SourceAnalysis-DualPivotQuickSort

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1 前言

Java中Arrays.sort排序方法对于基本类型的排序采用的是双轴快速排序。本篇博客主要分析JDK源码

2 常量

/**
 * 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;

• MAX_RUN_COUNT：run数量的最大值，如果超过这个值，那么便认为数组不是特别的有序
• MAX_RUN_LENGTH：run长度的最大值
• QUICKSORT_THRESHOLD：如果数组长度小于这个值，那么快排的效率会比归并排序的效率要高
• INSERTION_SORT_THRESHOLD：如果数组长度小于这个值，那么插入排序的效率会比快排的效率要高
• COUNTING_SORT_THRESHOLD_FOR_BYTE：对于字节数组而言，如果长度大于这个数值，那么用计数排序效率比较高，因为额外的内存空间是确定的，256
• COUNTING_SORT_THRESHOLD_FOR_SHORT_OR_CHAR：对于char或者short而言，如果数组长度大于这个数值，那么计数排序效率会比较高

从这几个常量就能清晰地看出，DualPivotQuickSort对于不同的基本类型的排序方式是不同的，对于short、char以及byte等16bit或者8bit的基本类型而言，计数排序不会造成特别大的空间开销。而对于其他基本类型，比如int，long，float，double等。则会根据数组的长度来选择相应的排序算法，包括插入排序、快速排序和归并排序

3 方法

3.1 sort

参数说明

• a：待排序的数组
• left：左边界，闭区间
• right：右边界，闭区间
• work：
• workBase：
• workLen

/**
 * 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
    //待排序序列长度小于QUICKSORT_THRESHOLD，则采用快速排序
    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用于保存所有run的起始下标
    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) {
        if (a[k] < a[k + 1]) { //ascending
            //找到整个递增序列
            while (++k <= right && a[k - 1] <= a[k]);
        } else if (a[k] > a[k + 1]) { //descending
            //找到整个递减序列
            while (++k <= right && a[k - 1] >= a[k]);
            //反转这个递减序列，使其变为递增
            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]; ) {
                //当发现相等的元素超过MAX_RUN_LENGTH时，采用快速排序？可是为什么要这样做呢？
                if (--m == 0) {
                    sort(a, left, right, true);
                    return;
                }
            }
        }

        /*
         * The array is not highly structured,
         * use Quicksort instead of merge sort.
         */
        //当发现run非常多，意味着这些run都是一些很短的片段，大致上可以判定为无规律，即无序状态。这个时候采用快速排序
        if (++count == MAX_RUN_COUNT) {
            sort(a, left, right, true);
            return;
        }
    }

    //Check special cases
    //Implementation note: variable "right" is increased by 1.
    //注意，这里会使得right增加1
    if (run[count] == right++) { //The last run contains one element
        run[++count] = right;
    } else if (count == 1) { //The array is already sorted
        //说明整个数组就是有序的，直接返回就行了
        return;
    }

    //下面开始归并排序

    //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
    //进行数轮合并操作，直至run的数量为1
    for (int last; count > 1; count = last) {
        //每一轮两两进行合并，每一轮合并后的run数量就是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];
                }
            }
            //将合并后的序列挪到前面的位置，因此last在进行完一轮合并后指向的就是最后一个run
            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;
    }
}

下面给一个流程图

flowchart TD
    st(["开始"])
    en1(["结束"])
    en2(["结束"])
    en3(["结束"])
    cond1{"QUICKSORT_THRESHOLD ?"}
    op1["利用sort方法进行排序"]
    cond2{"重复元素较少 ?"}
    cond3{"存在有序片段 ?"} 
    op2["利用sort方法进行排序"]
    op3["类似于Timsort的归并排序"]

    st --> cond1
    cond1 --> |no| op1
    op1 --> en1
    cond1 --> |yes| cond2
    cond2 --> |yes| cond3
    cond2 --> |no| op2
    cond3 --> |yes| op3
    cond3 --> |no| op2
    op2 --> en2
    op3 --> en3

3.2 sort

双轴快速排序

参数说明

• a：待排序数组
• left：待排序序列范围的左边界
• right：待排序序列范围的右边界
• leftmost：指定范围是否在数组最左边

/**
 * 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
    //此时需要采取插入排序算法
    if (length < INSERTION_SORT_THRESHOLD) {
        if (leftmost) {
            /*
             * 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
            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.
             */
            //首先跳过前面的升序部分
            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.
             */

             /*
             * 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
             */

            //left=k+1，且left每次迭代会递增2，一次在自增部分递增，另一次在条件部分递增
            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
                //pair-插入排序的优化点之一：比a1大的元素只需要挪动一次即可
                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;
            }

            //由于上面循环条件的缘故，最后一个元素是没有被插入的，因为left每次递增2
            int last = a[right];

            //对最后一个元素进行一次简单的插入排序
            while (last < a[--right]) {
                a[right + 1] = a[right];
            }
            a[right + 1] = last;
        }
        return;
    }

    //Inexpensive approximation of length / 7
    //length / 7 的一种低复杂度的实现, 近似值(length * 9 / 64 + 1)
    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.
     */
    //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
    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
    //以下就是将e1~e5这5个元素进行排序，利用的是冒泡排序
    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
    //中间区间的首元素下标
    int less  = left;  //The index of the first element of center part
    //右边区间的首元素下标
    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.
         */
        //利用第2与第4个元素作为双轴，注意到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.
         */
        //下面这两个循环会直接跳过left与right这两个位置的元素，因此需要将left于right放到中间的某两个位置上，而e2和e4位置上的元素已经被保存为pivot1和pivot2了，因此可以将left和right的元素放在这两个位置
        //因此排序的部分就是a[left...right]中除了pivot1和pivot2的所有元素
        //最终pivot1和pivot2将会被置于一个正确的位置
        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，即找到center part部分
        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
                //ak比pivot1小，因此放入left part中
                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
                //ak比pivot2大，因此放入right part中
                //按理来说，ad应该放入greate-1这个位置，即位置k与位置greate-1交换，但是greate-1的元素处于"?-part"，因此不妨在这个时候从右往左找到第一个<=pivot2的元素
                while (a[great] > pivot2) {
                    if (great-- == k) {
                        break outer;
                    }
                }
                //此时a[great] <= pivot2

                //如果这个时候a[great] < pivot1，那么应该放入left part中去
                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[i] = b; i--;"的效率比"a[i--] = b;"的效率高
                a[great] = ak;
                --great;
            }
        }

        //Swap pivots into their final positions
        //由于排序的部分是a[left+1...right-1]，而原范围确是a[left...right]，而那两个被扣掉的元素就是pivot1和pivot2，因此，经过一些交换，将pivot1和pivot2插入到a[left...right]范围中去，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
        //递归调用sort方法排序left-part和right-part
        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center part包含的元素过多，超过数组长度的 4/7，那么需要进行一些优化处理
         * 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.
             */
            //跳过那些与pivot1相同的元素
            while (a[less] == pivot1) {
                ++less;
            }

            //跳过那些与pivot2相同的元素
            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right part中去

                    //由于great位置上的元素处于?-part，因此不妨在这里就进行处理，即决定它是否位于right-part中。下面的循环向左找到第一个不等于pivot2的元素
                    while (a[great] == pivot2) {
                        if (great-- == k) {
                            break outer;
                        }
                    }
                    if (a[great] == pivot1) { //a[great] < pivot2
                        //首先将center-part的第一个元素挪到k位置，for循环递增后，k也属于center-part
                        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].
                         */
                        //然后center-part的第一个元素放置pivot1，在递增less，那么这个位置就属于left-part了
                        a[less] = pivot1;
                        ++less;
                    } else { //pivot1 < a[great] < pivot2
                        //将a[great]放置到k位置上，在k递增后，这个位置就变为center-part的最后一个元素
                        a[k] = a[great];
                    }
                    a[great] = ak;
                    --great;
                }
            }
        }

        //Sort center part recursively
        //递归调用sort排序中间部分即可
        sort(a, less, great, false);

    } else { //Partitioning with one pivot
        //用单轴3-way进行partition，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.
         */
        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
                //放到left-part中，并修改边界
                a[k] = a[less];
                a[less] = ak;
                ++less;
            } else { //a[k] > pivot - Move a[k] to right part
                //放到right-part中

                //由于great位置上的元素位于?-part，因此不妨在这里进行处理，即决定它位于right-part、left-part还是center-part
                while (a[great] > pivot) {
                    --great;
                }

                //如果位于left-part
                if (a[great] < pivot) { //a[great] <= pivot
                    a[k] = a[less];
                    a[less] = a[great];
                    ++less;
                } else { //a[great] == pivot
                    //如果位于center-part
                    /*
                     * 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.
         */
        //递归调用sort进行排序即可
        sort(a, left, less - 1, leftmost);
        sort(a, great + 1, right, false);
    }
}

下面给一个流程图

flowchart TD
    st(["开始"])
    en1(["结束"])
    en2(["结束"])
    en3(["结束"])

    cond1{"INSERTION_SORT_THRESHOLD ?"}
    cond2{"leftmost ?"}

    op1["经典插入排序"]
    op2["pair-插入排序"]

    op3["找出均匀分布的5个位置e1-e5"]
    op4["冒泡排序e1-e5"]

    cond3{"e1-e5严格单调 ?"}
    op5["以e2和e4为双轴进行第一次双轴partition"]
    op6["left-part(小于pivot1的部分)递归sort"]
    op7["right-part(大于pivot2的部分)递归sort"]

    cond4{"center-part过大 ?"}

    op8["进行第二次双轴partition"]
    op9["center-part递归sort"]

    op10["以e3位单轴进行3-way-partition"]
    op11["left-part(小于pivot的部分)递归sort"]
    op12["right-part(大于pivot的部分)递归sort"]

    st --> cond1
    cond1 --> |no| cond2
    cond2 --> |yes| op1
    cond2 --> |no| op2
    op1 --> en1
    op2 --> en1
    cond1 --> |yes| op3
    op3 --> op4
    op4 --> cond3
    cond3 --> |yes| op5
    op5 --> op6
    op6 --> op7
    op7 --> cond4
    cond4 --> |yes| op8
    op8 --> op9
    cond4 --> |no| op9
    op9 --> en2
    cond3 --> |no| op10
    op10 --> op11
    op11 --> op12
    op12 --> en3

4 参考

• DualPivotQuickSort 双轴快速排序 源码 笔记
• QUICKSORTING - 3-WAY AND DUAL PIVOT