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. Oracle designates this
8 * particular file as subject to the "Classpath" exception as provided
9 * by Oracle 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 Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
22 * or visit www.oracle.com if you need additional information or have any
23 * 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<>(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 }