总是检查一个最终运行是一个适当的结束符，否则我们有一个未终止的尾随运行，以处理下游。
        return;
    }
    right++;
    if (run[count] < right) {
        // 极端情况:最后的运行不是终结者。
        //这可能发生：如果最后一次运行是一个等于运行，或有一个单元素运行在结尾。通过在末尾添加适当的终止符来解决这个问题。
        //注意，我们以(right + 1)结束，在前面加1。
        run[++count] = right;
    }

    // 确定合并的交替基
    byte odd = 0;
    for (int n = 1; (n <<= 1) < count; odd ^= 1);

    // 使用或创建临时数组b进行合并
    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;
    }

    // 归并
    for (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;
    }
}

//上文快速排序入口

//如果要排序的数组长度小于这个值常量，插入排序优先于快速排序。
private static final int INSERTION_SORT_THRESHOLD = 47;

/**
* 按双重Povit快速排序 对数组的指定范围进行排序
*
* @param a 待排序数组
* @param left 待排数组左索引
* @param right 待排数组右索引
* @param leftmost 指示这部分是否为范围的最左边
*/
private static void sort(int[] a, int left, int right, boolean leftmost) {
    int length = right - left + 1;

    // 小于47，使用插入排序
    if (length < INSERTION_SORT_THRESHOLD) {
        if (leftmost) {
           //传统的(没有哨兵的)插入排序，为服务器虚拟机优化，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 {
            //跳过最长升序序列。
            do {
                if (left >= right) {
                    return;
                }
            } while (a[++left] >= a[left - 1]);

            //来自相邻部分的每一个元素都起着作用，因此这允许我们避免每次迭代的左范围检查。
            //此外,我们使用更优化的算法，所谓的对插入排序，它更快(在快速排序的上下文中)比传统的插入排序实现。
            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;
    }

    //////////////以下为双轴快排
    // 长度/7的近似
    int seventh = (length >> 3) + (length >> 6) + 1;

    // 对五个均匀间隔的元素进行排序 ，在范围的中心元素。这些元素将被用于枢轴选择如下所述。
    //间隔的选择 根据经验，这些因素决定很好地发挥作用 各种各样的输入。
    int e3 = (left + right) >>> 1; // The midpoint
    int e2 = e3 - seventh;
    int e1 = e2 - seventh;
    int e4 = e3 + seventh;
    int e5 = e4 + seventh;

    // 使用插入排序对这些元素进行排序
    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;  // The index of the first element of center part
    int great = right; // The index before the first element of right part

    if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
        //使用五个排序元素中的第二个和第四个作为枢轴。 
        //这些值是第一个和的廉价近似值. 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);

        /*
            * 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;
            }
        }

        // 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; ) {
                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;
                }
            }
        }

        // 对中心部分进行递归排序
        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.
            */
        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;
            }
        }

        //对左右部分递归排序。所有元素从中心部分是相等的，因此，已经排序。
        sort(a, left, less - 1, leftmost);
        sort(a, great + 1, right, false);
    }
}