从常用的Arrays.sort()源码学习快排和归并排序

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两岸猿声啼不住，轻舟已过万重山。——《早发白帝城》·李白

前言

在做Java的某些数组排序问题时，我们常常用到sort方法，那么现在就来具体看看sort里面藏了什么乾坤，它到底是如何运行的。

从Arrays.sort(int[] a)进入

public static void sort(int[] a) {
    DualPivotQuicksort.sort(a, 0, a.length - 1, null, 0, 0);
    /*各参数
    0				——>left,左下标
    a.length - 1	——>right，右下标
    null			——>work[],
    0				——>workBase
    0				——>workLen
    */
}

/**
 * If the length of an array to be sorted is less than this
 * constant, Quicksort is used in preference to merge sort.

	小于286时快排更加优先于归并排序
 */
private static final int QUICKSORT_THRESHOLD = 286;

归并排序

static void sort(int[] a, int left, int right,
                 int[] work, int workBase, int workLen) {
    // Use Quicksort on small arrays	
    // 顾名思义，数组大小比较小的时候采用快排，那到底是多小算小呢？
    if (right - left < QUICKSORT_THRESHOLD) {// 小于快排边界的时候算小，往上看，边界=286
        sort(a, left, right, true);// 此时采用快排<——>快排在下一节
        return;
    }

    /*
     * Index run[i] is the start of i-th run 	索引run[i]是第i次运行的开始
     * (ascending or descending sequence).	升序或降序
     */
    int[] run = new int[MAX_RUN_COUNT + 1];
    int count = 0; run[0] = left;

    // Check if the array is nearly sorted
    for (int k = left; k < right; run[count] = k) {
        // Equal items in the beginning of the sequence
        while (k < right && a[k] == a[k + 1])
            k++;
        if (k == right) break;  // Sequence finishes with equal items
        if (a[k] < a[k + 1]) { // ascending
            while (++k <= right && a[k - 1] <= a[k]);
        } else if (a[k] > a[k + 1]) { // descending
            while (++k <= right && a[k - 1] >= a[k]);
            // Transform into an ascending sequence
            for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
                int t = a[lo]; a[lo] = a[hi]; a[hi] = t;
            }
        }

        // Merge a transformed descending sequence followed by an
        // ascending sequence
        if (run[count] > left && a[run[count]] >= a[run[count] - 1]) {
            count--;
        }

        /*
         * The array is not highly structured,
         * use Quicksort instead of merge sort.
         */
        if (++count == MAX_RUN_COUNT) {
            sort(a, left, right, true);
            return;
        }
    }

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

    if (count == 0) {
        // A single equal run
        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.
        return;
    }
    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
    byte odd = 0;
    for (int n = 1; (n <<= 1) < count; odd ^= 1);

    // 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;
    }

    // Merging
    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;
    }
}

快速排序

private static void sort(int[] a, int left, int right, boolean leftmost) {
    int length = right - left + 1;

    // Use insertion sort on tiny arrays
    if (length < INSERTION_SORT_THRESHOLD) {
        if (leftmost) {
            /*
             * 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 {
            /*
             * 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.
             */
            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 / 7
    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.
     */
    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
    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]) {
        /*
         * 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);

        /*
         * 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;
                }
            }
        }

        // 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.
         */
        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;
            }
        }

        /*
         * 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);
    }
}