1 /*

        2  * Copyright 2009 Google Inc.  All Rights Reserved.

        3  * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.

        4  *

        5  * This code is free software; you can redistribute it and/or modify it

        6  * under the terms of the GNU General Public License version 2 only, as

        7  * published by the Free Software Foundation.  Sun designates this

        8  * particular file as subject to the "Classpath" exception as provided

        9  * by Sun in the LICENSE file that accompanied this code.

       10  *

       11  * This code is distributed in the hope that it will be useful, but WITHOUT

       12  * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or

       13  * FITNESS FOR A PARTICULAR PURPOSE.  See the GNU General Public License

       14  * version 2 for more details (a copy is included in the LICENSE file that

       15  * accompanied this code).

       16  *

       17  * You should have received a copy of the GNU General Public License version

       18  * 2 along with this work; if not, write to the Free Software Foundation,

       19  * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.

       20  *

       21  * Please contact Sun Microsystems, Inc., 4150 Network Circle, Santa Clara,

       22  * CA 95054 USA or visit www.sun.com if you need additional information or

       23  * have any questions.

       24  */

       25 

       26 package java.util;

       27 

       28 /**

       29  * A stable, adaptive, iterative mergesort that requires far fewer than

       30  * n lg(n) comparisons when running on partially sorted arrays, while

       31  * offering performance comparable to a traditional mergesort when run

       32  * on random arrays.  Like all proper mergesorts, this sort is stable and

       33  * runs O(n log n) time (worst case).  In the worst case, this sort requires

       34  * temporary storage space for n/2 object references; in the best case,

       35  * it requires only a small constant amount of space.

       36  *

       37  * This implementation was adapted from Tim Peters's list sort for

       38  * Python, which is described in detail here:

       39  *

       40  *   http://svn.python.org/projects/python/trunk/Objects/listsort.txt

       41  *

       42  * Tim's C code may be found here:

       43  *

       44  *   http://svn.python.org/projects/python/trunk/Objects/listobject.c

       45  *

       46  * The underlying techniques are described in this paper (and may have

       47  * even earlier origins):

       48  *

       49  *  "Optimistic Sorting and Information Theoretic Complexity"

       50  *  Peter McIlroy

       51  *  SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),

       52  *  pp 467-474, Austin, Texas, 25-27 January 1993.

       53  *

       54  * While the API to this class consists solely of static methods, it is

       55  * (privately) instantiable; a TimSort instance holds the state of an ongoing

       56  * sort, assuming the input array is large enough to warrant the full-blown

       57  * TimSort. Small arrays are sorted in place, using a binary insertion sort.

       58  *

       59  * @author Josh Bloch

       60  */

       61 class TimSort<T> {

       62     /**

       63      * This is the minimum sized sequence that will be merged.  Shorter

       64      * sequences will be lengthened by calling binarySort.  If the entire

       65      * array is less than this length, no merges will be performed.

       66      *

       67      * This constant should be a power of two.  It was 64 in Tim Peter's C

       68      * implementation, but 32 was empirically determined to work better in

       69      * this implementation.  In the unlikely event that you set this constant

       70      * to be a number that's not a power of two, you'll need to change the

       71      * {@link #minRunLength} computation.

       72      *

       73      * If you decrease this constant, you must change the stackLen

       74      * computation in the TimSort constructor, or you risk an

       75      * ArrayOutOfBounds exception.  See listsort.txt for a discussion

       76      * of the minimum stack length required as a function of the length

       77      * of the array being sorted and the minimum merge sequence length.

       78      */

       79     private static final int MIN_MERGE = 32;

       80 

       81     /**

       82      * The array being sorted.

       83      */

       84     private final T[] a;

       85 

       86     /**

       87      * The comparator for this sort.

       88      */

       89     private final Comparator<? super T> c;

       90 

       91     /**

       92      * When we get into galloping mode, we stay there until both runs win less

       93      * often than MIN_GALLOP consecutive times.

       94      */

       95     private static final int  MIN_GALLOP = 7;

       96 

       97     /**

       98      * This controls when we get *into* galloping mode.  It is initialized

       99      * to MIN_GALLOP.  The mergeLo and mergeHi methods nudge it higher for

      100      * random data, and lower for highly structured data.

      101      */

      102     private int minGallop = MIN_GALLOP;

      103 

      104     /**

      105      * Maximum initial size of tmp array, which is used for merging.  The array

      106      * can grow to accommodate demand.

      107      *

      108      * Unlike Tim's original C version, we do not allocate this much storage

      109      * when sorting smaller arrays.  This change was required for performance.

      110      */

      111     private static final int INITIAL_TMP_STORAGE_LENGTH = 256;

      112 

      113     /**

      114      * Temp storage for merges.

      115      */

      116     private T[] tmp; // Actual runtime type will be Object[], regardless of T

      117 

      118     /**

      119      * A stack of pending runs yet to be merged.  Run i starts at

      120      * address base[i] and extends for len[i] elements.  It's always

      121      * true (so long as the indices are in bounds) that:

      122      *

      123      *     runBase[i] + runLen[i] == runBase[i + 1]

      124      *

      125      * so we could cut the storage for this, but it's a minor amount,

      126      * and keeping all the info explicit simplifies the code.

      127      */

      128     private int stackSize = 0;  // Number of pending runs on stack

      129     private final int[] runBase;

      130     private final int[] runLen;

      131 

      132     /**

      133      * Creates a TimSort instance to maintain the state of an ongoing sort.

      134      *

      135      * @param a the array to be sorted

      136      * @param c the comparator to determine the order of the sort

      137      */

      138     private TimSort(T[] a, Comparator<? super T> c) {

      139         this.a = a;

      140         this.c = c;

      141 

      142         // Allocate temp storage (which may be increased later if necessary)

      143         int len = a.length;

      144         @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})

      145         T[] newArray = (T[]) new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ?

      146                                         len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];

      147         tmp = newArray;

      148 

      149         /*

      150          * Allocate runs-to-be-merged stack (which cannot be expanded).  The

      151          * stack length requirements are described in listsort.txt.  The C

      152          * version always uses the same stack length (85), but this was

      153          * measured to be too expensive when sorting "mid-sized" arrays (e.g.,

      154          * 100 elements) in Java.  Therefore, we use smaller (but sufficiently

      155          * large) stack lengths for smaller arrays.  The "magic numbers" in the

      156          * computation below must be changed if MIN_MERGE is decreased.  See

      157          * the MIN_MERGE declaration above for more information.

      158          */

      159         int stackLen = (len <    120  ?  5 :

      160                         len <   1542  ? 10 :

      161                         len < 119151  ? 19 : 40);

      162         runBase = new int[stackLen];

      163         runLen = new int[stackLen];

      164     }

      165 

      166     /*

      167      * The next two methods (which are package private and static) constitute

      168      * the entire API of this class.  Each of these methods obeys the contract

      169      * of the public method with the same signature in java.util.Arrays.

      170      */

      171 

      172     static <T> void sort(T[] a, Comparator<? super T> c) {

      173         sort(a, 0, a.length, c);

      174     }

      175 

      176     static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c) {

      177         if (c == null) {

      178             Arrays.sort(a, lo, hi);

      179             return;

      180         }

      181 

      182         rangeCheck(a.length, lo, hi);

      183         int nRemaining  = hi - lo;

      184         if (nRemaining < 2)

      185             return;  // Arrays of size 0 and 1 are always sorted

      186 

      187         // If array is small, do a "mini-TimSort" with no merges

      188         if (nRemaining < MIN_MERGE) {

      189             int initRunLen = countRunAndMakeAscending(a, lo, hi, c);

      190             binarySort(a, lo, hi, lo + initRunLen, c);

      191             return;

      192         }

      193 

      194         /**

      195          * March over the array once, left to right, finding natural runs,

      196          * extending short natural runs to minRun elements, and merging runs

      197          * to maintain stack invariant.

      198          */

      199         TimSort<T> ts = new TimSort<T>(a, c);

      200         int minRun = minRunLength(nRemaining);

      201         do {

      202             // Identify next run

      203             int runLen = countRunAndMakeAscending(a, lo, hi, c);

      204 

      205             // If run is short, extend to min(minRun, nRemaining)

      206             if (runLen < minRun) {

      207                 int force = nRemaining <= minRun ? nRemaining : minRun;

      208                 binarySort(a, lo, lo + force, lo + runLen, c);

      209                 runLen = force;

      210             }

      211 

      212             // Push run onto pending-run stack, and maybe merge

      213             ts.pushRun(lo, runLen);

      214             ts.mergeCollapse();

      215 

      216             // Advance to find next run

      217             lo += runLen;

      218             nRemaining -= runLen;

      219         } while (nRemaining != 0);

      220 

      221         // Merge all remaining runs to complete sort

      222         assert lo == hi;

      223         ts.mergeForceCollapse();

      224         assert ts.stackSize == 1;

      225     }

      226 

      227     /**

      228      * Sorts the specified portion of the specified array using a binary

      229      * insertion sort.  This is the best method for sorting small numbers

      230      * of elements.  It requires O(n log n) compares, but O(n^2) data

      231      * movement (worst case).

      232      *

      233      * If the initial part of the specified range is already sorted,

      234      * this method can take advantage of it: the method assumes that the

      235      * elements from index {@code lo}, inclusive, to {@code start},

      236      * exclusive are already sorted.

      237      *

      238      * @param a the array in which a range is to be sorted

      239      * @param lo the index of the first element in the range to be sorted

      240      * @param hi the index after the last element in the range to be sorted

      241      * @param start the index of the first element in the range that is

      242      *        not already known to be sorted (@code lo <= start <= hi}

      243      * @param c comparator to used for the sort

      244      */

      245     @SuppressWarnings("fallthrough")

      246     private static <T> void binarySort(T[] a, int lo, int hi, int start,

      247                                        Comparator<? super T> c) {

      248         assert lo <= start && start <= hi;

      249         if (start == lo)

      250             start++;

      251         for ( ; start < hi; start++) {

      252             T pivot = a[start];

      253 

      254             // Set left (and right) to the index where a[start] (pivot) belongs

      255             int left = lo;

      256             int right = start;

      257             assert left <= right;

      258             /*

      259              * Invariants:

      260              *   pivot >= all in [lo, left).

      261              *   pivot <  all in [right, start).

      262              */

      263             while (left < right) {

      264                 int mid = (left + right) >>> 1;

      265                 if (c.compare(pivot, a[mid]) < 0)

      266                     right = mid;

      267                 else

      268                     left = mid + 1;

      269             }

      270             assert left == right;

      271 

      272             /*

      273              * The invariants still hold: pivot >= all in [lo, left) and

      274              * pivot < all in [left, start), so pivot belongs at left.  Note

      275              * that if there are elements equal to pivot, left points to the

      276              * first slot after them -- that's why this sort is stable.

      277              * Slide elements over to make room to make room for pivot.

      278              */

      279             int n = start - left;  // The number of elements to move

      280             // Switch is just an optimization for arraycopy in default case

      281             switch(n) {

      282                 case 2:  a[left + 2] = a[left + 1];

      283                 case 1:  a[left + 1] = a[left];

      284                          break;

      285                 default: System.arraycopy(a, left, a, left + 1, n);

      286             }

      287             a[left] = pivot;

      288         }

      289     }

      290 

      291     /**

      292      * Returns the length of the run beginning at the specified position in

      293      * the specified array and reverses the run if it is descending (ensuring

      294      * that the run will always be ascending when the method returns).

      295      *

      296      * A run is the longest ascending sequence with:

      297      *

      298      *    a[lo] <= a[lo + 1] <= a[lo + 2] <= ...

      299      *

      300      * or the longest descending sequence with:

      301      *

      302      *    a[lo] >  a[lo + 1] >  a[lo + 2] >  ...

      303      *

      304      * For its intended use in a stable mergesort, the strictness of the

      305      * definition of "descending" is needed so that the call can safely

      306      * reverse a descending sequence without violating stability.

      307      *

      308      * @param a the array in which a run is to be counted and possibly reversed

      309      * @param lo index of the first element in the run

      310      * @param hi index after the last element that may be contained in the run.

      311               It is required that @code{lo < hi}.

      312      * @param c the comparator to used for the sort

      313      * @return  the length of the run beginning at the specified position in

      314      *          the specified array

      315      */

      316     private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,

      317                                                     Comparator<? super T> c) {

      318         assert lo < hi;

      319         int runHi = lo + 1;

      320         if (runHi == hi)

      321             return 1;

      322 

      323         // Find end of run, and reverse range if descending

      324         if (c.compare(a[runHi++], a[lo]) < 0) { // Descending

      325             while(runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)

      326                 runHi++;

      327             reverseRange(a, lo, runHi);

      328         } else {                              // Ascending

      329             while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)

      330                 runHi++;

      331         }

      332 

      333         return runHi - lo;

      334     }

      335 

      336     /**

      337      * Reverse the specified range of the specified array.

      338      *

      339      * @param a the array in which a range is to be reversed

      340      * @param lo the index of the first element in the range to be reversed

      341      * @param hi the index after the last element in the range to be reversed

      342      */

      343     private static void reverseRange(Object[] a, int lo, int hi) {

      344         hi--;

      345         while (lo < hi) {

      346             Object t = a[lo];

      347             a[lo++] = a[hi];

      348             a[hi--] = t;

      349         }

      350     }

      351 

      352     /**

      353      * Returns the minimum acceptable run length for an array of the specified

      354      * length. Natural runs shorter than this will be extended with

      355      * {@link #binarySort}.

      356      *

      357      * Roughly speaking, the computation is:

      358      *

      359      *  If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).

      360      *  Else if n is an exact power of 2, return MIN_MERGE/2.

      361      *  Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k

      362      *   is close to, but strictly less than, an exact power of 2.

      363      *

      364      * For the rationale, see listsort.txt.

      365      *

      366      * @param n the length of the array to be sorted

      367      * @return the length of the minimum run to be merged

      368      */

      369     private static int minRunLength(int n) {

      370         assert n >= 0;

      371         int r = 0;      // Becomes 1 if any 1 bits are shifted off

      372         while (n >= MIN_MERGE) {

      373             r |= (n & 1);

      374             n >>= 1;

      375         }

      376         return n + r;

      377     }

      378 

      379     /**

      380      * Pushes the specified run onto the pending-run stack.

      381      *

      382      * @param runBase index of the first element in the run

      383      * @param runLen  the number of elements in the run

      384      */

      385     private void pushRun(int runBase, int runLen) {

      386         this.runBase[stackSize] = runBase;

      387         this.runLen[stackSize] = runLen;

      388         stackSize++;

      389     }

      390 

      391     /**

      392      * Examines the stack of runs waiting to be merged and merges adjacent runs

      393      * until the stack invariants are reestablished:

      394      *

      395      *     1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]

      396      *     2. runLen[i - 2] > runLen[i - 1]

      397      *

      398      * This method is called each time a new run is pushed onto the stack,

      399      * so the invariants are guaranteed to hold for i < stackSize upon

      400      * entry to the method.

      401      */

      402     private void mergeCollapse() {

      403         while (stackSize > 1) {

      404             int n = stackSize - 2;

      405             if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {

      406                 if (runLen[n - 1] < runLen[n + 1])

      407                     n--;

      408                 mergeAt(n);

      409             } else if (runLen[n] <= runLen[n + 1]) {

      410                 mergeAt(n);

      411             } else {

      412                 break; // Invariant is established

      413             }

      414         }

      415     }

      416 

      417     /**

      418      * Merges all runs on the stack until only one remains.  This method is

      419      * called once, to complete the sort.

      420      */

      421     private void mergeForceCollapse() {

      422         while (stackSize > 1) {

      423             int n = stackSize - 2;

      424             if (n > 0 && runLen[n - 1] < runLen[n + 1])

      425                 n--;

      426             mergeAt(n);

      427         }

      428     }

      429 

      430     /**

      431      * Merges the two runs at stack indices i and i+1.  Run i must be

      432      * the penultimate or antepenultimate run on the stack.  In other words,

      433      * i must be equal to stackSize-2 or stackSize-3.

      434      *

      435      * @param i stack index of the first of the two runs to merge

      436      */

      437     private void mergeAt(int i) {

      438         assert stackSize >= 2;

      439         assert i >= 0;

      440         assert i == stackSize - 2 || i == stackSize - 3;

      441 

      442         int base1 = runBase[i];

      443         int len1 = runLen[i];

      444         int base2 = runBase[i + 1];

      445         int len2 = runLen[i + 1];

      446         assert len1 > 0 && len2 > 0;

      447         assert base1 + len1 == base2;

      448 

      449         /*

      450          * Record the length of the combined runs; if i is the 3rd-last

      451          * run now, also slide over the last run (which isn't involved

      452          * in this merge).  The current run (i+1) goes away in any case.

      453          */

      454         runLen[i] = len1 + len2;

      455         if (i == stackSize - 3) {

      456             runBase[i + 1] = runBase[i + 2];

      457             runLen[i + 1] = runLen[i + 2];

      458         }

      459         stackSize--;

      460 

      461         /*

      462          * Find where the first element of run2 goes in run1. Prior elements

      463          * in run1 can be ignored (because they're already in place).

      464          */

      465         int k = gallopRight(a[base2], a, base1, len1, 0, c);

      466         assert k >= 0;

      467         base1 += k;

      468         len1 -= k;

      469         if (len1 == 0)

      470             return;

      471 

      472         /*

      473          * Find where the last element of run1 goes in run2. Subsequent elements

      474          * in run2 can be ignored (because they're already in place).

      475          */

      476         len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);

      477         assert len2 >= 0;

      478         if (len2 == 0)

      479             return;

      480 

      481         // Merge remaining runs, using tmp array with min(len1, len2) elements

      482         if (len1 <= len2)

      483             mergeLo(base1, len1, base2, len2);

      484         else

      485             mergeHi(base1, len1, base2, len2);

      486     }

      487 

      488     /**

      489      * Locates the position at which to insert the specified key into the

      490      * specified sorted range; if the range contains an element equal to key,

      491      * returns the index of the leftmost equal element.

      492      *

      493      * @param key the key whose insertion point to search for

      494      * @param a the array in which to search

      495      * @param base the index of the first element in the range

      496      * @param len the length of the range; must be > 0

      497      * @param hint the index at which to begin the search, 0 <= hint < n.

      498      *     The closer hint is to the result, the faster this method will run.

      499      * @param c the comparator used to order the range, and to search

      500      * @return the int k,  0 <= k <= n such that a[b + k - 1] < key <= a[b + k],

      501      *    pretending that a[b - 1] is minus infinity and a[b + n] is infinity.

      502      *    In other words, key belongs at index b + k; or in other words,

      503      *    the first k elements of a should precede key, and the last n - k

      504      *    should follow it.

      505      */

      506     private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint,

      507                                       Comparator<? super T> c) {

      508         assert len > 0 && hint >= 0 && hint < len;

      509         int lastOfs = 0;

      510         int ofs = 1;

      511         if (c.compare(key, a[base + hint]) > 0) {

      512             // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]

      513             int maxOfs = len - hint;

      514             while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) {

      515                 lastOfs = ofs;

      516                 ofs = (ofs << 1) + 1;

      517                 if (ofs <= 0)   // int overflow

      518                     ofs = maxOfs;

      519             }

      520             if (ofs > maxOfs)

      521                 ofs = maxOfs;

      522 

      523             // Make offsets relative to base

      524             lastOfs += hint;

      525             ofs += hint;

      526         } else { // key <= a[base + hint]

      527             // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]

      528             final int maxOfs = hint + 1;

      529             while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) {

      530                 lastOfs = ofs;

      531                 ofs = (ofs << 1) + 1;

      532                 if (ofs <= 0)   // int overflow

      533                     ofs = maxOfs;

      534             }

      535             if (ofs > maxOfs)

      536                 ofs = maxOfs;

      537 

      538             // Make offsets relative to base

      539             int tmp = lastOfs;

      540             lastOfs = hint - ofs;

      541             ofs = hint - tmp;

      542         }

      543         assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

      544 

      545         /*

      546          * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere

      547          * to the right of lastOfs but no farther right than ofs.  Do a binary

      548          * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].

      549          */

      550         lastOfs++;

      551         while (lastOfs < ofs) {

      552             int m = lastOfs + ((ofs - lastOfs) >>> 1);

      553 

      554             if (c.compare(key, a[base + m]) > 0)

      555                 lastOfs = m + 1;  // a[base + m] < key

      556             else

      557                 ofs = m;          // key <= a[base + m]

      558         }

      559         assert lastOfs == ofs;    // so a[base + ofs - 1] < key <= a[base + ofs]

      560         return ofs;

      561     }

      562 

      563     /**

      564      * Like gallopLeft, except that if the range contains an element equal to

      565      * key, gallopRight returns the index after the rightmost equal element.

      566      *

      567      * @param key the key whose insertion point to search for

      568      * @param a the array in which to search

      569      * @param base the index of the first element in the range

      570      * @param len the length of the range; must be > 0

      571      * @param hint the index at which to begin the search, 0 <= hint < n.

      572      *     The closer hint is to the result, the faster this method will run.

      573      * @param c the comparator used to order the range, and to search

      574      * @return the int k,  0 <= k <= n such that a[b + k - 1] <= key < a[b + k]

      575      */

      576     private static <T> int gallopRight(T key, T[] a, int base, int len,

      577                                        int hint, Comparator<? super T> c) {

      578         assert len > 0 && hint >= 0 && hint < len;

      579 

      580         int ofs = 1;

      581         int lastOfs = 0;

      582         if (c.compare(key, a[base + hint]) < 0) {

      583             // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]

      584             int maxOfs = hint + 1;

      585             while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) {

      586                 lastOfs = ofs;

      587                 ofs = (ofs << 1) + 1;

      588                 if (ofs <= 0)   // int overflow

      589                     ofs = maxOfs;

      590             }

      591             if (ofs > maxOfs)

      592                 ofs = maxOfs;

      593 

      594             // Make offsets relative to b

      595             int tmp = lastOfs;

      596             lastOfs = hint - ofs;

      597             ofs = hint - tmp;

      598         } else { // a[b + hint] <= key

      599             // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]

      600             int maxOfs = len - hint;

      601             while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) {

      602                 lastOfs = ofs;

      603                 ofs = (ofs << 1) + 1;

      604                 if (ofs <= 0)   // int overflow

      605                     ofs = maxOfs;

      606             }

      607             if (ofs > maxOfs)

      608                 ofs = maxOfs;

      609 

      610             // Make offsets relative to b

      611             lastOfs += hint;

      612             ofs += hint;

      613         }

      614         assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

      615 

      616         /*

      617          * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to

      618          * the right of lastOfs but no farther right than ofs.  Do a binary

      619          * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].

      620          */

      621         lastOfs++;

      622         while (lastOfs < ofs) {

      623             int m = lastOfs + ((ofs - lastOfs) >>> 1);

      624 

      625             if (c.compare(key, a[base + m]) < 0)

      626                 ofs = m;          // key < a[b + m]

      627             else

      628                 lastOfs = m + 1;  // a[b + m] <= key

      629         }

      630         assert lastOfs == ofs;    // so a[b + ofs - 1] <= key < a[b + ofs]

      631         return ofs;

      632     }

      633 

      634     /**

      635      * Merges two adjacent runs in place, in a stable fashion.  The first

      636      * element of the first run must be greater than the first element of the

      637      * second run (a[base1] > a[base2]), and the last element of the first run

      638      * (a[base1 + len1-1]) must be greater than all elements of the second run.

      639      *

      640      * For performance, this method should be called only when len1 <= len2;

      641      * its twin, mergeHi should be called if len1 >= len2.  (Either method

      642      * may be called if len1 == len2.)

      643      *

      644      * @param base1 index of first element in first run to be merged

      645      * @param len1  length of first run to be merged (must be > 0)

      646      * @param base2 index of first element in second run to be merged

      647      *        (must be aBase + aLen)

      648      * @param len2  length of second run to be merged (must be > 0)

      649      */

      650     private void mergeLo(int base1, int len1, int base2, int len2) {

      651         assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

      652 

      653         // Copy first run into temp array

      654         T[] a = this.a; // For performance

      655         T[] tmp = ensureCapacity(len1);

      656         System.arraycopy(a, base1, tmp, 0, len1);

      657 

      658         int cursor1 = 0;       // Indexes into tmp array

      659         int cursor2 = base2;   // Indexes int a

      660         int dest = base1;      // Indexes int a

      661 

      662         // Move first element of second run and deal with degenerate cases

      663         a[dest++] = a[cursor2++];

      664         if (--len2 == 0) {

      665             System.arraycopy(tmp, cursor1, a, dest, len1);

      666             return;

      667         }

      668         if (len1 == 1) {

      669             System.arraycopy(a, cursor2, a, dest, len2);

      670             a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge

      671             return;

      672         }

      673 

      674         Comparator<? super T> c = this.c;  // Use local variable for performance

      675         int minGallop = this.minGallop;    //  "    "       "     "      "

      676     outer:

      677         while (true) {

      678             int count1 = 0; // Number of times in a row that first run won

      679             int count2 = 0; // Number of times in a row that second run won

      680 

      681             /*

      682              * Do the straightforward thing until (if ever) one run starts

      683              * winning consistently.

      684              */

      685             do {

      686                 assert len1 > 1 && len2 > 0;

      687                 if (c.compare(a[cursor2], tmp[cursor1]) < 0) {

      688                     a[dest++] = a[cursor2++];

      689                     count2++;

      690                     count1 = 0;

      691                     if (--len2 == 0)

      692                         break outer;

      693                 } else {

      694                     a[dest++] = tmp[cursor1++];

      695                     count1++;

      696                     count2 = 0;

      697                     if (--len1 == 1)

      698                         break outer;

      699                 }

      700             } while ((count1 | count2) < minGallop);

      701 

      702             /*

      703              * One run is winning so consistently that galloping may be a

      704              * huge win. So try that, and continue galloping until (if ever)

      705              * neither run appears to be winning consistently anymore.

      706              */

      707             do {

      708                 assert len1 > 1 && len2 > 0;

      709                 count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);

      710                 if (count1 != 0) {

      711                     System.arraycopy(tmp, cursor1, a, dest, count1);

      712                     dest += count1;

      713                     cursor1 += count1;

      714                     len1 -= count1;

      715                     if (len1 <= 1) // len1 == 1 || len1 == 0

      716                         break outer;

      717                 }

      718                 a[dest++] = a[cursor2++];

      719                 if (--len2 == 0)

      720                     break outer;

      721 

      722                 count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);

      723                 if (count2 != 0) {

      724                     System.arraycopy(a, cursor2, a, dest, count2);

      725                     dest += count2;

      726                     cursor2 += count2;

      727                     len2 -= count2;

      728                     if (len2 == 0)

      729                         break outer;

      730                 }

      731                 a[dest++] = tmp[cursor1++];

      732                 if (--len1 == 1)

      733                     break outer;

      734                 minGallop--;

      735             } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);

      736             if (minGallop < 0)

      737                 minGallop = 0;

      738             minGallop += 2;  // Penalize for leaving gallop mode

      739         }  // End of "outer" loop

      740         this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

      741 

      742         if (len1 == 1) {

      743             assert len2 > 0;

      744             System.arraycopy(a, cursor2, a, dest, len2);

      745             a[dest + len2] = tmp[cursor1]; //  Last elt of run 1 to end of merge

      746         } else if (len1 == 0) {

      747             throw new IllegalArgumentException(

      748                 "Comparison method violates its general contract!");

      749         } else {

      750             assert len2 == 0;

      751             assert len1 > 1;

      752             System.arraycopy(tmp, cursor1, a, dest, len1);

      753         }

      754     }

      755 

      756     /**

      757      * Like mergeLo, except that this method should be called only if

      758      * len1 >= len2; mergeLo should be called if len1 <= len2.  (Either method

      759      * may be called if len1 == len2.)

      760      *

      761      * @param base1 index of first element in first run to be merged

      762      * @param len1  length of first run to be merged (must be > 0)

      763      * @param base2 index of first element in second run to be merged

      764      *        (must be aBase + aLen)

      765      * @param len2  length of second run to be merged (must be > 0)

      766      */

      767     private void mergeHi(int base1, int len1, int base2, int len2) {

      768         assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

      769 

      770         // Copy second run into temp array

      771         T[] a = this.a; // For performance

      772         T[] tmp = ensureCapacity(len2);

      773         System.arraycopy(a, base2, tmp, 0, len2);

      774 

      775         int cursor1 = base1 + len1 - 1;  // Indexes into a

      776         int cursor2 = len2 - 1;          // Indexes into tmp array

      777         int dest = base2 + len2 - 1;     // Indexes into a

      778 

      779         // Move last element of first run and deal with degenerate cases

      780         a[dest--] = a[cursor1--];

      781         if (--len1 == 0) {

      782             System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);

      783             return;

      784         }

      785         if (len2 == 1) {

      786             dest -= len1;

      787             cursor1 -= len1;

      788             System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);

      789             a[dest] = tmp[cursor2];

      790             return;

      791         }

      792 

      793         Comparator<? super T> c = this.c;  // Use local variable for performance

      794         int minGallop = this.minGallop;    //  "    "       "     "      "

      795     outer:

      796         while (true) {

      797             int count1 = 0; // Number of times in a row that first run won

      798             int count2 = 0; // Number of times in a row that second run won

      799 

      800             /*

      801              * Do the straightforward thing until (if ever) one run

      802              * appears to win consistently.

      803              */

      804             do {

      805                 assert len1 > 0 && len2 > 1;

      806                 if (c.compare(tmp[cursor2], a[cursor1]) < 0) {

      807                     a[dest--] = a[cursor1--];

      808                     count1++;

      809                     count2 = 0;

      810                     if (--len1 == 0)

      811                         break outer;

      812                 } else {

      813                     a[dest--] = tmp[cursor2--];

      814                     count2++;

      815                     count1 = 0;

      816                     if (--len2 == 1)

      817                         break outer;

      818                 }

      819             } while ((count1 | count2) < minGallop);

      820 

      821             /*

      822              * One run is winning so consistently that galloping may be a

      823              * huge win. So try that, and continue galloping until (if ever)

      824              * neither run appears to be winning consistently anymore.

      825              */

      826             do {

      827                 assert len1 > 0 && len2 > 1;

      828                 count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);

      829                 if (count1 != 0) {

      830                     dest -= count1;

      831                     cursor1 -= count1;

      832                     len1 -= count1;

      833                     System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);

      834                     if (len1 == 0)

      835                         break outer;

      836                 }

      837                 a[dest--] = tmp[cursor2--];

      838                 if (--len2 == 1)

      839                     break outer;

      840 

      841                 count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);

      842                 if (count2 != 0) {

      843                     dest -= count2;

      844                     cursor2 -= count2;

      845                     len2 -= count2;

      846                     System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);

      847                     if (len2 <= 1)  // len2 == 1 || len2 == 0

      848                         break outer;

      849                 }

      850                 a[dest--] = a[cursor1--];

      851                 if (--len1 == 0)

      852                     break outer;

      853                 minGallop--;

      854             } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);

      855             if (minGallop < 0)

      856                 minGallop = 0;

      857             minGallop += 2;  // Penalize for leaving gallop mode

      858         }  // End of "outer" loop

      859         this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

      860 

      861         if (len2 == 1) {

      862             assert len1 > 0;

      863             dest -= len1;

      864             cursor1 -= len1;

      865             System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);

      866             a[dest] = tmp[cursor2];  // Move first elt of run2 to front of merge

      867         } else if (len2 == 0) {

      868             throw new IllegalArgumentException(

      869                 "Comparison method violates its general contract!");

      870         } else {

      871             assert len1 == 0;

      872             assert len2 > 0;

      873             System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);

      874         }

      875     }

      876 

      877     /**

      878      * Ensures that the external array tmp has at least the specified

      879      * number of elements, increasing its size if necessary.  The size

      880      * increases exponentially to ensure amortized linear time complexity.

      881      *

      882      * @param minCapacity the minimum required capacity of the tmp array

      883      * @return tmp, whether or not it grew

      884      */

      885     private T[] ensureCapacity(int minCapacity) {

      886         if (tmp.length < minCapacity) {

      887             // Compute smallest power of 2 > minCapacity

      888             int newSize = minCapacity;

      889             newSize |= newSize >> 1;

      890             newSize |= newSize >> 2;

      891             newSize |= newSize >> 4;

      892             newSize |= newSize >> 8;

      893             newSize |= newSize >> 16;

      894             newSize++;

      895 

      896             if (newSize < 0) // Not bloody likely!

      897                 newSize = minCapacity;

      898             else

      899                 newSize = Math.min(newSize, a.length >>> 1);

      900 

      901             @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})

      902             T[] newArray = (T[]) new Object[newSize];

      903             tmp = newArray;

      904         }

      905         return tmp;

      906     }

      907 

      908     /**

      909      * Checks that fromIndex and toIndex are in range, and throws an

      910      * appropriate exception if they aren't.

      911      *

      912      * @param arrayLen the length of the array

      913      * @param fromIndex the index of the first element of the range

      914      * @param toIndex the index after the last element of the range

      915      * @throws IllegalArgumentException if fromIndex > toIndex

      916      * @throws ArrayIndexOutOfBoundsException if fromIndex < 0

      917      *         or toIndex > arrayLen

      918      */

      919     private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) {

      920         if (fromIndex > toIndex)

      921             throw new IllegalArgumentException("fromIndex(" + fromIndex +

      922                        ") > toIndex(" + toIndex+")");

      923         if (fromIndex < 0)

      924             throw new ArrayIndexOutOfBoundsException(fromIndex);

      925         if (toIndex > arrayLen)

      926             throw new ArrayIndexOutOfBoundsException(toIndex);

      927     }

      928 }