TimSort.java revision 02f409d949c9d50806809b827ec503765bed34fd
1/* 2 * Copyright (c) 2009, 2013, Oracle and/or its affiliates. All rights reserved. 3 * Copyright 2009 Google Inc. All Rights Reserved. 4 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. 5 * 6 * This code is free software; you can redistribute it and/or modify it 7 * under the terms of the GNU General Public License version 2 only, as 8 * published by the Free Software Foundation. Oracle designates this 9 * particular file as subject to the "Classpath" exception as provided 10 * by Oracle in the LICENSE file that accompanied this code. 11 * 12 * This code is distributed in the hope that it will be useful, but WITHOUT 13 * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or 14 * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License 15 * version 2 for more details (a copy is included in the LICENSE file that 16 * accompanied this code). 17 * 18 * You should have received a copy of the GNU General Public License version 19 * 2 along with this work; if not, write to the Free Software Foundation, 20 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. 21 * 22 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA 23 * or visit www.oracle.com if you need additional information or have any 24 * questions. 25 */ 26 27package java.util; 28 29/** 30 * A stable, adaptive, iterative mergesort that requires far fewer than 31 * n lg(n) comparisons when running on partially sorted arrays, while 32 * offering performance comparable to a traditional mergesort when run 33 * on random arrays. Like all proper mergesorts, this sort is stable and 34 * runs O(n log n) time (worst case). In the worst case, this sort requires 35 * temporary storage space for n/2 object references; in the best case, 36 * it requires only a small constant amount of space. 37 * 38 * This implementation was adapted from Tim Peters's list sort for 39 * Python, which is described in detail here: 40 * 41 * http://svn.python.org/projects/python/trunk/Objects/listsort.txt 42 * 43 * Tim's C code may be found here: 44 * 45 * http://svn.python.org/projects/python/trunk/Objects/listobject.c 46 * 47 * The underlying techniques are described in this paper (and may have 48 * even earlier origins): 49 * 50 * "Optimistic Sorting and Information Theoretic Complexity" 51 * Peter McIlroy 52 * SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms), 53 * pp 467-474, Austin, Texas, 25-27 January 1993. 54 * 55 * While the API to this class consists solely of static methods, it is 56 * (privately) instantiable; a TimSort instance holds the state of an ongoing 57 * sort, assuming the input array is large enough to warrant the full-blown 58 * TimSort. Small arrays are sorted in place, using a binary insertion sort. 59 * 60 * @author Josh Bloch 61 */ 62class TimSort<T> { 63 /** 64 * This is the minimum sized sequence that will be merged. Shorter 65 * sequences will be lengthened by calling binarySort. If the entire 66 * array is less than this length, no merges will be performed. 67 * 68 * This constant should be a power of two. It was 64 in Tim Peter's C 69 * implementation, but 32 was empirically determined to work better in 70 * this implementation. In the unlikely event that you set this constant 71 * to be a number that's not a power of two, you'll need to change the 72 * {@link #minRunLength} computation. 73 * 74 * If you decrease this constant, you must change the stackLen 75 * computation in the TimSort constructor, or you risk an 76 * ArrayOutOfBounds exception. See listsort.txt for a discussion 77 * of the minimum stack length required as a function of the length 78 * of the array being sorted and the minimum merge sequence length. 79 */ 80 private static final int MIN_MERGE = 32; 81 82 /** 83 * The array being sorted. 84 */ 85 private final T[] a; 86 87 /** 88 * The comparator for this sort. 89 */ 90 private final Comparator<? super T> c; 91 92 /** 93 * When we get into galloping mode, we stay there until both runs win less 94 * often than MIN_GALLOP consecutive times. 95 */ 96 private static final int MIN_GALLOP = 7; 97 98 /** 99 * This controls when we get *into* galloping mode. It is initialized 100 * to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for 101 * random data, and lower for highly structured data. 102 */ 103 private int minGallop = MIN_GALLOP; 104 105 /** 106 * Maximum initial size of tmp array, which is used for merging. The array 107 * can grow to accommodate demand. 108 * 109 * Unlike Tim's original C version, we do not allocate this much storage 110 * when sorting smaller arrays. This change was required for performance. 111 */ 112 private static final int INITIAL_TMP_STORAGE_LENGTH = 256; 113 114 /** 115 * Temp storage for merges. 116 */ 117 private T[] tmp; // Actual runtime type will be Object[], regardless of T 118 119 /** 120 * A stack of pending runs yet to be merged. Run i starts at 121 * address base[i] and extends for len[i] elements. It's always 122 * true (so long as the indices are in bounds) that: 123 * 124 * runBase[i] + runLen[i] == runBase[i + 1] 125 * 126 * so we could cut the storage for this, but it's a minor amount, 127 * and keeping all the info explicit simplifies the code. 128 */ 129 private int stackSize = 0; // Number of pending runs on stack 130 private final int[] runBase; 131 private final int[] runLen; 132 133 /** 134 * Creates a TimSort instance to maintain the state of an ongoing sort. 135 * 136 * @param a the array to be sorted 137 * @param c the comparator to determine the order of the sort 138 */ 139 private TimSort(T[] a, Comparator<? super T> c) { 140 this.a = a; 141 this.c = c; 142 143 // Allocate temp storage (which may be increased later if necessary) 144 int len = a.length; 145 @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) 146 T[] newArray = (T[]) new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ? 147 len >>> 1 : INITIAL_TMP_STORAGE_LENGTH]; 148 tmp = newArray; 149 150 /* 151 * Allocate runs-to-be-merged stack (which cannot be expanded). The 152 * stack length requirements are described in listsort.txt. The C 153 * version always uses the same stack length (85), but this was 154 * measured to be too expensive when sorting "mid-sized" arrays (e.g., 155 * 100 elements) in Java. Therefore, we use smaller (but sufficiently 156 * large) stack lengths for smaller arrays. The "magic numbers" in the 157 * computation below must be changed if MIN_MERGE is decreased. See 158 * the MIN_MERGE declaration above for more information. 159 */ 160 int stackLen = (len < 120 ? 5 : 161 len < 1542 ? 10 : 162 len < 119151 ? 24 : 40); 163 runBase = new int[stackLen]; 164 runLen = new int[stackLen]; 165 } 166 167 /* 168 * The next two methods (which are package private and static) constitute 169 * the entire API of this class. Each of these methods obeys the contract 170 * of the public method with the same signature in java.util.Arrays. 171 */ 172 173 static <T> void sort(T[] a, Comparator<? super T> c) { 174 sort(a, 0, a.length, c); 175 } 176 177 static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c) { 178 if (c == null) { 179 Arrays.sort(a, lo, hi); 180 return; 181 } 182 183 rangeCheck(a.length, lo, hi); 184 int nRemaining = hi - lo; 185 if (nRemaining < 2) 186 return; // Arrays of size 0 and 1 are always sorted 187 188 // If array is small, do a "mini-TimSort" with no merges 189 if (nRemaining < MIN_MERGE) { 190 int initRunLen = countRunAndMakeAscending(a, lo, hi, c); 191 binarySort(a, lo, hi, lo + initRunLen, c); 192 return; 193 } 194 195 /** 196 * March over the array once, left to right, finding natural runs, 197 * extending short natural runs to minRun elements, and merging runs 198 * to maintain stack invariant. 199 */ 200 TimSort<T> ts = new TimSort<>(a, c); 201 int minRun = minRunLength(nRemaining); 202 do { 203 // Identify next run 204 int runLen = countRunAndMakeAscending(a, lo, hi, c); 205 206 // If run is short, extend to min(minRun, nRemaining) 207 if (runLen < minRun) { 208 int force = nRemaining <= minRun ? nRemaining : minRun; 209 binarySort(a, lo, lo + force, lo + runLen, c); 210 runLen = force; 211 } 212 213 // Push run onto pending-run stack, and maybe merge 214 ts.pushRun(lo, runLen); 215 ts.mergeCollapse(); 216 217 // Advance to find next run 218 lo += runLen; 219 nRemaining -= runLen; 220 } while (nRemaining != 0); 221 222 // Merge all remaining runs to complete sort 223 assert lo == hi; 224 ts.mergeForceCollapse(); 225 assert ts.stackSize == 1; 226 } 227 228 /** 229 * Sorts the specified portion of the specified array using a binary 230 * insertion sort. This is the best method for sorting small numbers 231 * of elements. It requires O(n log n) compares, but O(n^2) data 232 * movement (worst case). 233 * 234 * If the initial part of the specified range is already sorted, 235 * this method can take advantage of it: the method assumes that the 236 * elements from index {@code lo}, inclusive, to {@code start}, 237 * exclusive are already sorted. 238 * 239 * @param a the array in which a range is to be sorted 240 * @param lo the index of the first element in the range to be sorted 241 * @param hi the index after the last element in the range to be sorted 242 * @param start the index of the first element in the range that is 243 * not already known to be sorted ({@code lo <= start <= hi}) 244 * @param c comparator to used for the sort 245 */ 246 @SuppressWarnings("fallthrough") 247 private static <T> void binarySort(T[] a, int lo, int hi, int start, 248 Comparator<? super T> c) { 249 assert lo <= start && start <= hi; 250 if (start == lo) 251 start++; 252 for ( ; start < hi; start++) { 253 T pivot = a[start]; 254 255 // Set left (and right) to the index where a[start] (pivot) belongs 256 int left = lo; 257 int right = start; 258 assert left <= right; 259 /* 260 * Invariants: 261 * pivot >= all in [lo, left). 262 * pivot < all in [right, start). 263 */ 264 while (left < right) { 265 int mid = (left + right) >>> 1; 266 if (c.compare(pivot, a[mid]) < 0) 267 right = mid; 268 else 269 left = mid + 1; 270 } 271 assert left == right; 272 273 /* 274 * The invariants still hold: pivot >= all in [lo, left) and 275 * pivot < all in [left, start), so pivot belongs at left. Note 276 * that if there are elements equal to pivot, left points to the 277 * first slot after them -- that's why this sort is stable. 278 * Slide elements over to make room for pivot. 279 */ 280 int n = start - left; // The number of elements to move 281 // Switch is just an optimization for arraycopy in default case 282 switch (n) { 283 case 2: a[left + 2] = a[left + 1]; 284 case 1: a[left + 1] = a[left]; 285 break; 286 default: System.arraycopy(a, left, a, left + 1, n); 287 } 288 a[left] = pivot; 289 } 290 } 291 292 /** 293 * Returns the length of the run beginning at the specified position in 294 * the specified array and reverses the run if it is descending (ensuring 295 * that the run will always be ascending when the method returns). 296 * 297 * A run is the longest ascending sequence with: 298 * 299 * a[lo] <= a[lo + 1] <= a[lo + 2] <= ... 300 * 301 * or the longest descending sequence with: 302 * 303 * a[lo] > a[lo + 1] > a[lo + 2] > ... 304 * 305 * For its intended use in a stable mergesort, the strictness of the 306 * definition of "descending" is needed so that the call can safely 307 * reverse a descending sequence without violating stability. 308 * 309 * @param a the array in which a run is to be counted and possibly reversed 310 * @param lo index of the first element in the run 311 * @param hi index after the last element that may be contained in the run. 312 It is required that {@code lo < hi}. 313 * @param c the comparator to used for the sort 314 * @return the length of the run beginning at the specified position in 315 * the specified array 316 */ 317 private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi, 318 Comparator<? super T> c) { 319 assert lo < hi; 320 int runHi = lo + 1; 321 if (runHi == hi) 322 return 1; 323 324 // Find end of run, and reverse range if descending 325 if (c.compare(a[runHi++], a[lo]) < 0) { // Descending 326 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0) 327 runHi++; 328 reverseRange(a, lo, runHi); 329 } else { // Ascending 330 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0) 331 runHi++; 332 } 333 334 return runHi - lo; 335 } 336 337 /** 338 * Reverse the specified range of the specified array. 339 * 340 * @param a the array in which a range is to be reversed 341 * @param lo the index of the first element in the range to be reversed 342 * @param hi the index after the last element in the range to be reversed 343 */ 344 private static void reverseRange(Object[] a, int lo, int hi) { 345 hi--; 346 while (lo < hi) { 347 Object t = a[lo]; 348 a[lo++] = a[hi]; 349 a[hi--] = t; 350 } 351 } 352 353 /** 354 * Returns the minimum acceptable run length for an array of the specified 355 * length. Natural runs shorter than this will be extended with 356 * {@link #binarySort}. 357 * 358 * Roughly speaking, the computation is: 359 * 360 * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff). 361 * Else if n is an exact power of 2, return MIN_MERGE/2. 362 * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k 363 * is close to, but strictly less than, an exact power of 2. 364 * 365 * For the rationale, see listsort.txt. 366 * 367 * @param n the length of the array to be sorted 368 * @return the length of the minimum run to be merged 369 */ 370 private static int minRunLength(int n) { 371 assert n >= 0; 372 int r = 0; // Becomes 1 if any 1 bits are shifted off 373 while (n >= MIN_MERGE) { 374 r |= (n & 1); 375 n >>= 1; 376 } 377 return n + r; 378 } 379 380 /** 381 * Pushes the specified run onto the pending-run stack. 382 * 383 * @param runBase index of the first element in the run 384 * @param runLen the number of elements in the run 385 */ 386 private void pushRun(int runBase, int runLen) { 387 this.runBase[stackSize] = runBase; 388 this.runLen[stackSize] = runLen; 389 stackSize++; 390 } 391 392 /** 393 * Examines the stack of runs waiting to be merged and merges adjacent runs 394 * until the stack invariants are reestablished: 395 * 396 * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] 397 * 2. runLen[i - 2] > runLen[i - 1] 398 * 399 * This method is called each time a new run is pushed onto the stack, 400 * so the invariants are guaranteed to hold for i < stackSize upon 401 * entry to the method. 402 */ 403 private void mergeCollapse() { 404 while (stackSize > 1) { 405 int n = stackSize - 2; 406 if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) { 407 if (runLen[n - 1] < runLen[n + 1]) 408 n--; 409 mergeAt(n); 410 } else if (runLen[n] <= runLen[n + 1]) { 411 mergeAt(n); 412 } else { 413 break; // Invariant is established 414 } 415 } 416 } 417 418 /** 419 * Merges all runs on the stack until only one remains. This method is 420 * called once, to complete the sort. 421 */ 422 private void mergeForceCollapse() { 423 while (stackSize > 1) { 424 int n = stackSize - 2; 425 if (n > 0 && runLen[n - 1] < runLen[n + 1]) 426 n--; 427 mergeAt(n); 428 } 429 } 430 431 /** 432 * Merges the two runs at stack indices i and i+1. Run i must be 433 * the penultimate or antepenultimate run on the stack. In other words, 434 * i must be equal to stackSize-2 or stackSize-3. 435 * 436 * @param i stack index of the first of the two runs to merge 437 */ 438 private void mergeAt(int i) { 439 assert stackSize >= 2; 440 assert i >= 0; 441 assert i == stackSize - 2 || i == stackSize - 3; 442 443 int base1 = runBase[i]; 444 int len1 = runLen[i]; 445 int base2 = runBase[i + 1]; 446 int len2 = runLen[i + 1]; 447 assert len1 > 0 && len2 > 0; 448 assert base1 + len1 == base2; 449 450 /* 451 * Record the length of the combined runs; if i is the 3rd-last 452 * run now, also slide over the last run (which isn't involved 453 * in this merge). The current run (i+1) goes away in any case. 454 */ 455 runLen[i] = len1 + len2; 456 if (i == stackSize - 3) { 457 runBase[i + 1] = runBase[i + 2]; 458 runLen[i + 1] = runLen[i + 2]; 459 } 460 stackSize--; 461 462 /* 463 * Find where the first element of run2 goes in run1. Prior elements 464 * in run1 can be ignored (because they're already in place). 465 */ 466 int k = gallopRight(a[base2], a, base1, len1, 0, c); 467 assert k >= 0; 468 base1 += k; 469 len1 -= k; 470 if (len1 == 0) 471 return; 472 473 /* 474 * Find where the last element of run1 goes in run2. Subsequent elements 475 * in run2 can be ignored (because they're already in place). 476 */ 477 len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c); 478 assert len2 >= 0; 479 if (len2 == 0) 480 return; 481 482 // Merge remaining runs, using tmp array with min(len1, len2) elements 483 if (len1 <= len2) 484 mergeLo(base1, len1, base2, len2); 485 else 486 mergeHi(base1, len1, base2, len2); 487 } 488 489 /** 490 * Locates the position at which to insert the specified key into the 491 * specified sorted range; if the range contains an element equal to key, 492 * returns the index of the leftmost equal element. 493 * 494 * @param key the key whose insertion point to search for 495 * @param a the array in which to search 496 * @param base the index of the first element in the range 497 * @param len the length of the range; must be > 0 498 * @param hint the index at which to begin the search, 0 <= hint < n. 499 * The closer hint is to the result, the faster this method will run. 500 * @param c the comparator used to order the range, and to search 501 * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k], 502 * pretending that a[b - 1] is minus infinity and a[b + n] is infinity. 503 * In other words, key belongs at index b + k; or in other words, 504 * the first k elements of a should precede key, and the last n - k 505 * should follow it. 506 */ 507 private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint, 508 Comparator<? super T> c) { 509 assert len > 0 && hint >= 0 && hint < len; 510 int lastOfs = 0; 511 int ofs = 1; 512 if (c.compare(key, a[base + hint]) > 0) { 513 // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs] 514 int maxOfs = len - hint; 515 while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) { 516 lastOfs = ofs; 517 ofs = (ofs << 1) + 1; 518 if (ofs <= 0) // int overflow 519 ofs = maxOfs; 520 } 521 if (ofs > maxOfs) 522 ofs = maxOfs; 523 524 // Make offsets relative to base 525 lastOfs += hint; 526 ofs += hint; 527 } else { // key <= a[base + hint] 528 // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs] 529 final int maxOfs = hint + 1; 530 while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) { 531 lastOfs = ofs; 532 ofs = (ofs << 1) + 1; 533 if (ofs <= 0) // int overflow 534 ofs = maxOfs; 535 } 536 if (ofs > maxOfs) 537 ofs = maxOfs; 538 539 // Make offsets relative to base 540 int tmp = lastOfs; 541 lastOfs = hint - ofs; 542 ofs = hint - tmp; 543 } 544 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; 545 546 /* 547 * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere 548 * to the right of lastOfs but no farther right than ofs. Do a binary 549 * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs]. 550 */ 551 lastOfs++; 552 while (lastOfs < ofs) { 553 int m = lastOfs + ((ofs - lastOfs) >>> 1); 554 555 if (c.compare(key, a[base + m]) > 0) 556 lastOfs = m + 1; // a[base + m] < key 557 else 558 ofs = m; // key <= a[base + m] 559 } 560 assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs] 561 return ofs; 562 } 563 564 /** 565 * Like gallopLeft, except that if the range contains an element equal to 566 * key, gallopRight returns the index after the rightmost equal element. 567 * 568 * @param key the key whose insertion point to search for 569 * @param a the array in which to search 570 * @param base the index of the first element in the range 571 * @param len the length of the range; must be > 0 572 * @param hint the index at which to begin the search, 0 <= hint < n. 573 * The closer hint is to the result, the faster this method will run. 574 * @param c the comparator used to order the range, and to search 575 * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k] 576 */ 577 private static <T> int gallopRight(T key, T[] a, int base, int len, 578 int hint, Comparator<? super T> c) { 579 assert len > 0 && hint >= 0 && hint < len; 580 581 int ofs = 1; 582 int lastOfs = 0; 583 if (c.compare(key, a[base + hint]) < 0) { 584 // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs] 585 int maxOfs = hint + 1; 586 while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) { 587 lastOfs = ofs; 588 ofs = (ofs << 1) + 1; 589 if (ofs <= 0) // int overflow 590 ofs = maxOfs; 591 } 592 if (ofs > maxOfs) 593 ofs = maxOfs; 594 595 // Make offsets relative to b 596 int tmp = lastOfs; 597 lastOfs = hint - ofs; 598 ofs = hint - tmp; 599 } else { // a[b + hint] <= key 600 // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs] 601 int maxOfs = len - hint; 602 while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) { 603 lastOfs = ofs; 604 ofs = (ofs << 1) + 1; 605 if (ofs <= 0) // int overflow 606 ofs = maxOfs; 607 } 608 if (ofs > maxOfs) 609 ofs = maxOfs; 610 611 // Make offsets relative to b 612 lastOfs += hint; 613 ofs += hint; 614 } 615 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; 616 617 /* 618 * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to 619 * the right of lastOfs but no farther right than ofs. Do a binary 620 * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs]. 621 */ 622 lastOfs++; 623 while (lastOfs < ofs) { 624 int m = lastOfs + ((ofs - lastOfs) >>> 1); 625 626 if (c.compare(key, a[base + m]) < 0) 627 ofs = m; // key < a[b + m] 628 else 629 lastOfs = m + 1; // a[b + m] <= key 630 } 631 assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs] 632 return ofs; 633 } 634 635 /** 636 * Merges two adjacent runs in place, in a stable fashion. The first 637 * element of the first run must be greater than the first element of the 638 * second run (a[base1] > a[base2]), and the last element of the first run 639 * (a[base1 + len1-1]) must be greater than all elements of the second run. 640 * 641 * For performance, this method should be called only when len1 <= len2; 642 * its twin, mergeHi should be called if len1 >= len2. (Either method 643 * may be called if len1 == len2.) 644 * 645 * @param base1 index of first element in first run to be merged 646 * @param len1 length of first run to be merged (must be > 0) 647 * @param base2 index of first element in second run to be merged 648 * (must be aBase + aLen) 649 * @param len2 length of second run to be merged (must be > 0) 650 */ 651 private void mergeLo(int base1, int len1, int base2, int len2) { 652 assert len1 > 0 && len2 > 0 && base1 + len1 == base2; 653 654 // Copy first run into temp array 655 T[] a = this.a; // For performance 656 T[] tmp = ensureCapacity(len1); 657 System.arraycopy(a, base1, tmp, 0, len1); 658 659 int cursor1 = 0; // Indexes into tmp array 660 int cursor2 = base2; // Indexes int a 661 int dest = base1; // Indexes int a 662 663 // Move first element of second run and deal with degenerate cases 664 a[dest++] = a[cursor2++]; 665 if (--len2 == 0) { 666 System.arraycopy(tmp, cursor1, a, dest, len1); 667 return; 668 } 669 if (len1 == 1) { 670 System.arraycopy(a, cursor2, a, dest, len2); 671 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge 672 return; 673 } 674 675 Comparator<? super T> c = this.c; // Use local variable for performance 676 int minGallop = this.minGallop; // " " " " " 677 outer: 678 while (true) { 679 int count1 = 0; // Number of times in a row that first run won 680 int count2 = 0; // Number of times in a row that second run won 681 682 /* 683 * Do the straightforward thing until (if ever) one run starts 684 * winning consistently. 685 */ 686 do { 687 assert len1 > 1 && len2 > 0; 688 if (c.compare(a[cursor2], tmp[cursor1]) < 0) { 689 a[dest++] = a[cursor2++]; 690 count2++; 691 count1 = 0; 692 if (--len2 == 0) 693 break outer; 694 } else { 695 a[dest++] = tmp[cursor1++]; 696 count1++; 697 count2 = 0; 698 if (--len1 == 1) 699 break outer; 700 } 701 } while ((count1 | count2) < minGallop); 702 703 /* 704 * One run is winning so consistently that galloping may be a 705 * huge win. So try that, and continue galloping until (if ever) 706 * neither run appears to be winning consistently anymore. 707 */ 708 do { 709 assert len1 > 1 && len2 > 0; 710 count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c); 711 if (count1 != 0) { 712 System.arraycopy(tmp, cursor1, a, dest, count1); 713 dest += count1; 714 cursor1 += count1; 715 len1 -= count1; 716 if (len1 <= 1) // len1 == 1 || len1 == 0 717 break outer; 718 } 719 a[dest++] = a[cursor2++]; 720 if (--len2 == 0) 721 break outer; 722 723 count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c); 724 if (count2 != 0) { 725 System.arraycopy(a, cursor2, a, dest, count2); 726 dest += count2; 727 cursor2 += count2; 728 len2 -= count2; 729 if (len2 == 0) 730 break outer; 731 } 732 a[dest++] = tmp[cursor1++]; 733 if (--len1 == 1) 734 break outer; 735 minGallop--; 736 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); 737 if (minGallop < 0) 738 minGallop = 0; 739 minGallop += 2; // Penalize for leaving gallop mode 740 } // End of "outer" loop 741 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field 742 743 if (len1 == 1) { 744 assert len2 > 0; 745 System.arraycopy(a, cursor2, a, dest, len2); 746 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge 747 } else if (len1 == 0) { 748 throw new IllegalArgumentException( 749 "Comparison method violates its general contract!"); 750 } else { 751 assert len2 == 0; 752 assert len1 > 1; 753 System.arraycopy(tmp, cursor1, a, dest, len1); 754 } 755 } 756 757 /** 758 * Like mergeLo, except that this method should be called only if 759 * len1 >= len2; mergeLo should be called if len1 <= len2. (Either method 760 * may be called if len1 == len2.) 761 * 762 * @param base1 index of first element in first run to be merged 763 * @param len1 length of first run to be merged (must be > 0) 764 * @param base2 index of first element in second run to be merged 765 * (must be aBase + aLen) 766 * @param len2 length of second run to be merged (must be > 0) 767 */ 768 private void mergeHi(int base1, int len1, int base2, int len2) { 769 assert len1 > 0 && len2 > 0 && base1 + len1 == base2; 770 771 // Copy second run into temp array 772 T[] a = this.a; // For performance 773 T[] tmp = ensureCapacity(len2); 774 System.arraycopy(a, base2, tmp, 0, len2); 775 776 int cursor1 = base1 + len1 - 1; // Indexes into a 777 int cursor2 = len2 - 1; // Indexes into tmp array 778 int dest = base2 + len2 - 1; // Indexes into a 779 780 // Move last element of first run and deal with degenerate cases 781 a[dest--] = a[cursor1--]; 782 if (--len1 == 0) { 783 System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2); 784 return; 785 } 786 if (len2 == 1) { 787 dest -= len1; 788 cursor1 -= len1; 789 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); 790 a[dest] = tmp[cursor2]; 791 return; 792 } 793 794 Comparator<? super T> c = this.c; // Use local variable for performance 795 int minGallop = this.minGallop; // " " " " " 796 outer: 797 while (true) { 798 int count1 = 0; // Number of times in a row that first run won 799 int count2 = 0; // Number of times in a row that second run won 800 801 /* 802 * Do the straightforward thing until (if ever) one run 803 * appears to win consistently. 804 */ 805 do { 806 assert len1 > 0 && len2 > 1; 807 if (c.compare(tmp[cursor2], a[cursor1]) < 0) { 808 a[dest--] = a[cursor1--]; 809 count1++; 810 count2 = 0; 811 if (--len1 == 0) 812 break outer; 813 } else { 814 a[dest--] = tmp[cursor2--]; 815 count2++; 816 count1 = 0; 817 if (--len2 == 1) 818 break outer; 819 } 820 } while ((count1 | count2) < minGallop); 821 822 /* 823 * One run is winning so consistently that galloping may be a 824 * huge win. So try that, and continue galloping until (if ever) 825 * neither run appears to be winning consistently anymore. 826 */ 827 do { 828 assert len1 > 0 && len2 > 1; 829 count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c); 830 if (count1 != 0) { 831 dest -= count1; 832 cursor1 -= count1; 833 len1 -= count1; 834 System.arraycopy(a, cursor1 + 1, a, dest + 1, count1); 835 if (len1 == 0) 836 break outer; 837 } 838 a[dest--] = tmp[cursor2--]; 839 if (--len2 == 1) 840 break outer; 841 842 count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c); 843 if (count2 != 0) { 844 dest -= count2; 845 cursor2 -= count2; 846 len2 -= count2; 847 System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2); 848 if (len2 <= 1) // len2 == 1 || len2 == 0 849 break outer; 850 } 851 a[dest--] = a[cursor1--]; 852 if (--len1 == 0) 853 break outer; 854 minGallop--; 855 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); 856 if (minGallop < 0) 857 minGallop = 0; 858 minGallop += 2; // Penalize for leaving gallop mode 859 } // End of "outer" loop 860 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field 861 862 if (len2 == 1) { 863 assert len1 > 0; 864 dest -= len1; 865 cursor1 -= len1; 866 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); 867 a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge 868 } else if (len2 == 0) { 869 throw new IllegalArgumentException( 870 "Comparison method violates its general contract!"); 871 } else { 872 assert len1 == 0; 873 assert len2 > 0; 874 System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2); 875 } 876 } 877 878 /** 879 * Ensures that the external array tmp has at least the specified 880 * number of elements, increasing its size if necessary. The size 881 * increases exponentially to ensure amortized linear time complexity. 882 * 883 * @param minCapacity the minimum required capacity of the tmp array 884 * @return tmp, whether or not it grew 885 */ 886 private T[] ensureCapacity(int minCapacity) { 887 if (tmp.length < minCapacity) { 888 // Compute smallest power of 2 > minCapacity 889 int newSize = minCapacity; 890 newSize |= newSize >> 1; 891 newSize |= newSize >> 2; 892 newSize |= newSize >> 4; 893 newSize |= newSize >> 8; 894 newSize |= newSize >> 16; 895 newSize++; 896 897 if (newSize < 0) // Not bloody likely! 898 newSize = minCapacity; 899 else 900 newSize = Math.min(newSize, a.length >>> 1); 901 902 @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) 903 T[] newArray = (T[]) new Object[newSize]; 904 tmp = newArray; 905 } 906 return tmp; 907 } 908 909 /** 910 * Checks that fromIndex and toIndex are in range, and throws an 911 * appropriate exception if they aren't. 912 * 913 * @param arrayLen the length of the array 914 * @param fromIndex the index of the first element of the range 915 * @param toIndex the index after the last element of the range 916 * @throws IllegalArgumentException if fromIndex > toIndex 917 * @throws ArrayIndexOutOfBoundsException if fromIndex < 0 918 * or toIndex > arrayLen 919 */ 920 private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) { 921 if (fromIndex > toIndex) 922 throw new IllegalArgumentException("fromIndex(" + fromIndex + 923 ") > toIndex(" + toIndex+")"); 924 if (fromIndex < 0) 925 throw new ArrayIndexOutOfBoundsException(fromIndex); 926 if (toIndex > arrayLen) 927 throw new ArrayIndexOutOfBoundsException(toIndex); 928 } 929} 930