TimSort 在 JDK10 中的实现

• Timsort 是一种稳定高效的排序算法，混合了合并排序和二分插入排序，旨在很好地处理多种真实数据。它由Tim Peters于2002年实施使用在Python编程语言中。该算法查找已经排序的数据的子序列，并使用该知识更有效地对其余部分进行排序。从版本2.3开始，Timsort一直是Python的标准排序算法。如今，Timsort 已是是 Python、 Java、 Android平台 和 GNU Octave 的默认排序算法。
• 大致思路是：因为二分插入排序处理短序列的效率比较高，所以就把待排序的长序列切分成多个短序列，每个短序列独立执行二分插入排序，最后再对排好序的短序列做归并排序。在归并排序时，通过检测序列中已经排好序的片段，巧妙地把这些片段直接作为排序结果，跳过归并操作，从而减少了归并排序时的迭代次数。
• 复杂度比较(借用网上的图，原图地址)：

• jdk/lib/src.zip/java.base/java/util/TimSort.java

package java.util;

/**
 * A stable, adaptive, iterative mergesort that requires far fewer than
 * n lg(n) comparisons when running on partially sorted arrays, while
 * offering performance comparable to a traditional mergesort when run
 * on random arrays.  Like all proper mergesorts, this sort is stable and
 * runs O(n log n) time (worst case).  In the worst case, this sort requires
 * temporary storage space for n/2 object references; in the best case,
 * it requires only a small constant amount of space.
 *
 * This implementation was adapted from Tim Peters's list sort for
 * Python, which is described in detail here:
 *
 *   http://svn.python.org/projects/python/trunk/Objects/listsort.txt
 *
 * Tim's C code may be found here:
 *
 *   http://svn.python.org/projects/python/trunk/Objects/listobject.c
 *
 * The underlying techniques are described in this paper (and may have
 * even earlier origins):
 *
 *  "Optimistic Sorting and Information Theoretic Complexity"
 *  Peter McIlroy
 *  SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),
 *  pp 467-474, Austin, Texas, 25-27 January 1993.
 *
 * While the API to this class consists solely of static methods, it is
 * (privately) instantiable; a TimSort instance holds the state of an ongoing
 * sort, assuming the input array is large enough to warrant the full-blown
 * TimSort. Small arrays are sorted in place, using a binary insertion sort.
 *
 * @author Josh Bloch
 */
class TimSort<T> {
    /**
     * This is the minimum sized sequence that will be merged.  Shorter
     * sequences will be lengthened by calling binarySort.  If the entire
     * array is less than this length, no merges will be performed.
     *
     * This constant should be a power of two.  It was 64 in Tim Peter's C
     * implementation, but 32 was empirically determined to work better in
     * this implementation.  In the unlikely event that you set this constant
     * to be a number that's not a power of two, you'll need to change the
     * {@link #minRunLength} computation.
     *
     * If you decrease this constant, you must change the stackLen
     * computation in the TimSort constructor, or you risk an
     * ArrayOutOfBounds exception.  See listsort.txt for a discussion
     * of the minimum stack length required as a function of the length
     * of the array being sorted and the minimum merge sequence length.
     */
    private static final int MIN_MERGE = 32;

    /**
     * The array being sorted.
     */
    private final T[] a;

    /**
     * The comparator for this sort.
     * 比较器
     */
    private final Comparator<? super T> c;

    /**
     * When we get into galloping mode, we stay there until both runs win less
     * often than MIN_GALLOP consecutive times.
     */
    private static final int  MIN_GALLOP = 7;

    /**
     * This controls when we get *into* galloping mode.  It is initialized
     * to MIN_GALLOP.  The mergeLo and mergeHi methods nudge it higher for
     * random data, and lower for highly structured data.
     */
    private int minGallop = MIN_GALLOP;

    /**
     * Maximum initial size of tmp array, which is used for merging.  The array
     * can grow to accommodate demand.
     * 归并排序中临时数组的初始长度
     *
     * Unlike Tim's original C version, we do not allocate this much storage
     * when sorting smaller arrays.  This change was required for performance.
     */
    private static final int INITIAL_TMP_STORAGE_LENGTH = 256;

    /**
     * Temp storage for merges. A workspace array may optionally be
     * provided in constructor, and if so will be used as long as it
     * is big enough.
     * 归并排序的临时数组，tmpBase是数组中有效数据的起始下标，tmpLen是有效数据的长度
     */
    private T[] tmp;
    private int tmpBase; // base of tmp array slice 
    private int tmpLen;  // length of tmp array slice

    /**
     * A stack of pending runs yet to be merged.  Run i starts at
     * address base[i] and extends for len[i] elements.  It's always
     * true (so long as the indices are in bounds) that:
     *
     *     runBase[i] + runLen[i] == runBase[i + 1]
     *
     * so we could cut the storage for this, but it's a minor amount,
     * and keeping all the info explicit simplifies the code.
     */
    private int stackSize = 0;  // Number of pending runs on stack
    private final int[] runBase;
    private final int[] runLen;

    /**
     * Creates a TimSort instance to maintain the state of an ongoing sort.
     * 构造函数是私有的，单例模式
     *
     * @param a the array to be sorted
     * @param c the comparator to determine the order of the sort
     * @param work a workspace array (slice)
     * @param workBase origin of usable space in work array
     * @param workLen usable size of work array
     */
    private TimSort(T[] a, Comparator<? super T> c, T[] work, int workBase, int workLen) {
        this.a = a;
        this.c = c;

        // Allocate temp storage (which may be increased later if necessary)
        // 确定临时数组的初始长度，如果低于默认值256的2倍，则空间大小为原始数组a的长度的一半，否则为默认长度256
        int len = a.length;
        int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ?
            len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
        // 排序需要的临时数组长度是tlen，如果给定的work数组无法容纳tlen长度的数据，就创建新的临时数组，否则就直接把work数组当临时数组用
        // work数组貌似没什么用，只有在ArraysParallelSortHelpers.java中会设置work，绝大多数的类调用TimSort时work参数都是null，所以没必要看了
        if (work == null || workLen < tlen || workBase + tlen > work.length) {
            // 屏蔽泛型数组转型产生的警告
            @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
            T[] newArray = (T[])java.lang.reflect.Array.newInstance
                (a.getClass().getComponentType(), tlen);
            tmp = newArray;
            tmpBase = 0;
            tmpLen = tlen;
        }
        else {
            tmp = work;
            tmpBase = workBase;
            tmpLen = workLen;
        }

        /*
         * Allocate runs-to-be-merged stack (which cannot be expanded).  The
         * stack length requirements are described in listsort.txt.  The C
         * version always uses the same stack length (85), but this was
         * measured to be too expensive when sorting "mid-sized" arrays (e.g.,
         * 100 elements) in Java.  Therefore, we use smaller (but sufficiently
         * large) stack lengths for smaller arrays.  The "magic numbers" in the
         * computation below must be changed if MIN_MERGE is decreased.  See
         * the MIN_MERGE declaration above for more information.
         * The maximum value of 49 allows for an array up to length
         * Integer.MAX_VALUE-4, if array is filled by the worst case stack size
         * increasing scenario. More explanations are given in section 4 of:
         * http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf

         * 存储run的栈，不能在运行时动态扩展。一个长数组会被切分成多个短数组分别排序，run存储的就是每个子数组的长度和在原数组中的起始下标，在最后归并的时候用来定位子数组
         */
        int stackLen = (len <    120  ?  5 :
                        len <   1542  ? 10 :
                        len < 119151  ? 24 : 49);
        runBase = new int[stackLen];
        runLen = new int[stackLen];
    }

    /*
     * The next method (package private and static) constitutes the
     * entire API of this class.
     * sort方法是TimSort类对外暴露的唯一方法，其余都是私有方法
     */

    /**
     * Sorts the given range, using the given workspace array slice
     * for temp storage when possible. This method is designed to be
     * invoked from public methods (in class Arrays) after performing
     * any necessary array bounds checks and expanding parameters into
     * the required forms.
     *
     * 对数组a中的[lo, hi)片段做排序
     * @param a the array to be sorted
     * @param lo the index of the first element, inclusive, to be sorted
     * @param hi the index of the last element, exclusive, to be sorted
     * @param c the comparator to use
     * @param work a workspace array (slice)
     * @param workBase origin of usable space in work array
     * @param workLen usable size of work array
     * @since 1.8
     */
    static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c,
                         T[] work, int workBase, int workLen) {
        assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length;
        // 待排序的元素个数
        int nRemaining  = hi - lo;
        // 待排序元素只有0个或1个，不需要排序
        if (nRemaining < 2)
            return;  // Arrays of size 0 and 1 are always sorted

        // If array is small, do a "mini-TimSort" with no merges
        // 如果待排序元素少于归并排序的最小元素数量，就执行不包含归并排序的版本
        if (nRemaining < MIN_MERGE) {
            // 获取从a[lo]开始的最长有序连续片段的长度，如果片段是递减的会被反转成递增，得到的initRunLen就是从a[lo]开始已经排好序的长度
            int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
            // 调用二分插入排序，对a[lo + initRunLen]到a[hi]的所有元素执行插入
            binarySort(a, lo, hi, lo + initRunLen, c);
            return;
        }

        /**
         * March over the array once, left to right, finding natural runs,
         * extending short natural runs to minRun elements, and merging runs
         * to maintain stack invariant.
         */
        // 创建TimSort实例，保存中间结果
        TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen);
        // 切分子数组的最小长度
        int minRun = minRunLength(nRemaining);
        do {
            // Identify next run
            // a[lo]开始已经排好序的长度
            int runLen = countRunAndMakeAscending(a, lo, hi, c);

            // If run is short, extend to min(minRun, nRemaining)
            // 切分子数组，如果剩余长度不足最小长度，就只排序剩余长度的子数组。
            // 为什么minRun会被解释成minimum length呢？当nRemaining <= minRun时，子数组长度就是nRemaining，说明minRun不是长度的下限，而是上限。
            if (runLen < minRun) {
                int force = nRemaining <= minRun ? nRemaining : minRun;
                // 对切分出的a[lo, lo+force]片段执行二分插入排序，从a[lo+runLen]元素开始执行插入
                binarySort(a, lo, lo + force, lo + runLen, c);
                runLen = force;
            }

            // Push run onto pending-run stack, and maybe merge
            // 记录当前子数组的run信息
            ts.pushRun(lo, runLen);
            // 每生成一个run后，尝试做归并
            ts.mergeCollapse();

            // Advance to find next run
            // 把lo后移到下个待切分的子数组头部
            lo += runLen;
            nRemaining -= runLen;
        } while (nRemaining != 0);

        // Merge all remaining runs to complete sort
        // 所有切分都完成后，lo一定等于hi
        assert lo == hi;
        // 做最终的归并
        ts.mergeForceCollapse();
        // 归并完成后只有一个run
        assert ts.stackSize == 1;
    }

    /**
     * Sorts the specified portion of the specified array using a binary
     * insertion sort.  This is the best method for sorting small numbers
     * of elements.  It requires O(n log n) compares, but O(n^2) data
     * movement (worst case).
     * 二分插入排序，对于小片段排序是最优算法
     * 在调用binarySort前已经调用过countRunAndMakeAscending，确保[lo, start)的片段已经排好序了
     *
     * If the initial part of the specified range is already sorted,
     * this method can take advantage of it: the method assumes that the
     * elements from index {@code lo}, inclusive, to {@code start},
     * exclusive are already sorted.
     *
     * @param a the array in which a range is to be sorted
     * @param lo the index of the first element in the range to be sorted
     * @param hi the index after the last element in the range to be sorted
     * @param start the index of the first element in the range that is
     *        not already known to be sorted ({@code lo <= start <= hi})
     * @param c comparator to used for the sort
     */
    // 屏蔽与switch陈述式中遗漏break相关的警告
    @SuppressWarnings("fallthrough")
    private static <T> void binarySort(T[] a, int lo, int hi, int start,
                                       Comparator<? super T> c) {
        // 感觉这种assert毫无意义
        assert lo <= start && start <= hi;
        if (start == lo)
            start++;
        //从start开始，逐个进行插入，start是递增的
        for ( ; start < hi; start++) {
            // pivot是当前待插入的元素
            T pivot = a[start];

            // Set left (and right) to the index where a[start] (pivot) belongs
            // a[lo, start)片段是排好序的，a[start]是待排序的，所以每一轮的插入实际就是把a[start]插入到a[lo, start)中，插入的结果就是a[lo, start]是排好序的，因为start是循环变量，所以最后一轮插入完成后，排好序的a[lo, start]就是a[lo, hi]
            // 在调用binarySort前确定a[lo, start)排好序，就省掉了a[lo+1]到a[start-1]这些元素的插入操作，对于有很多有序片段的数组排序效率很高
            int left = lo;
            int right = start;
            assert left <= right;
            /*
             * Invariants:
             *   pivot >= all in [lo, left).
             *   pivot <  all in [right, start).
             */
            // 二分查找确定插入位置
            while (left < right) {
                int mid = (left + right) >>> 1;
                if (c.compare(pivot, a[mid]) < 0)
                    right = mid;
                else
                    left = mid + 1;
            }
            assert left == right;

            /*
             * The invariants still hold: pivot >= all in [lo, left) and
             * pivot < all in [left, start), so pivot belongs at left.  Note
             * that if there are elements equal to pivot, left points to the
             * first slot after them -- that's why this sort is stable.
             * Slide elements over to make room for pivot.
             */
            // 插入位置是left，所以包含a[left]在内的后面所有元素都要后移一位
            int n = start - left;  // The number of elements to move
            // Switch is just an optimization for arraycopy in default case
            // 一旦case匹配，会无条件执行后面所有case直到遇到break，所以case2和case1要么都执行要么都不执行
            switch (n) {
                // 要移动的元素小于等于2个，直接用赋值方式移动
                case 2:  a[left + 2] = a[left + 1];
                case 1:  a[left + 1] = a[left];
                         break;
                // 要移动的元素大于2个，用arraycopy把a[left, start-1]写到a[left+1, start]，直接复制内存
                default: System.arraycopy(a, left, a, left + 1, n);
            }
            // 插入pivot
            a[left] = pivot;
        }
    }

    /**
     * Returns the length of the run beginning at the specified position in
     * the specified array and reverses the run if it is descending (ensuring
     * that the run will always be ascending when the method returns).
     *
     * A run is the longest ascending sequence with:
     *
     *    a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
     *
     * or the longest descending sequence with:
     *
     *    a[lo] >  a[lo + 1] >  a[lo + 2] >  ...
     *
     * For its intended use in a stable mergesort, the strictness of the
     * definition of "descending" is needed so that the call can safely
     * reverse a descending sequence without violating stability.
     *
     * @param a the array in which a run is to be counted and possibly reversed
     * @param lo index of the first element in the run
     * @param hi index after the last element that may be contained in the run.
              It is required that {@code lo < hi}.
     * @param c the comparator to used for the sort
     * @return  the length of the run beginning at the specified position in
     *          the specified array

     * 返回从a[lo]开始的最长有序连续片段的长度
     */
    private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,
                                                    Comparator<? super T> c) {
        assert lo < hi;
        // hi-lo<2 的情况在调用countRunAndMakeAscending前已经判断过了，这里没必要
        int runHi = lo + 1;
        if (runHi == hi)
            return 1;

        // Find end of run, and reverse range if descending
        if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
            while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
                runHi++;
            // 默认的排序结果是递增，所以如果从a[lo]开始是递减的，要把最长递减片段反转
            // 因为runHi++了，所以反转的范围是[lo, runHi)
            reverseRange(a, lo, runHi);
        } else {                              // Ascending
            while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
                runHi++;
        }

        return runHi - lo;
    }

    /**
     * Reverse the specified range of the specified array.
     * 反转数组a中[lo, hi)的片段，因为调用reverseRange前hi多++了一次，所以反转不包含a[hi]
     * @param a the array in which a range is to be reversed
     * @param lo the index of the first element in the range to be reversed
     * @param hi the index after the last element in the range to be reversed
     */
    private static void reverseRange(Object[] a, int lo, int hi) {
        hi--;
        // 反转过程就是逐层交换两端的值
        while (lo < hi) {
            Object t = a[lo];
            a[lo++] = a[hi];
            a[hi--] = t;
        }
    }

    /**
     * Returns the minimum acceptable run length for an array of the specified
     * length. Natural runs shorter than this will be extended with
     * {@link #binarySort}.

     * 根据待排序的元素数量，返回参与归并的子数组的最小长度，也就是执行二分插入排序的子数组的最小长度。因为二分插入排序适合小片段排序，所以对于长的数组，要拆分成多个子数组，分别执行二分插入排序，最后再归并
     * 调用minRunLength前已经判断过nRemaining < MIN_MERGE的情况，所以这里的n一定是大于等于MIN_MERGE的。如果n是2的幂次，返回的就是MIN_MERGE/2，否则返回的是[MIN_MERGE/2, MIN_MERGE]内的一个值
     * Roughly speaking, the computation is:
     *
     *  If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
     *  Else if n is an exact power of 2, return MIN_MERGE/2.
     *  Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
     *   is close to, but strictly less than, an exact power of 2.
     *
     * For the rationale, see listsort.txt.
     *
     * @param n the length of the array to be sorted
     * @return the length of the minimum run to be merged
     */
    private static int minRunLength(int n) {
        assert n >= 0;
        int r = 0;      // Becomes 1 if any 1 bits are shifted off
        while (n >= MIN_MERGE) {
            r |= (n & 1);
            n >>= 1;
        }
        return n + r;
    }

    /**
     * Pushes the specified run onto the pending-run stack.
     *
     * @param runBase index of the first element in the run
     * @param runLen  the number of elements in the run
     * stack不能扩容，所以这里会有越界的风险
     */
    private void pushRun(int runBase, int runLen) {
        this.runBase[stackSize] = runBase;
        this.runLen[stackSize] = runLen;
        stackSize++;
    }

    /**
     * Examines the stack of runs waiting to be merged and merges adjacent runs
     * until the stack invariants are reestablished:
     *
     *     1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
     *     2. runLen[i - 2] > runLen[i - 1]
     *
     * This method is called each time a new run is pushed onto the stack,
     * so the invariants are guaranteed to hold for i < stackSize upon
     * entry to the method.

     * 每次生成一个新的run后，都要尝试做一次归并，让stack中的run序列满足上面两个式子，大概意思就是保证越靠前的run长度越长。因为归并排序的最优情况就是先归并短序列，再归并长序列，对于这里的stack，最终的归并是从后往前按顺序归并的，如果run的分布长短交错，就会经常遇到一个长序列和一个短序列做归并，效率非常低，所以每生成一个run后都要尝试做归并，尽量把相邻的短序列提早归并掉，来保证最终做归并时待归并的序列越来越长。
     * 因为每次插入新的run后，stackSize最后都加一，所以最后一个run的下标是stackSize-1，从后往前归并时先归并下标是stackSize-2和stackSize-1的两个run，也就是调用mergeAt(stackSize-2)
     */
    private void mergeCollapse() {
        while (stackSize > 1) {
            int n = stackSize - 2;
            if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
                if (runLen[n - 1] < runLen[n + 1])
                    n--;
                mergeAt(n);
            } else if (runLen[n] <= runLen[n + 1]) {
                mergeAt(n);
            } else {
                break; // Invariant is established
            }
        }
    }

    /**
     * Merges all runs on the stack until only one remains.  This method is
     * called once, to complete the sort.
     * 最终的归并，在stack中从后往前按顺序归并所有的run
     */
    private void mergeForceCollapse() {
        while (stackSize > 1) {
            int n = stackSize - 2;
            if (n > 0 && runLen[n - 1] < runLen[n + 1])
                n--;
            mergeAt(n);
        }
    }

    /**
     * Merges the two runs at stack indices i and i+1.  Run i must be
     * the penultimate or antepenultimate run on the stack.  In other words,
     * i must be equal to stackSize-2 or stackSize-3.

     * 归并下标是i和i+1的两个run，要保证i是stackSize-2或stackSize-3，也就是保证归并的两个run在stack的尾部，在调用mergeCollapse时可能会跳过最后一个run，归并倒数第二个和第三个run，所以i也可以是stackSize-3。因为每次归并完stackSize都会减小，所以i和stackSize的关系会始终如此。
     * 归并的结果就是两个子数组原地排好序，同时在stack中两个run合并成一个
     * @param i stack index of the first of the two runs to merge
     */
    private void mergeAt(int i) {
        assert stackSize >= 2;
        assert i >= 0;
        assert i == stackSize - 2 || i == stackSize - 3;

        int base1 = runBase[i];
        int len1 = runLen[i];
        int base2 = runBase[i + 1];
        int len2 = runLen[i + 1];
        // 为什么要加一堆assert，debug忘删了？
        assert len1 > 0 && len2 > 0;
        assert base1 + len1 == base2;

        /*
         * Record the length of the combined runs; if i is the 3rd-last
         * run now, also slide over the last run (which isn't involved
         * in this merge).  The current run (i+1) goes away in any case.
         *
         * 归并之前，先把runBase、runLen、stackSize更新到归并后的结果，反正当前的值已经保存在临时变量中了
         */
        runLen[i] = len1 + len2;
        if (i == stackSize - 3) {
            runBase[i + 1] = runBase[i + 2];
            runLen[i + 1] = runLen[i + 2];
        }
        stackSize--;

        /*
         * Find where the first element of run2 goes in run1. Prior elements
         * in run1 can be ignored (because they're already in place).
         */
        // 查找run2的第一个元素a[base2]在run1中的插入位置，有相同元素时插入位置靠后
        int k = gallopRight(a[base2], a, base1, len1, 0, c);
        assert k >= 0;
        // run2的第一个元素插入到run1的第k个位置，所以归并后a[base1, base1+k)一定在开头，不需要参与归并的过程，直接略过
        // 之所以有相同元素时插入位置靠后，就是保证略过的元素尽量多，提高效率
        base1 += k;
        len1 -= k;
        // 如果略过了整个run1，说明排序的结果就是整个run2都在run1的后面，已经排好了，不需要再执行归并
        if (len1 == 0)
            return;

        /*
         * Find where the last element of run1 goes in run2. Subsequent elements
         * in run2 can be ignored (because they're already in place).
         */
        // 查找run1的最后一个元素a[base1 + len1 - 1]在run2中的插入位置，有相同元素时插入位置靠前
        len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
        // run1的最后一个元素插入到run2的第len2个位置，所以归并后a(base2+len2, base2+oldlen2]一定在末尾，不需要参与归并的过程，直接略过
        // 之所以有相同元素时插入位置靠前，也是保证略过的元素尽量多
        assert len2 >= 0;
        if (len2 == 0)
            return;

        // Merge remaining runs, using tmp array with min(len1, len2) elements
        // run1和run2挨着，要想原地排序就一定要把其中一个run放入临时数组，空出一些位置来存放排好序的元素，否则就需要各种复杂的原地交换操作。
        // 为了节省空间，选择把run1和run2中较短的一个放入临时数组。如果run1放入临时数组，空出的位置就在前面，所以mergeLo的归并顺序就是从base1往后按从小到大的顺序存放归并的元素。如果run2放入临时数组，空出的位置就在后面，所以mergeHi的归并顺序就是从base2+len2往前按从大到小的顺序存放归并的元素
        if (len1 <= len2)
            mergeLo(base1, len1, base2, len2);
        else
            mergeHi(base1, len1, base2, len2);
    }

    /**
     * Locates the position at which to insert the specified key into the
     * specified sorted range; if the range contains an element equal to key,
     * returns the index of the leftmost equal element.
     * 把key插入到base和len对应的run中，返回插入位置。
     * 如果遇到与key相同的元素，就会插到相同元素的前面，因为判定条件中左侧是严格大于，右侧是小于等于
     * hint是开始查找的位置，是子数组从0开始的下标，0 <= hint < n，所以是从a[base + hint]开始比较
     * 返回值是子数组从0开始的下标，而不是原数组中的下标
     * 为什么不直接二分查找呢？效率会差很多？
     *
     * @param key the key whose insertion point to search for
     * @param a the array in which to search
     * @param base the index of the first element in the range
     * @param len the length of the range; must be > 0
     * @param hint the index at which to begin the search, 0 <= hint < n.
     *     The closer hint is to the result, the faster this method will run.
     * @param c the comparator used to order the range, and to search
     * @return the int k,  0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
     *    pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
     *    In other words, key belongs at index b + k; or in other words,
     *    the first k elements of a should precede key, and the last n - k
     *    should follow it.
     */
    private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint,
                                      Comparator<? super T> c) {
        assert len > 0 && hint >= 0 && hint < len;
        int lastOfs = 0;
        int ofs = 1;
        // key > a[base + hint]
        if (c.compare(key, a[base + hint]) > 0) {
            // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
            int maxOfs = len - hint;
            // 从a[base+hint+1]开始向右遍历，直到 key <= a[base+hint+ofs]，lastOfs是上一轮迭代的ofs，所以就有a[base+hint+lastOfs] < key <= a[base+hint+ofs]
            while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) {
                lastOfs = ofs;
                // 偏移ofs是按照1、3、7、15这样指数级增长的，因为a[base, base+len]已经是有序的了，所以不怕步子太大漏掉
                ofs = (ofs << 1) + 1;
                // 如果ofs整型溢出了，就直接设为maxOfs，把右边界拉到a[base+hint+len]
                if (ofs <= 0)   // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs)
                ofs = maxOfs;

            // Make offsets relative to base
            // ofs是从hint开始算的偏移量，这里加上hint以后，ofs就变成子数组的下标了，后面就不需要hint了
            lastOfs += hint;
            ofs += hint;
        } else { // key <= a[base + hint]
            // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
            final int maxOfs = hint + 1;
            // 从a[base+hint-1]开始向左遍历，直到 key > a[base+hint-ofs]，lastOfs是上一轮迭代的ofs，所以就有a[base+hint-ofs] < key <= a[base+hint-lastOfs]
            while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0)   // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs)
                ofs = maxOfs;

            // Make offsets relative to base
            // 因为是向左遍历，lastOfs应该在ofs的右边，这里为了和上面的结果保持一致，交换了lastOfs和ofs，让lastOfs在ofs的左边
            int tmp = lastOfs;
            lastOfs = hint - ofs;
            ofs = hint - tmp;
        }
        // lastOfs不可能是负数吧？
        assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

        /*
         * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
         * to the right of lastOfs but no farther right than ofs.  Do a binary
         * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
         *
         * 到此已经确定了a[base+lastOfs] < key <= a[base+ofs]，在这个范围里进行二分查找确定key的插入位置
         */
        lastOfs++;
        while (lastOfs < ofs) {
            int m = lastOfs + ((ofs - lastOfs) >>> 1);

            if (c.compare(key, a[base + m]) > 0)
                lastOfs = m + 1;  // a[base + m] < key
            else
                ofs = m;          // key <= a[base + m]
        }
        assert lastOfs == ofs;    // so a[base + ofs - 1] < key <= a[base + ofs]
        return ofs;
    }

    /**
     * Like gallopLeft, except that if the range contains an element equal to
     * key, gallopRight returns the index after the rightmost equal element.
     * 
     * 和gallopLeft差不多，区别是如果遇到与key相同的元素，就会插到相同元素的后面，因为判定条件中左侧是大于等于，右侧是严格小于
     * @param key the key whose insertion point to search for
     * @param a the array in which to search
     * @param base the index of the first element in the range
     * @param len the length of the range; must be > 0
     * @param hint the index at which to begin the search, 0 <= hint < n.
     *     The closer hint is to the result, the faster this method will run.
     * @param c the comparator used to order the range, and to search
     * @return the int k,  0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
     */
    private static <T> int gallopRight(T key, T[] a, int base, int len,
                                       int hint, Comparator<? super T> c) {
        assert len > 0 && hint >= 0 && hint < len;

        int ofs = 1;
        int lastOfs = 0;
        if (c.compare(key, a[base + hint]) < 0) {
            // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
            int maxOfs = hint + 1;
            while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0)   // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs)
                ofs = maxOfs;

            // Make offsets relative to b
            int tmp = lastOfs;
            lastOfs = hint - ofs;
            ofs = hint - tmp;
        } else { // a[b + hint] <= key
            // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
            int maxOfs = len - hint;
            while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0)   // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs)
                ofs = maxOfs;

            // Make offsets relative to b
            lastOfs += hint;
            ofs += hint;
        }
        assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

        /*
         * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
         * the right of lastOfs but no farther right than ofs.  Do a binary
         * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
         */
        lastOfs++;
        while (lastOfs < ofs) {
            int m = lastOfs + ((ofs - lastOfs) >>> 1);

            if (c.compare(key, a[base + m]) < 0)
                ofs = m;          // key < a[b + m]
            else
                lastOfs = m + 1;  // a[b + m] <= key
        }
        assert lastOfs == ofs;    // so a[b + ofs - 1] <= key < a[b + ofs]
        return ofs;
    }

    /**
     * Merges two adjacent runs in place, in a stable fashion.  The first
     * element of the first run must be greater than the first element of the
     * second run (a[base1] > a[base2]), and the last element of the first run
     * (a[base1 + len1-1]) must be greater than all elements of the second run.
     *
     * For performance, this method should be called only when len1 <= len2;
     * its twin, mergeHi should be called if len1 >= len2.  (Either method
     * may be called if len1 == len2.)
     *
     * @param base1 index of first element in first run to be merged
     * @param len1  length of first run to be merged (must be > 0)
     * @param base2 index of first element in second run to be merged
     *        (must be aBase + aLen)
     * @param len2  length of second run to be merged (must be > 0)
     */
    private void mergeLo(int base1, int len1, int base2, int len2) {
        assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

        // Copy first run into temp array
        T[] a = this.a; // For performance
        // 检查临时数组的长度，必要时扩容
        T[] tmp = ensureCapacity(len1);
        int cursor1 = tmpBase; // Indexes into tmp array
        int cursor2 = base2;   // Indexes int a
        int dest = base1;      // Indexes int a
        // 把run1写入临时数组
        System.arraycopy(a, base1, tmp, cursor1, len1);

        // Move first element of second run and deal with degenerate cases
        // 因为run1前面一部分被略过了，剩下的部分里一定是run2的第一个元素在开头
        a[dest++] = a[cursor2++];
        // 如果run2只有一个元素，插入到头部，剩下的run1直接后移一位，排序就完成了
        if (--len2 == 0) {
            System.arraycopy(tmp, cursor1, a, dest, len1);
            return;
        }
        // 因为run2后面一部分被略过了，剩下的部分里一定是run1的最后一个元素在末尾
        // 如果run1只有一个元素，插入到末尾，剩下的run2直接前移一位，排序就完成了
        // 直接return了，所以len1归不归零都无所谓
        if (len1 == 1) {
            System.arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
            return;
        }

        Comparator<? super T> c = this.c;  // Use local variable for performance
        // 用新的变量，本次对阈值minGallop的修改不会影响下次归并
        int minGallop = this.minGallop;    //  "    "       "     "      "
    outer:
        while (true) {
            int count1 = 0; // run1连续比run2小多少次
            int count2 = 0; // run2连续比run1小多少次

            /*
             * Do the straightforward thing until (if ever) one run starts
             * winning consistently.
             */
            // 正常的归并操作，但是如果其中一个run的头部连续minGallop次都比另一个run小，就先中止当前的归并，退出循环做一些优化
            do {
                assert len1 > 1 && len2 > 0;
                if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
                    a[dest++] = a[cursor2++];
                    count2++;
                    count1 = 0;
                    //如果其中一个run空了，表示归并完成，退出最外层循环
                    if (--len2 == 0)
                        break outer;
                } else {
                    a[dest++] = tmp[cursor1++];
                    count1++;
                    count2 = 0;
                    if (--len1 == 1)
                        break outer;
                }
                // count1和count2总有一个是0，或运算得到的就是不为0的那个
            } while ((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a
             * huge win. So try that, and continue galloping until (if ever)
             * neither run appears to be winning consistently anymore.
             */
            // 如果runA的头部连续minGallop次比runB小，那么很可能其后面很长一部分都比runB的头部小。这时逐个元素进行归并效率就比较低。所以调用gallop方法，找到runB的头部在runA中插入的位置，使用arraycopy直接把runA中该位置之前的片段一次性复制到a中排好序的位置上，比归并的效率高
            do {
                assert len1 > 1 && len2 > 0;
                count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
                if (count1 != 0) {
                    System.arraycopy(tmp, cursor1, a, dest, count1);
                    dest += count1;
                    cursor1 += count1;
                    len1 -= count1;
                    // run1最后一个元素一定是放在末尾，只剩一个元素就不需要归并了，如果run1空了，说明出了bug
                    if (len1 <= 1) // len1 == 1 || len1 == 0
                        break outer;
                }
                a[dest++] = a[cursor2++];
                if (--len2 == 0)
                    break outer;
                // 找完run1的头部序列，顺便找run2的头部序列，因为如果run1头部很多元素比run2小，在除去这些元素后，就很有可能run2的头部有很多元素比run1大。
                // 又不是找尾部序列，为什么要调用gallopLeft？
                count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
                if (count2 != 0) {
                    System.arraycopy(a, cursor2, a, dest, count2);
                    dest += count2;
                    cursor2 += count2;
                    len2 -= count2;
                    if (len2 == 0)
                        break outer;
                }
                a[dest++] = tmp[cursor1++];
                if (--len1 == 1)
                    break outer;
                // 动态调整阈值，可能是根据经验觉得会有效吧
                minGallop--;
            // 如果连续性还是很大，就继续调用gallop替代归并
            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if (minGallop < 0)
                minGallop = 0;
            // 跳出循环，说明连续性没那么高了，可以适当提高阈值
            minGallop += 2;  // Penalize for leaving gallop mode
        }  // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

        // 跳出循环只有两种情况，要么run1只剩一个元素，要么run2空了
        if (len1 == 1) {
            assert len2 > 0;
            System.arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; //  Last elt of run 1 to end of merge
        } else if (len1 == 0) {
            // run1最后一个元素一定是放在末尾，如果run1空了，说明出了bug
            throw new IllegalArgumentException(
                "Comparison method violates its general contract!");
        } else {
            assert len2 == 0;
            assert len1 > 1;
            System.arraycopy(tmp, cursor1, a, dest, len1);
        }
    }

    /**
     * Like mergeLo, except that this method should be called only if
     * len1 >= len2; mergeLo should be called if len1 <= len2.  (Either method
     * may be called if len1 == len2.)
     *
     * @param base1 index of first element in first run to be merged
     * @param len1  length of first run to be merged (must be > 0)
     * @param base2 index of first element in second run to be merged
     *        (must be aBase + aLen)
     * @param len2  length of second run to be merged (must be > 0)
     */
    private void mergeHi(int base1, int len1, int base2, int len2) {
        assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

        // Copy second run into temp array
        T[] a = this.a; // For performance
        T[] tmp = ensureCapacity(len2);
        int tmpBase = this.tmpBase;
        System.arraycopy(a, base2, tmp, tmpBase, len2);

        int cursor1 = base1 + len1 - 1;  // Indexes into a
        int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array
        int dest = base2 + len2 - 1;     // Indexes into a

        // Move last element of first run and deal with degenerate cases
        a[dest--] = a[cursor1--];
        if (--len1 == 0) {
            System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2);
            return;
        }
        if (len2 == 1) {
            dest -= len1;
            cursor1 -= len1;
            System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
            a[dest] = tmp[cursor2];
            return;
        }

        Comparator<? super T> c = this.c;  // Use local variable for performance
        int minGallop = this.minGallop;    //  "    "       "     "      "
    outer:
        while (true) {
            int count1 = 0; // Number of times in a row that first run won
            int count2 = 0; // Number of times in a row that second run won

            /*
             * Do the straightforward thing until (if ever) one run
             * appears to win consistently.
             */
            do {
                assert len1 > 0 && len2 > 1;
                if (c.compare(tmp[cursor2], a[cursor1]) < 0) {
                    a[dest--] = a[cursor1--];
                    count1++;
                    count2 = 0;
                    if (--len1 == 0)
                        break outer;
                } else {
                    a[dest--] = tmp[cursor2--];
                    count2++;
                    count1 = 0;
                    if (--len2 == 1)
                        break outer;
                }
            } while ((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a
             * huge win. So try that, and continue galloping until (if ever)
             * neither run appears to be winning consistently anymore.
             */
            do {
                assert len1 > 0 && len2 > 1;
                count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
                if (count1 != 0) {
                    dest -= count1;
                    cursor1 -= count1;
                    len1 -= count1;
                    System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
                    if (len1 == 0)
                        break outer;
                }
                a[dest--] = tmp[cursor2--];
                if (--len2 == 1)
                    break outer;

                count2 = len2 - gallopLeft(a[cursor1], tmp, tmpBase, len2, len2 - 1, c);
                if (count2 != 0) {
                    dest -= count2;
                    cursor2 -= count2;
                    len2 -= count2;
                    System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
                    if (len2 <= 1)  // len2 == 1 || len2 == 0
                        break outer;
                }
                a[dest--] = a[cursor1--];
                if (--len1 == 0)
                    break outer;
                minGallop--;
            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if (minGallop < 0)
                minGallop = 0;
            minGallop += 2;  // Penalize for leaving gallop mode
        }  // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

        if (len2 == 1) {
            assert len1 > 0;
            dest -= len1;
            cursor1 -= len1;
            System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
            a[dest] = tmp[cursor2];  // Move first elt of run2 to front of merge
        } else if (len2 == 0) {
            throw new IllegalArgumentException(
                "Comparison method violates its general contract!");
        } else {
            assert len1 == 0;
            assert len2 > 0;
            System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2);
        }
    }

    /**
     * Ensures that the external array tmp has at least the specified
     * number of elements, increasing its size if necessary.  The size
     * increases exponentially to ensure amortized linear time complexity.
     * 保证临时数组tmp能够容纳下minCapacity个元素，不足时扩容
     *
     * @param minCapacity the minimum required capacity of the tmp array
     * @return tmp, whether or not it grew
     */
    private T[] ensureCapacity(int minCapacity) {
        // 只有长度不够才需要扩容，足够时tmp不需要修改
        if (tmpLen < minCapacity) {
            // Compute smallest power of 2 > minCapacity
            // 计算最小的大于minCapacity的2的幂，花里胡哨的位运算，相当于把minCapacity最高位的1后面全置为1，最后newSize++，让最高位的1左移一位，后面全变成0，就得到了2的幂。因为java的int是32位，所以到>> 16就截止了。
            int newSize = minCapacity;
            newSize |= newSize >> 1;
            newSize |= newSize >> 2;
            newSize |= newSize >> 4;
            newSize |= newSize >> 8;
            newSize |= newSize >> 16;
            newSize++;
            // 小于0是溢出了？
            if (newSize < 0) // Not bloody likely!
                newSize = minCapacity;
            else
                // minCapacity是归并的两个子数组里较短的子数组的长度，所以不可能超过a.length/2
                newSize = Math.min(newSize, a.length >>> 1);
            
            // 根据扩容后的长度重新创建临时数组
            @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
            T[] newArray = (T[])java.lang.reflect.Array.newInstance
                (a.getClass().getComponentType(), newSize);
            tmp = newArray;
            tmpLen = newSize;
            tmpBase = 0;
        }
        // 为什么tmpLen和tmpBase可以直接修改，tmp要作为返回值返回，因为tmp是数组？？？
        return tmp;
    }
}