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;
     }
 }