TimSort Java源码个人解读

/*JDK 1.8
 */

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.
     */
    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.
     *
     * 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.
     */
    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.
     *
     * @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)
        int len = a.length;
        // 确定临时数组的长度， 如果低于默认值256的2倍， 则空间大小为原始数组a的长度乘以2， 否则为默认长度
        int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ?
            len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
        if (work == null || workLen < tlen || workBase + tlen > work.length) {
            @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
            T[] newArray = (T[])java.lang.reflect.Array.newInstance
                (a.getClass().getComponentType(), tlen);
            tmp = newArray;
            tmpBase = 0;
            tmpLen = tlen;
        }
        else {
            // 当指定的work数组不为空， 且workLen大于计算出的tlen的长度， 并且work数组的有效长度大于tlen的长度时， 使用指定的临时数组
            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
         */
        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.
     */

    /**
     * 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.
     *
     * @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;

        int nRemaining  = hi - lo;        //总共的待排序的元素个数
        if (nRemaining < 2)
            return;  // Arrays of size 0 and 1 are always sorted

        // If array is small, do a "mini-TimSort" with no merges
        // 当元素个数小于7时， 使用折半插入排序， 因为插入排序对于元素个数少的数组更快。
        if (nRemaining < MIN_MERGE) {
            // initRunLen是初始的有序的元素的个数，而从索引位置lo + initRunLen开始， 后面为乱序的元素。 一种优化方法， lo + initRunLen之前的不用排序了
            int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
            // 折半插入排序
            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.
         */
         // 新建一个TimSort实例， 存储运行时的状态， 比如临时的run(一个run即一个有序的数组)
        TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen);
        // 查找一个minRun值， 小于minRun时使用折半插入排序，MIN_MERGE/2 <= minRun <= MIN_MERGE
        int minRun = minRunLength(nRemaining);
        do {
            // Identify next run
            // 找到下一个有序的数组的长度
            int runLen = countRunAndMakeAscending(a, lo, hi, c);

            // If run is short, extend to min(minRun, nRemaining)
            if (runLen < minRun) {
                // 使用折半插入排序扩展数组， 使之达到minRun， 因为元素个数小于minRun， 折半插入排序更快速
                int force = nRemaining <= minRun ? nRemaining : minRun;
                binarySort(a, lo, lo + force, lo + runLen, c);
                runLen = force;
            }

            // Push run onto pending-run stack, and maybe merge
            // 把这个有序数组压入栈中
            ts.pushRun(lo, runLen);
            /**
            * 判断当 
            *    runLen[i - 3] <= runLen[i - 2] + runLen[i - 1]
            *        且 runLen[i-3] < runLen[i-1]时
             *   或
            *    runLen[i - 2] <= runLen[i - 1]
            *    合并较小的两个有序数组， 以达到最大的平衡（即每个数组大小基本相同）
            */
            ts.mergeCollapse();

            // Advance to find next run
            lo += runLen;
            nRemaining -= runLen;
        } while (nRemaining != 0);

        // Merge all remaining runs to complete sort
        assert lo == hi;
        //合并剩余数组
        ts.mergeForceCollapse();
        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).
     *
     * 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
     */
    @SuppressWarnings("fallthrough")
    private static <T> void binarySort(T[] a, int lo, int hi, int start,
                                       Comparator<? super T> c) {
        assert lo <= start && start <= hi;
        // start之前的有序元素直接略过
        if (start == lo)
            start++;
        // 从start到hi， 使用折半插入排序进行数组排序
        for ( ; start < hi; start++) {
            //待插入的元素
            T pivot = a[start];

            // Set left (and right) to the index where a[start] (pivot) belongs
            // 从left到right， 找到插入位置
            int left = lo;
            int right = start;
            assert left <= right;
            /*
             * Invariants:
             *   pivot >= all in [lo, left).
             *   pivot <  all in [right, start).
             */
            while (left < right) {
                int mid = (left + right) >>> 1;
                if (c.compare(pivot, a[mid]) < 0)
                    right = mid;
                else
                    left = mid + 1;
            }
            // left即为最终的插入位置， 因为start>=lo && start <=hi， 所以最终一定会找到一个位置使得pivot>=a[mid]， 因此最终一定是pivot >= right， 因此最终为left的位置， 即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.
             */
            int n = start - left;  // The number of elements to move
            // Switch is just an optimization for arraycopy in default case
            switch (n) {
                case 2:  a[left + 2] = a[left + 1]; // 如果待移动元素个数小于等于2则直接移动
                case 1:  a[left + 1] = a[left];
                         break;
                default: System.arraycopy(a, left, a, left + 1, n);  // 从left开始往后移， 然后把start位置的元素插入到原来的left的位置
            }
            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
     */
    private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,
                                                    Comparator<? super T> c) {
        assert lo < hi;
        int runHi = lo + 1;
        if (runHi == hi)
            return 1;  // lo < hi， 且lo + 1 = hi， 因此是有序且升序的， 直接返回

        // Find end of run, and reverse range if descending
        if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
            // 如果是降序的， 找到最长的有序降序序列的长度， 并且把序列倒置， 使之升序
            while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
                runHi++;
            reverseRange(a, lo, runHi);
        } else {                              // Ascending
            // 如果是升序的， 同样找到最长的有序序列的长度
            while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
                runHi++;
        }

        // 返回有序序列的长度
        return runHi - lo;
    }

    /**
     * Reverse the specified range of the specified array.
     *
     * @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--;
        // 首尾倒置
        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}.
     *
     * 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) {
            // n&1是判断n是能否被2整除， 如果不能被2整除， 最后一个Bit位一定是1,则1&1为1， r = r | 1 为1
            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
     */
     // 把有序序列起始位置和长度放入栈中
    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.
     */
     // 判断合并栈顶的三个元素中较小的两个， 或如果第二个元素比第一个小， 则合并， 使栈中所有的序列大小达到近似相等
    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.
     */
     // 最后合并栈中所有的序列， 直到最后只剩一个有序序列
    private void mergeForceCollapse() {
        while (stackSize > 1) {
            int n = stackSize - 2;
            // 如果runLen[n-1] < runLen[n+1]， 则合并较小的较小的runBase[n-1]和runBase[n]， 否则合并runBase[n]和runBase[n+1]
            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.
     *
     * @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];
        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.
         */
        runLen[i] = len1 + len2;
        // 如果i是栈顶倒数第三个元素， 则最后i+1一定会合进i数组， 因此i+1的位置替换成i+2
        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).
         */
         // 找到run2的首元素在run1中的位置
        int k = gallopRight(a[base2], a, base1, len1, 0, c);
        assert k >= 0;
        // 忽略k之前的序列， 因为已经有序， 减少比较次数
        base1 += k;
        len1 -= k;
        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).
         */
         // 找打run1的尾元素在run2中的位置
        len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
        assert len2 >= 0;
        // len2 == 0， 说明run1和run2已经是一个整体有序的序列了， 直接返回。
        if (len2 == 0)
            return;

        // Merge remaining runs, using tmp array with min(len1, len2) elements
        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.
     *
     * @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;
        if (c.compare(key, a[base + hint]) > 0) {
            // 查找区间[base + hint, len], 查找使得a[base+hint+lastOfs] < key <= a[base+hint+ofs]成立的ofs值。
            int maxOfs = len - hint;
            // 向右查找， 最大可能Ofs为maxOfs-1， 即len - hint - 1， 即a[base + len - 1]
            while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) {
                // 记录上一次ofs的值， 存到lastOfs中
                lastOfs = ofs;
                // ofs乘以2再加1
                ofs = (ofs << 1) + 1;
                // 整数溢出
                if (ofs <= 0)
                    ofs = maxOfs;
            }
            
            // ofs最大值为maxOfs
            if (ofs > maxOfs)
                ofs = maxOfs;

            // Make offsets relative to base
            // 之前的ofs和lastOfs都是相对于hint位置的， 现在把它重置为相对于base的位置
            lastOfs += hint;
            ofs += hint;
        } else { // key <= a[base + hint]
            // 从base+hint向前搜索，查找区间[base, base + hint]， 直到找到ofs值使得a[base+hint-ofs] < key <= a[base+hint-lastOfs]
            // maxOfs为hint+1， 而ofs < maxOfs, 因此当ofs = maxOfs -1 时， 比较到的最左边的元素为a[base + hint - hint] == a[base]
            final int maxOfs = hint + 1;
            while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) {
                lastOfs = ofs;
                // ofs乘以2再加1
                ofs = (ofs << 1) + 1;
                // 正整数溢出
                if (ofs <= 0)
                    ofs = maxOfs;
            }
            // 最大为maxOfs
            if (ofs > maxOfs)
                ofs = maxOfs;

            // 重置ofs和lastOfs为相对于base的位置索引
            int tmp = lastOfs;
            lastOfs = hint - ofs;
            ofs = hint - tmp;
        }
        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].
         */
        lastOfs++;
        // 查找准确位置
        while (lastOfs < ofs) {
            // 中间位置
            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.
     *
     * @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) {
            // 从base + hint位置向前搜索区间[base, base + hint]， 使得a[b+hint - ofs] <= key < a[b+hint - lastOfs]
            int maxOfs = hint + 1;
            while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) {
                // 记录上次查找位置
                lastOfs = ofs;
                // 乘以2 加1
                ofs = (ofs << 1) + 1;
                // 正整数溢出
                if (ofs <= 0)
                    ofs = maxOfs;
            }
            // 最大为maxOfs
            if (ofs > maxOfs)
                ofs = maxOfs;

            // 重置ofs和lastOfs为相对于base的位置索引
            int tmp = lastOfs;
            lastOfs = hint - ofs;
            ofs = hint - tmp;
        } else { // a[b + hint] <= key
            // 搜索区间[base + hint, base + len -1]使得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;
                // 正整数溢出
                if (ofs <= 0)  
                    ofs = maxOfs;
            }
            if (ofs > maxOfs)
                ofs = maxOfs;

            // 重置ofs和lastOfs为相对于base的位置索引
            lastOfs += hint;
            ofs += hint;
        }
        assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

        lastOfs++;
        // 查找key的准确位置
        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
        }
        // 最终会找到m使得k >= a[base + m]， 而此时lastOfs == ofs且lastOfs = m +1， 则ofs = m +1， 因此a[base + ofs] > k >= a[base + ofs -1]， ofs即m+1， ofs - 1即为m， 因此ofs位置的值大于key
        assert lastOfs == ofs;    // so a[b + ofs - 1] <= key < a[b + ofs]
        return ofs;
    }

    /**
    *基于以上gallopRight方法最后查找key的索引的解释， 因此run1的第一个元素一定大于run2的第一个元素，     *而run2中run1最后一个元素所在索引位置之后的值也被忽略掉， 因此run1的最后一个元素大于run2中的所有元素的值。*/
    
    
    /**
     * 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
        T[] tmp = ensureCapacity(len1);
        int cursor1 = tmpBase; // Indexes into tmp array
        int cursor2 = base2;   // Indexes int a
        // 目标位置从base1的索引开始， 因为base1在base2之前， 下面会将base1的内容放入临时数组， 这样run1中的内容就可以覆盖了
        int dest = base1;      // Indexes int a
        // 把第一个序列的内容放入临时数组tmp中
        System.arraycopy(a, base1, tmp, cursor1, len1);

        // Move first element of second run and deal with degenerate cases
        // 因为run1的第一个元素大于run2的第一个元素， 因此将run2的第一个元素先放进
        a[dest++] = a[cursor2++];
        // 如果run2只有一个元素， 则将run1中剩余的元素放入正确位置后返回
        if (--len2 == 0) {
            System.arraycopy(tmp, cursor1, a, dest, len1);
            return;
        }
        // 如果run1只有一个元素， 因为run1的最后一个元素大于run2中的所有元素， 因此先将run2中的元素放入正确位置， 然后将run1的唯一的一个元素放入最后一个位置， 然后返回
        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
        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 starts
             * winning consistently.
             */
            do {
                assert len1 > 1 && len2 > 0;
                if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
                    a[dest++] = a[cursor2++];
                    count2++;
                    count1 = 0;
                    if (--len2 == 0)
                        break outer;
                } else {
                    a[dest++] = tmp[cursor1++];
                    count1++;
                    count2 = 0;
                    if (--len1 == 1)
                        break outer;
                }
                // 当每个序列中的连续放入目标位置的元素个数小于minGallop时， 这样分别拷贝就可以了
            } 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.
             */
             // 因为两个run序列的大小是近似相等的， 如果一个序列连续超过minGallop个数的元素被放入目标位置， 则另一个有接近大小的连续序列等待被放入正确位置，切换成Gallopping模式
            do {
                assert len1 > 1 && len2 > 0;
                //查找run1第一个大于run2中第一个元素的元素的位置索引
                count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
                if (count1 != 0) {
                    // run1中count1之前的元素全部放入目标序列
                    System.arraycopy(tmp, cursor1, a, dest, count1);
                    // 移动索引位置
                    dest += count1;
                    cursor1 += count1;
                    len1 -= count1;
                    if (len1 <= 1) // len1 == 1 || len1 == 0
                        break outer;
                }
                // 移动run2的第一个元素到目标序列中
                a[dest++] = a[cursor2++];
                // 如果run2中没有其他元素则跳出
                if (--len2 == 0)
                    break outer;

                // 查找run2中第一个小于等于run1当前元素的元素的位置索引
                count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
                if (count2 != 0) {
                    // 拷贝count2之后的元素到目标序列
                    System.arraycopy(a, cursor2, a, dest, count2);
                    dest += count2;
                    cursor2 += count2;
                    len2 -= count2;
                    // run2中没有其他元素则跳出
                    if (len2 == 0)
                        break outer;
                }
                
                // 此时run2中的第一个元素大于等于run1中的第一个元素， 拷贝run1中的第一个元素到目标序列
                a[dest++] = tmp[cursor1++];
                // 如果run1中只有一个元素则跳出
                if (--len1 == 1)
                    break outer;
                // 动态调整minGallop的值
                minGallop--;
            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if (minGallop < 0)
                minGallop = 0;
            // 调整minGallop的值， 使得在有序序列不多的情况下不用Gallopping模式
            minGallop += 2;  // Penalize for leaving gallop mode
        }  // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

        // run1中只有一个元素
        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) {
            // 因为run1中的最后一个元素大于run2中的所有元素， 因此这种情况不存在
            throw new IllegalArgumentException(
                "Comparison method violates its general contract!");
        } else {
            // run2中已经没有元素
            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;
        // 将run2中的所有元素放入临时数组tmp中
        System.arraycopy(a, base2, tmp, tmpBase, len2);

        int cursor1 = base1 + len1 - 1;  // Indexes into a
        int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array
        // 从后往前插入元素
        int dest = base2 + len2 - 1;     // Indexes into a

        // Move last element of first run and deal with degenerate cases
        // run1的最后一个元素导入目标位置
        a[dest--] = a[cursor1--];
        // 如果run1中只有一个元素， 将run2中的剩余元素放入目标位置（从后往前）
        if (--len1 == 0) {
            System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2);
            return;
        }
        if (len2 == 1) {
            // run2中只有一个元素， 因为run1的第一个元素大于run2的第一个元素， 因此， run2中唯一的一个元素小于run1中所有的元素， 因此将run1中的元素全部放入目标位置， 最后将唯一的run2中的一个元素放入第一个位置
            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) {
                       // 从后往前放入目标位置
                    a[dest--] = a[cursor1--];
                    count1++;
                    count2 = 0;
                    // run1中没有了元素
                    if (--len1 == 0)
                        break outer;
                } else {
                    a[dest--] = tmp[cursor2--];
                    count2++;
                    count1 = 0;
                    // run2中只有一个剩余元素
                    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;
                // 找到大于run2当前位置的元素的run1中元素， 因为是从后往前查找， 因此找到的位置比如k1，k1之后的所有元素大于run1, run2中的剩余所有元素
                count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
                if (count1 != 0) {
                    dest -= count1;
                    cursor1 -= count1;
                    len1 -= count1;
                    // 拷贝run1中的大的元素
                    System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
                    // run1中没有元素， 跳出
                    if (len1 == 0)
                        break outer;
                }
                // run2的当前元素拷贝到目标位置
                a[dest--] = tmp[cursor2--];
                if (--len2 == 1)
                    break outer;

                // 找到run2中大于等于run1当前元素的元素的位置索引比如K2， 则k2之后的所有元素大于run1, run2中的剩余元素
                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;
                }
                // 拷贝run1的当前元素到目标位置， 因为a[cursior1]大于等于run2中的剩余元素
                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

        // run2中只有一个元素
        if (len2 == 1) {
            assert len1 > 0;
            dest -= len1;
            cursor1 -= len1;
            // 拷贝run1中的所有元素到目标位置
            System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
            // run2的最后一个元素放入第一个位置
            a[dest] = tmp[cursor2];  // Move first elt of run2 to front of merge
        } else if (len2 == 0) {
            // 因为run2的第一个元素小于run1， run2中的所有元素， 因此run2不可能最后为空
            throw new IllegalArgumentException(
                "Comparison method violates its general contract!");
        } else {
            assert len1 == 0;
            assert len2 > 0;
            // 拷贝run2中的剩余元素
            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.
     *
     * @param minCapacity the minimum required capacity of the tmp array
     * @return tmp, whether or not it grew
     */
    private T[] ensureCapacity(int minCapacity) {
        if (tmpLen < minCapacity) {
            // Compute smallest power of 2 > minCapacity
            int newSize = minCapacity;
            newSize |= newSize >> 1;
            newSize |= newSize >> 2;
            newSize |= newSize >> 4;
            newSize |= newSize >> 8;
            newSize |= newSize >> 16;
            newSize++;

            if (newSize < 0) // Not bloody likely!
                newSize = minCapacity;
            else
                newSize = Math.min(newSize, a.length >>> 1);

            @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
            T[] newArray = (T[])java.lang.reflect.Array.newInstance
                (a.getClass().getComponentType(), newSize);
            tmp = newArray;
            tmpLen = newSize;
            tmpBase = 0;
        }
        return tmp;
    }
}