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 * This is a near duplicate of {@link TimSort}, modified for use with
30 * arrays of objects that implement {@link Comparable}, instead of using
31 * explicit comparators.
32 *
33 * <p>If you are using an optimizing VM, you may find that ComparableTimSort
34 * offers no performance benefit over TimSort in conjunction with a
35 * comparator that simply returns {@code ((Comparable)first).compareTo(Second)}.
36 * If this is the case, you are better off deleting ComparableTimSort to
37 * eliminate the code duplication. (See Arrays.java for details.)
38 *
39 * @author Josh Bloch
40 */
41 class ComparableTimSort {
42 /**
43 * This is the minimum sized sequence that will be merged. Shorter
44 * sequences will be lengthened by calling binarySort. If the entire
45 * array is less than this length, no merges will be performed.
46 *
47 * This constant should be a power of two. It was 64 in Tim Peter's C
48 * implementation, but 32 was empirically determined to work better in
49 * this implementation. In the unlikely event that you set this constant
50 * to be a number that's not a power of two, you'll need to change the
51 * {@link #minRunLength} computation.
52 *
53 * If you decrease this constant, you must change the stackLen
54 * computation in the TimSort constructor, or you risk an
55 * ArrayOutOfBounds exception. See listsort.txt for a discussion
56 * of the minimum stack length required as a function of the length
57 * of the array being sorted and the minimum merge sequence length.
58 */
59 private static final int MIN_MERGE = 32;
60
61 /**
62 * The array being sorted.
63 */
64 private final Object[] a;
65
66 /**
67 * When we get into galloping mode, we stay there until both runs win less
68 * often than MIN_GALLOP consecutive times.
69 */
70 private static final int MIN_GALLOP = 7;
71
72 /**
73 * This controls when we get *into* galloping mode. It is initialized
74 * to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
75 * random data, and lower for highly structured data.
76 */
77 private int minGallop = MIN_GALLOP;
78
79 /**
80 * Maximum initial size of tmp array, which is used for merging. The array
81 * can grow to accommodate demand.
82 *
83 * Unlike Tim's original C version, we do not allocate this much storage
84 * when sorting smaller arrays. This change was required for performance.
85 */
86 private static final int INITIAL_TMP_STORAGE_LENGTH = 256;
87
88 /**
89 * Temp storage for merges.
90 */
91 private Object[] tmp;
92
93 /**
94 * A stack of pending runs yet to be merged. Run i starts at
95 * address base[i] and extends for len[i] elements. It's always
96 * true (so long as the indices are in bounds) that:
97 *
98 * runBase[i] + runLen[i] == runBase[i + 1]
99 *
100 * so we could cut the storage for this, but it's a minor amount,
101 * and keeping all the info explicit simplifies the code.
102 */
103 private int stackSize = 0; // Number of pending runs on stack
104 private final int[] runBase;
105 private final int[] runLen;
106
107 /**
108 * Creates a TimSort instance to maintain the state of an ongoing sort.
109 *
110 * @param a the array to be sorted
111 */
112 private ComparableTimSort(Object[] a) {
113 this.a = a;
114
115 // Allocate temp storage (which may be increased later if necessary)
116 int len = a.length;
117 Object[] newArray = new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ?
118 len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];
119 tmp = newArray;
120
121 /*
122 * Allocate runs-to-be-merged stack (which cannot be expanded). The
123 * stack length requirements are described in listsort.txt. The C
124 * version always uses the same stack length (85), but this was
125 * measured to be too expensive when sorting "mid-sized" arrays (e.g.,
126 * 100 elements) in Java. Therefore, we use smaller (but sufficiently
127 * large) stack lengths for smaller arrays. The "magic numbers" in the
128 * computation below must be changed if MIN_MERGE is decreased. See
129 * the MIN_MERGE declaration above for more information.
130 */
131 int stackLen = (len < 120 ? 5 :
132 len < 1542 ? 10 :
133 len < 119151 ? 19 : 40);
134 runBase = new int[stackLen];
135 runLen = new int[stackLen];
136 }
137
138 /*
139 * The next two methods (which are package private and static) constitute
140 * the entire API of this class. Each of these methods obeys the contract
141 * of the public method with the same signature in java.util.Arrays.
142 */
143
144 static void sort(Object[] a) {
145 sort(a, 0, a.length);
146 }
147
148 static void sort(Object[] a, int lo, int hi) {
149 rangeCheck(a.length, lo, hi);
150 int nRemaining = hi - lo;
151 if (nRemaining < 2)
152 return; // Arrays of size 0 and 1 are always sorted
153
154 // If array is small, do a "mini-TimSort" with no merges
155 if (nRemaining < MIN_MERGE) {
156 int initRunLen = countRunAndMakeAscending(a, lo, hi);
157 binarySort(a, lo, hi, lo + initRunLen);
158 return;
159 }
160
161 /**
162 * March over the array once, left to right, finding natural runs,
163 * extending short natural runs to minRun elements, and merging runs
164 * to maintain stack invariant.
165 */
166 ComparableTimSort ts = new ComparableTimSort(a);
167 int minRun = minRunLength(nRemaining);
168 do {
169 // Identify next run
170 int runLen = countRunAndMakeAscending(a, lo, hi);
171
172 // If run is short, extend to min(minRun, nRemaining)
173 if (runLen < minRun) {
174 int force = nRemaining <= minRun ? nRemaining : minRun;
175 binarySort(a, lo, lo + force, lo + runLen);
176 runLen = force;
177 }
178
179 // Push run onto pending-run stack, and maybe merge
180 ts.pushRun(lo, runLen);
181 ts.mergeCollapse();
182
183 // Advance to find next run
184 lo += runLen;
185 nRemaining -= runLen;
186 } while (nRemaining != 0);
187
188 // Merge all remaining runs to complete sort
189 assert lo == hi;
190 ts.mergeForceCollapse();
191 assert ts.stackSize == 1;
192 }
193
194 /**
195 * Sorts the specified portion of the specified array using a binary
196 * insertion sort. This is the best method for sorting small numbers
197 * of elements. It requires O(n log n) compares, but O(n^2) data
198 * movement (worst case).
199 *
200 * If the initial part of the specified range is already sorted,
201 * this method can take advantage of it: the method assumes that the
202 * elements from index {@code lo}, inclusive, to {@code start},
203 * exclusive are already sorted.
204 *
205 * @param a the array in which a range is to be sorted
206 * @param lo the index of the first element in the range to be sorted
207 * @param hi the index after the last element in the range to be sorted
208 * @param start the index of the first element in the range that is
209 * not already known to be sorted ({@code lo <= start <= hi})
210 */
211 @SuppressWarnings({"fallthrough", "rawtypes", "unchecked"})
212 private static void binarySort(Object[] a, int lo, int hi, int start) {
213 assert lo <= start && start <= hi;
214 if (start == lo)
215 start++;
216 for ( ; start < hi; start++) {
217 Comparable pivot = (Comparable) a[start];
218
219 // Set left (and right) to the index where a[start] (pivot) belongs
220 int left = lo;
221 int right = start;
222 assert left <= right;
223 /*
224 * Invariants:
225 * pivot >= all in [lo, left).
226 * pivot < all in [right, start).
227 */
228 while (left < right) {
229 int mid = (left + right) >>> 1;
230 if (pivot.compareTo(a[mid]) < 0)
231 right = mid;
232 else
233 left = mid + 1;
234 }
235 assert left == right;
236
237 /*
238 * The invariants still hold: pivot >= all in [lo, left) and
239 * pivot < all in [left, start), so pivot belongs at left. Note
240 * that if there are elements equal to pivot, left points to the
241 * first slot after them -- that's why this sort is stable.
242 * Slide elements over to make room for pivot.
243 */
244 int n = start - left; // The number of elements to move
245 // Switch is just an optimization for arraycopy in default case
246 switch (n) {
247 case 2: a[left + 2] = a[left + 1];
248 case 1: a[left + 1] = a[left];
249 break;
250 default: System.arraycopy(a, left, a, left + 1, n);
251 }
252 a[left] = pivot;
253 }
254 }
255
256 /**
257 * Returns the length of the run beginning at the specified position in
258 * the specified array and reverses the run if it is descending (ensuring
259 * that the run will always be ascending when the method returns).
260 *
261 * A run is the longest ascending sequence with:
262 *
263 * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
264 *
265 * or the longest descending sequence with:
266 *
267 * a[lo] > a[lo + 1] > a[lo + 2] > ...
268 *
269 * For its intended use in a stable mergesort, the strictness of the
270 * definition of "descending" is needed so that the call can safely
271 * reverse a descending sequence without violating stability.
272 *
273 * @param a the array in which a run is to be counted and possibly reversed
274 * @param lo index of the first element in the run
275 * @param hi index after the last element that may be contained in the run.
276 It is required that {@code lo < hi}.
277 * @return the length of the run beginning at the specified position in
278 * the specified array
279 */
280 @SuppressWarnings({"unchecked", "rawtypes"})
281 private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
282 assert lo < hi;
283 int runHi = lo + 1;
284 if (runHi == hi)
285 return 1;
286
287 // Find end of run, and reverse range if descending
288 if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending
289 while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0)
290 runHi++;
291 reverseRange(a, lo, runHi);
292 } else { // Ascending
293 while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0)
294 runHi++;
295 }
296
297 return runHi - lo;
298 }
299
300 /**
301 * Reverse the specified range of the specified array.
302 *
303 * @param a the array in which a range is to be reversed
304 * @param lo the index of the first element in the range to be reversed
305 * @param hi the index after the last element in the range to be reversed
306 */
307 private static void reverseRange(Object[] a, int lo, int hi) {
308 hi--;
309 while (lo < hi) {
310 Object t = a[lo];
311 a[lo++] = a[hi];
312 a[hi--] = t;
313 }
314 }
315
316 /**
317 * Returns the minimum acceptable run length for an array of the specified
318 * length. Natural runs shorter than this will be extended with
319 * {@link #binarySort}.
320 *
321 * Roughly speaking, the computation is:
322 *
323 * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
324 * Else if n is an exact power of 2, return MIN_MERGE/2.
325 * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
326 * is close to, but strictly less than, an exact power of 2.
327 *
328 * For the rationale, see listsort.txt.
329 *
330 * @param n the length of the array to be sorted
331 * @return the length of the minimum run to be merged
332 */
333 private static int minRunLength(int n) {
334 assert n >= 0;
335 int r = 0; // Becomes 1 if any 1 bits are shifted off
336 while (n >= MIN_MERGE) {
337 r |= (n & 1);
338 n >>= 1;
339 }
340 return n + r;
341 }
342
343 /**
344 * Pushes the specified run onto the pending-run stack.
345 *
346 * @param runBase index of the first element in the run
347 * @param runLen the number of elements in the run
348 */
349 private void pushRun(int runBase, int runLen) {
350 this.runBase[stackSize] = runBase;
351 this.runLen[stackSize] = runLen;
352 stackSize++;
353 }
354
355 /**
356 * Examines the stack of runs waiting to be merged and merges adjacent runs
357 * until the stack invariants are reestablished:
358 *
359 * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
360 * 2. runLen[i - 2] > runLen[i - 1]
361 *
362 * This method is called each time a new run is pushed onto the stack,
363 * so the invariants are guaranteed to hold for i < stackSize upon
364 * entry to the method.
365 */
366 private void mergeCollapse() {
367 while (stackSize > 1) {
368 int n = stackSize - 2;
369 if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
370 if (runLen[n - 1] < runLen[n + 1])
371 n--;
372 mergeAt(n);
373 } else if (runLen[n] <= runLen[n + 1]) {
374 mergeAt(n);
375 } else {
376 break; // Invariant is established
377 }
378 }
379 }
380
381 /**
382 * Merges all runs on the stack until only one remains. This method is
383 * called once, to complete the sort.
384 */
385 private void mergeForceCollapse() {
386 while (stackSize > 1) {
387 int n = stackSize - 2;
388 if (n > 0 && runLen[n - 1] < runLen[n + 1])
389 n--;
390 mergeAt(n);
391 }
392 }
393
394 /**
395 * Merges the two runs at stack indices i and i+1. Run i must be
396 * the penultimate or antepenultimate run on the stack. In other words,
397 * i must be equal to stackSize-2 or stackSize-3.
398 *
399 * @param i stack index of the first of the two runs to merge
400 */
401 @SuppressWarnings("unchecked")
402 private void mergeAt(int i) {
403 assert stackSize >= 2;
404 assert i >= 0;
405 assert i == stackSize - 2 || i == stackSize - 3;
406
407 int base1 = runBase[i];
408 int len1 = runLen[i];
409 int base2 = runBase[i + 1];
410 int len2 = runLen[i + 1];
411 assert len1 > 0 && len2 > 0;
412 assert base1 + len1 == base2;
413
414 /*
415 * Record the length of the combined runs; if i is the 3rd-last
416 * run now, also slide over the last run (which isn't involved
417 * in this merge). The current run (i+1) goes away in any case.
418 */
419 runLen[i] = len1 + len2;
420 if (i == stackSize - 3) {
421 runBase[i + 1] = runBase[i + 2];
422 runLen[i + 1] = runLen[i + 2];
423 }
424 stackSize--;
425
426 /*
427 * Find where the first element of run2 goes in run1. Prior elements
428 * in run1 can be ignored (because they're already in place).
429 */
430 int k = gallopRight((Comparable<Object>) a[base2], a, base1, len1, 0);
431 assert k >= 0;
432 base1 += k;
433 len1 -= k;
434 if (len1 == 0)
435 return;
436
437 /*
438 * Find where the last element of run1 goes in run2. Subsequent elements
439 * in run2 can be ignored (because they're already in place).
440 */
441 len2 = gallopLeft((Comparable<Object>) a[base1 + len1 - 1], a,
442 base2, len2, len2 - 1);
443 assert len2 >= 0;
444 if (len2 == 0)
445 return;
446
447 // Merge remaining runs, using tmp array with min(len1, len2) elements
448 if (len1 <= len2)
449 mergeLo(base1, len1, base2, len2);
450 else
451 mergeHi(base1, len1, base2, len2);
452 }
453
454 /**
455 * Locates the position at which to insert the specified key into the
456 * specified sorted range; if the range contains an element equal to key,
457 * returns the index of the leftmost equal element.
458 *
459 * @param key the key whose insertion point to search for
460 * @param a the array in which to search
461 * @param base the index of the first element in the range
462 * @param len the length of the range; must be > 0
463 * @param hint the index at which to begin the search, 0 <= hint < n.
464 * The closer hint is to the result, the faster this method will run.
465 * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
466 * pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
467 * In other words, key belongs at index b + k; or in other words,
468 * the first k elements of a should precede key, and the last n - k
469 * should follow it.
470 */
471 private static int gallopLeft(Comparable<Object> key, Object[] a,
472 int base, int len, int hint) {
473 assert len > 0 && hint >= 0 && hint < len;
474
475 int lastOfs = 0;
476 int ofs = 1;
477 if (key.compareTo(a[base + hint]) > 0) {
478 // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
479 int maxOfs = len - hint;
480 while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) > 0) {
481 lastOfs = ofs;
482 ofs = (ofs << 1) + 1;
483 if (ofs <= 0) // int overflow
484 ofs = maxOfs;
485 }
486 if (ofs > maxOfs)
487 ofs = maxOfs;
488
489 // Make offsets relative to base
490 lastOfs += hint;
491 ofs += hint;
492 } else { // key <= a[base + hint]
493 // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
494 final int maxOfs = hint + 1;
495 while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) <= 0) {
496 lastOfs = ofs;
497 ofs = (ofs << 1) + 1;
498 if (ofs <= 0) // int overflow
499 ofs = maxOfs;
500 }
501 if (ofs > maxOfs)
502 ofs = maxOfs;
503
504 // Make offsets relative to base
505 int tmp = lastOfs;
506 lastOfs = hint - ofs;
507 ofs = hint - tmp;
508 }
509 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
510
511 /*
512 * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
513 * to the right of lastOfs but no farther right than ofs. Do a binary
514 * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
515 */
516 lastOfs++;
517 while (lastOfs < ofs) {
518 int m = lastOfs + ((ofs - lastOfs) >>> 1);
519
520 if (key.compareTo(a[base + m]) > 0)
521 lastOfs = m + 1; // a[base + m] < key
522 else
523 ofs = m; // key <= a[base + m]
524 }
525 assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
526 return ofs;
527 }
528
529 /**
530 * Like gallopLeft, except that if the range contains an element equal to
531 * key, gallopRight returns the index after the rightmost equal element.
532 *
533 * @param key the key whose insertion point to search for
534 * @param a the array in which to search
535 * @param base the index of the first element in the range
536 * @param len the length of the range; must be > 0
537 * @param hint the index at which to begin the search, 0 <= hint < n.
538 * The closer hint is to the result, the faster this method will run.
539 * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
540 */
541 private static int gallopRight(Comparable<Object> key, Object[] a,
542 int base, int len, int hint) {
543 assert len > 0 && hint >= 0 && hint < len;
544
545 int ofs = 1;
546 int lastOfs = 0;
547 if (key.compareTo(a[base + hint]) < 0) {
548 // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
549 int maxOfs = hint + 1;
550 while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) < 0) {
551 lastOfs = ofs;
552 ofs = (ofs << 1) + 1;
553 if (ofs <= 0) // int overflow
554 ofs = maxOfs;
555 }
556 if (ofs > maxOfs)
557 ofs = maxOfs;
558
559 // Make offsets relative to b
560 int tmp = lastOfs;
561 lastOfs = hint - ofs;
562 ofs = hint - tmp;
563 } else { // a[b + hint] <= key
564 // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
565 int maxOfs = len - hint;
566 while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) >= 0) {
567 lastOfs = ofs;
568 ofs = (ofs << 1) + 1;
569 if (ofs <= 0) // int overflow
570 ofs = maxOfs;
571 }
572 if (ofs > maxOfs)
573 ofs = maxOfs;
574
575 // Make offsets relative to b
576 lastOfs += hint;
577 ofs += hint;
578 }
579 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
580
581 /*
582 * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
583 * the right of lastOfs but no farther right than ofs. Do a binary
584 * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
585 */
586 lastOfs++;
587 while (lastOfs < ofs) {
588 int m = lastOfs + ((ofs - lastOfs) >>> 1);
589
590 if (key.compareTo(a[base + m]) < 0)
591 ofs = m; // key < a[b + m]
592 else
593 lastOfs = m + 1; // a[b + m] <= key
594 }
595 assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
596 return ofs;
597 }
598
599 /**
600 * Merges two adjacent runs in place, in a stable fashion. The first
601 * element of the first run must be greater than the first element of the
602 * second run (a[base1] > a[base2]), and the last element of the first run
603 * (a[base1 + len1-1]) must be greater than all elements of the second run.
604 *
605 * For performance, this method should be called only when len1 <= len2;
606 * its twin, mergeHi should be called if len1 >= len2. (Either method
607 * may be called if len1 == len2.)
608 *
609 * @param base1 index of first element in first run to be merged
610 * @param len1 length of first run to be merged (must be > 0)
611 * @param base2 index of first element in second run to be merged
612 * (must be aBase + aLen)
613 * @param len2 length of second run to be merged (must be > 0)
614 */
615 @SuppressWarnings({"unchecked", "rawtypes"})
616 private void mergeLo(int base1, int len1, int base2, int len2) {
617 assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
618
619 // Copy first run into temp array
620 Object[] a = this.a; // For performance
621 Object[] tmp = ensureCapacity(len1);
622 System.arraycopy(a, base1, tmp, 0, len1);
623
624 int cursor1 = 0; // Indexes into tmp array
625 int cursor2 = base2; // Indexes int a
626 int dest = base1; // Indexes int a
627
628 // Move first element of second run and deal with degenerate cases
629 a[dest++] = a[cursor2++];
630 if (--len2 == 0) {
631 System.arraycopy(tmp, cursor1, a, dest, len1);
632 return;
633 }
634 if (len1 == 1) {
635 System.arraycopy(a, cursor2, a, dest, len2);
636 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
637 return;
638 }
639
640 int minGallop = this.minGallop; // Use local variable for performance
641 outer:
642 while (true) {
643 int count1 = 0; // Number of times in a row that first run won
644 int count2 = 0; // Number of times in a row that second run won
645
646 /*
647 * Do the straightforward thing until (if ever) one run starts
648 * winning consistently.
649 */
650 do {
651 assert len1 > 1 && len2 > 0;
652 if (((Comparable) a[cursor2]).compareTo(tmp[cursor1]) < 0) {
653 a[dest++] = a[cursor2++];
654 count2++;
655 count1 = 0;
656 if (--len2 == 0)
657 break outer;
658 } else {
659 a[dest++] = tmp[cursor1++];
660 count1++;
661 count2 = 0;
662 if (--len1 == 1)
663 break outer;
664 }
665 } while ((count1 | count2) < minGallop);
666
667 /*
668 * One run is winning so consistently that galloping may be a
669 * huge win. So try that, and continue galloping until (if ever)
670 * neither run appears to be winning consistently anymore.
671 */
672 do {
673 assert len1 > 1 && len2 > 0;
674 count1 = gallopRight((Comparable) a[cursor2], tmp, cursor1, len1, 0);
675 if (count1 != 0) {
676 System.arraycopy(tmp, cursor1, a, dest, count1);
677 dest += count1;
678 cursor1 += count1;
679 len1 -= count1;
680 if (len1 <= 1) // len1 == 1 || len1 == 0
681 break outer;
682 }
683 a[dest++] = a[cursor2++];
684 if (--len2 == 0)
685 break outer;
686
687 count2 = gallopLeft((Comparable) tmp[cursor1], a, cursor2, len2, 0);
688 if (count2 != 0) {
689 System.arraycopy(a, cursor2, a, dest, count2);
690 dest += count2;
691 cursor2 += count2;
692 len2 -= count2;
693 if (len2 == 0)
694 break outer;
695 }
696 a[dest++] = tmp[cursor1++];
697 if (--len1 == 1)
698 break outer;
699 minGallop--;
700 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
701 if (minGallop < 0)
702 minGallop = 0;
703 minGallop += 2; // Penalize for leaving gallop mode
704 } // End of "outer" loop
705 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
706
707 if (len1 == 1) {
708 assert len2 > 0;
709 System.arraycopy(a, cursor2, a, dest, len2);
710 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
711 } else if (len1 == 0) {
712 throw new IllegalArgumentException(
713 "Comparison method violates its general contract!");
714 } else {
715 assert len2 == 0;
716 assert len1 > 1;
717 System.arraycopy(tmp, cursor1, a, dest, len1);
718 }
719 }
720
721 /**
722 * Like mergeLo, except that this method should be called only if
723 * len1 >= len2; mergeLo should be called if len1 <= len2. (Either method
724 * may be called if len1 == len2.)
725 *
726 * @param base1 index of first element in first run to be merged
727 * @param len1 length of first run to be merged (must be > 0)
728 * @param base2 index of first element in second run to be merged
729 * (must be aBase + aLen)
730 * @param len2 length of second run to be merged (must be > 0)
731 */
732 @SuppressWarnings({"unchecked", "rawtypes"})
733 private void mergeHi(int base1, int len1, int base2, int len2) {
734 assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
735
736 // Copy second run into temp array
737 Object[] a = this.a; // For performance
738 Object[] tmp = ensureCapacity(len2);
739 System.arraycopy(a, base2, tmp, 0, len2);
740
741 int cursor1 = base1 + len1 - 1; // Indexes into a
742 int cursor2 = len2 - 1; // Indexes into tmp array
743 int dest = base2 + len2 - 1; // Indexes into a
744
745 // Move last element of first run and deal with degenerate cases
746 a[dest--] = a[cursor1--];
747 if (--len1 == 0) {
748 System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
749 return;
750 }
751 if (len2 == 1) {
752 dest -= len1;
753 cursor1 -= len1;
754 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
755 a[dest] = tmp[cursor2];
756 return;
757 }
758
759 int minGallop = this.minGallop; // Use local variable for performance
760 outer:
761 while (true) {
762 int count1 = 0; // Number of times in a row that first run won
763 int count2 = 0; // Number of times in a row that second run won
764
765 /*
766 * Do the straightforward thing until (if ever) one run
767 * appears to win consistently.
768 */
769 do {
770 assert len1 > 0 && len2 > 1;
771 if (((Comparable) tmp[cursor2]).compareTo(a[cursor1]) < 0) {
772 a[dest--] = a[cursor1--];
773 count1++;
774 count2 = 0;
775 if (--len1 == 0)
776 break outer;
777 } else {
778 a[dest--] = tmp[cursor2--];
779 count2++;
780 count1 = 0;
781 if (--len2 == 1)
782 break outer;
783 }
784 } while ((count1 | count2) < minGallop);
785
786 /*
787 * One run is winning so consistently that galloping may be a
788 * huge win. So try that, and continue galloping until (if ever)
789 * neither run appears to be winning consistently anymore.
790 */
791 do {
792 assert len1 > 0 && len2 > 1;
793 count1 = len1 - gallopRight((Comparable) tmp[cursor2], a, base1, len1, len1 - 1);
794 if (count1 != 0) {
795 dest -= count1;
796 cursor1 -= count1;
797 len1 -= count1;
798 System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
799 if (len1 == 0)
800 break outer;
801 }
802 a[dest--] = tmp[cursor2--];
803 if (--len2 == 1)
804 break outer;
805
806 count2 = len2 - gallopLeft((Comparable) a[cursor1], tmp, 0, len2, len2 - 1);
807 if (count2 != 0) {
808 dest -= count2;
809 cursor2 -= count2;
810 len2 -= count2;
811 System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
812 if (len2 <= 1)
813 break outer; // len2 == 1 || len2 == 0
814 }
815 a[dest--] = a[cursor1--];
816 if (--len1 == 0)
817 break outer;
818 minGallop--;
819 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
820 if (minGallop < 0)
821 minGallop = 0;
822 minGallop += 2; // Penalize for leaving gallop mode
823 } // End of "outer" loop
824 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
825
826 if (len2 == 1) {
827 assert len1 > 0;
828 dest -= len1;
829 cursor1 -= len1;
830 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
831 a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
832 } else if (len2 == 0) {
833 throw new IllegalArgumentException(
834 "Comparison method violates its general contract!");
835 } else {
836 assert len1 == 0;
837 assert len2 > 0;
838 System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
839 }
840 }
841
842 /**
843 * Ensures that the external array tmp has at least the specified
844 * number of elements, increasing its size if necessary. The size
845 * increases exponentially to ensure amortized linear time complexity.
846 *
847 * @param minCapacity the minimum required capacity of the tmp array
848 * @return tmp, whether or not it grew
849 */
850 private Object[] ensureCapacity(int minCapacity) {
851 if (tmp.length < minCapacity) {
852 // Compute smallest power of 2 > minCapacity
853 int newSize = minCapacity;
854 newSize |= newSize >> 1;
855 newSize |= newSize >> 2;
856 newSize |= newSize >> 4;
857 newSize |= newSize >> 8;
858 newSize |= newSize >> 16;
859 newSize++;
860
861 if (newSize < 0) // Not bloody likely!
862 newSize = minCapacity;
863 else
864 newSize = Math.min(newSize, a.length >>> 1);
865
866 Object[] newArray = new Object[newSize];
867 tmp = newArray;
868 }
869 return tmp;
870 }
871
872 /**
873 * Checks that fromIndex and toIndex are in range, and throws an
874 * appropriate exception if they aren't.
875 *
876 * @param arrayLen the length of the array
877 * @param fromIndex the index of the first element of the range
878 * @param toIndex the index after the last element of the range
879 * @throws IllegalArgumentException if fromIndex > toIndex
880 * @throws ArrayIndexOutOfBoundsException if fromIndex < 0
881 * or toIndex > arrayLen
882 */
883 private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) {
884 if (fromIndex > toIndex)
885 throw new IllegalArgumentException("fromIndex(" + fromIndex +
886 ") > toIndex(" + toIndex+")");
887 if (fromIndex < 0)
888 throw new ArrayIndexOutOfBoundsException(fromIndex);
889 if (toIndex > arrayLen)
890 throw new ArrayIndexOutOfBoundsException(toIndex);
891 }
892 }