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 27 package 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 */ 62 class 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. A workspace array may optionally be 116 * provided in constructor, and if so will be used as long as it 117 * is big enough. 118 */ 119 private T[] tmp; 120 private int tmpBase; // base of tmp array slice 121 private int tmpLen; // length of tmp array slice 122 123 /** 124 * A stack of pending runs yet to be merged. Run i starts at 125 * address base[i] and extends for len[i] elements. It's always 126 * true (so long as the indices are in bounds) that: 127 * 128 * runBase[i] + runLen[i] == runBase[i + 1] 129 * 130 * so we could cut the storage for this, but it's a minor amount, 131 * and keeping all the info explicit simplifies the code. 132 */ 133 private int stackSize = 0; // Number of pending runs on stack 134 private final int[] runBase; 135 private final int[] runLen; 136 137 /** 138 * Creates a TimSort instance to maintain the state of an ongoing sort. 139 * 140 * @param a the array to be sorted 141 * @param c the comparator to determine the order of the sort 142 * @param work a workspace array (slice) 143 * @param workBase origin of usable space in work array 144 * @param workLen usable size of work array 145 */ TimSort(T[] a, Comparator<? super T> c, T[] work, int workBase, int workLen)146 private TimSort(T[] a, Comparator<? super T> c, T[] work, int workBase, int workLen) { 147 this.a = a; 148 this.c = c; 149 150 // Allocate temp storage (which may be increased later if necessary) 151 int len = a.length; 152 int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ? 153 len >>> 1 : INITIAL_TMP_STORAGE_LENGTH; 154 if (work == null || workLen < tlen || workBase + tlen > work.length) { 155 @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) 156 T[] newArray = (T[])java.lang.reflect.Array.newInstance 157 (a.getClass().getComponentType(), tlen); 158 tmp = newArray; 159 tmpBase = 0; 160 tmpLen = tlen; 161 } 162 else { 163 tmp = work; 164 tmpBase = workBase; 165 tmpLen = workLen; 166 } 167 168 /* 169 * Allocate runs-to-be-merged stack (which cannot be expanded). The 170 * stack length requirements are described in listsort.txt. The C 171 * version always uses the same stack length (85), but this was 172 * measured to be too expensive when sorting "mid-sized" arrays (e.g., 173 * 100 elements) in Java. Therefore, we use smaller (but sufficiently 174 * large) stack lengths for smaller arrays. The "magic numbers" in the 175 * computation below must be changed if MIN_MERGE is decreased. See 176 * the MIN_MERGE declaration above for more information. 177 */ 178 int stackLen = (len < 120 ? 5 : 179 len < 1542 ? 10 : 180 len < 119151 ? 24 : 40); 181 runBase = new int[stackLen]; 182 runLen = new int[stackLen]; 183 } 184 185 /* 186 * The next method (package private and static) constitutes the 187 * entire API of this class. 188 */ 189 190 /** 191 * Sorts the given range, using the given workspace array slice 192 * for temp storage when possible. This method is designed to be 193 * invoked from public methods (in class Arrays) after performing 194 * any necessary array bounds checks and expanding parameters into 195 * the required forms. 196 * 197 * @param a the array to be sorted 198 * @param lo the index of the first element, inclusive, to be sorted 199 * @param hi the index of the last element, exclusive, to be sorted 200 * @param c the comparator to use 201 * @param work a workspace array (slice) 202 * @param workBase origin of usable space in work array 203 * @param workLen usable size of work array 204 * @since 1.8 205 */ sort(T[] a, int lo, int hi, Comparator<? super T> c, T[] work, int workBase, int workLen)206 static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c, 207 T[] work, int workBase, int workLen) { 208 assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length; 209 210 int nRemaining = hi - lo; 211 if (nRemaining < 2) 212 return; // Arrays of size 0 and 1 are always sorted 213 214 // If array is small, do a "mini-TimSort" with no merges 215 if (nRemaining < MIN_MERGE) { 216 int initRunLen = countRunAndMakeAscending(a, lo, hi, c); 217 binarySort(a, lo, hi, lo + initRunLen, c); 218 return; 219 } 220 221 /** 222 * March over the array once, left to right, finding natural runs, 223 * extending short natural runs to minRun elements, and merging runs 224 * to maintain stack invariant. 225 */ 226 TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen); 227 int minRun = minRunLength(nRemaining); 228 do { 229 // Identify next run 230 int runLen = countRunAndMakeAscending(a, lo, hi, c); 231 232 // If run is short, extend to min(minRun, nRemaining) 233 if (runLen < minRun) { 234 int force = nRemaining <= minRun ? nRemaining : minRun; 235 binarySort(a, lo, lo + force, lo + runLen, c); 236 runLen = force; 237 } 238 239 // Push run onto pending-run stack, and maybe merge 240 ts.pushRun(lo, runLen); 241 ts.mergeCollapse(); 242 243 // Advance to find next run 244 lo += runLen; 245 nRemaining -= runLen; 246 } while (nRemaining != 0); 247 248 // Merge all remaining runs to complete sort 249 assert lo == hi; 250 ts.mergeForceCollapse(); 251 assert ts.stackSize == 1; 252 } 253 254 /** 255 * Sorts the specified portion of the specified array using a binary 256 * insertion sort. This is the best method for sorting small numbers 257 * of elements. It requires O(n log n) compares, but O(n^2) data 258 * movement (worst case). 259 * 260 * If the initial part of the specified range is already sorted, 261 * this method can take advantage of it: the method assumes that the 262 * elements from index {@code lo}, inclusive, to {@code start}, 263 * exclusive are already sorted. 264 * 265 * @param a the array in which a range is to be sorted 266 * @param lo the index of the first element in the range to be sorted 267 * @param hi the index after the last element in the range to be sorted 268 * @param start the index of the first element in the range that is 269 * not already known to be sorted ({@code lo <= start <= hi}) 270 * @param c comparator to used for the sort 271 */ 272 @SuppressWarnings("fallthrough") binarySort(T[] a, int lo, int hi, int start, Comparator<? super T> c)273 private static <T> void binarySort(T[] a, int lo, int hi, int start, 274 Comparator<? super T> c) { 275 assert lo <= start && start <= hi; 276 if (start == lo) 277 start++; 278 for ( ; start < hi; start++) { 279 T pivot = a[start]; 280 281 // Set left (and right) to the index where a[start] (pivot) belongs 282 int left = lo; 283 int right = start; 284 assert left <= right; 285 /* 286 * Invariants: 287 * pivot >= all in [lo, left). 288 * pivot < all in [right, start). 289 */ 290 while (left < right) { 291 int mid = (left + right) >>> 1; 292 if (c.compare(pivot, a[mid]) < 0) 293 right = mid; 294 else 295 left = mid + 1; 296 } 297 assert left == right; 298 299 /* 300 * The invariants still hold: pivot >= all in [lo, left) and 301 * pivot < all in [left, start), so pivot belongs at left. Note 302 * that if there are elements equal to pivot, left points to the 303 * first slot after them -- that's why this sort is stable. 304 * Slide elements over to make room for pivot. 305 */ 306 int n = start - left; // The number of elements to move 307 // Switch is just an optimization for arraycopy in default case 308 switch (n) { 309 case 2: a[left + 2] = a[left + 1]; 310 case 1: a[left + 1] = a[left]; 311 break; 312 default: System.arraycopy(a, left, a, left + 1, n); 313 } 314 a[left] = pivot; 315 } 316 } 317 318 /** 319 * Returns the length of the run beginning at the specified position in 320 * the specified array and reverses the run if it is descending (ensuring 321 * that the run will always be ascending when the method returns). 322 * 323 * A run is the longest ascending sequence with: 324 * 325 * a[lo] <= a[lo + 1] <= a[lo + 2] <= ... 326 * 327 * or the longest descending sequence with: 328 * 329 * a[lo] > a[lo + 1] > a[lo + 2] > ... 330 * 331 * For its intended use in a stable mergesort, the strictness of the 332 * definition of "descending" is needed so that the call can safely 333 * reverse a descending sequence without violating stability. 334 * 335 * @param a the array in which a run is to be counted and possibly reversed 336 * @param lo index of the first element in the run 337 * @param hi index after the last element that may be contained in the run. 338 It is required that {@code lo < hi}. 339 * @param c the comparator to used for the sort 340 * @return the length of the run beginning at the specified position in 341 * the specified array 342 */ countRunAndMakeAscending(T[] a, int lo, int hi, Comparator<? super T> c)343 private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi, 344 Comparator<? super T> c) { 345 assert lo < hi; 346 int runHi = lo + 1; 347 if (runHi == hi) 348 return 1; 349 350 // Find end of run, and reverse range if descending 351 if (c.compare(a[runHi++], a[lo]) < 0) { // Descending 352 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0) 353 runHi++; 354 reverseRange(a, lo, runHi); 355 } else { // Ascending 356 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0) 357 runHi++; 358 } 359 360 return runHi - lo; 361 } 362 363 /** 364 * Reverse the specified range of the specified array. 365 * 366 * @param a the array in which a range is to be reversed 367 * @param lo the index of the first element in the range to be reversed 368 * @param hi the index after the last element in the range to be reversed 369 */ 370 private static void reverseRange(Object[] a, int lo, int hi) { 371 hi--; 372 while (lo < hi) { 373 Object t = a[lo]; 374 a[lo++] = a[hi]; 375 a[hi--] = t; 376 } 377 } 378 379 /** 380 * Returns the minimum acceptable run length for an array of the specified 381 * length. Natural runs shorter than this will be extended with 382 * {@link #binarySort}. 383 * 384 * Roughly speaking, the computation is: 385 * 386 * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff). 387 * Else if n is an exact power of 2, return MIN_MERGE/2. 388 * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k 389 * is close to, but strictly less than, an exact power of 2. 390 * 391 * For the rationale, see listsort.txt. 392 * 393 * @param n the length of the array to be sorted 394 * @return the length of the minimum run to be merged 395 */ 396 private static int minRunLength(int n) { 397 assert n >= 0; 398 int r = 0; // Becomes 1 if any 1 bits are shifted off 399 while (n >= MIN_MERGE) { 400 r |= (n & 1); 401 n >>= 1; 402 } 403 return n + r; 404 } 405 406 /** 407 * Pushes the specified run onto the pending-run stack. 408 * 409 * @param runBase index of the first element in the run 410 * @param runLen the number of elements in the run 411 */ 412 private void pushRun(int runBase, int runLen) { 413 this.runBase[stackSize] = runBase; 414 this.runLen[stackSize] = runLen; 415 stackSize++; 416 } 417 418 /** 419 * Examines the stack of runs waiting to be merged and merges adjacent runs 420 * until the stack invariants are reestablished: 421 * 422 * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] 423 * 2. runLen[i - 2] > runLen[i - 1] 424 * 425 * This method is called each time a new run is pushed onto the stack, 426 * so the invariants are guaranteed to hold for i < stackSize upon 427 * entry to the method. 428 */ 429 private void mergeCollapse() { 430 while (stackSize > 1) { 431 int n = stackSize - 2; 432 if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) { 433 if (runLen[n - 1] < runLen[n + 1]) 434 n--; 435 mergeAt(n); 436 } else if (runLen[n] <= runLen[n + 1]) { 437 mergeAt(n); 438 } else { 439 break; // Invariant is established 440 } 441 } 442 } 443 444 /** 445 * Merges all runs on the stack until only one remains. This method is 446 * called once, to complete the sort. 447 */ mergeForceCollapse()448 private void mergeForceCollapse() { 449 while (stackSize > 1) { 450 int n = stackSize - 2; 451 if (n > 0 && runLen[n - 1] < runLen[n + 1]) 452 n--; 453 mergeAt(n); 454 } 455 } 456 457 /** 458 * Merges the two runs at stack indices i and i+1. Run i must be 459 * the penultimate or antepenultimate run on the stack. In other words, 460 * i must be equal to stackSize-2 or stackSize-3. 461 * 462 * @param i stack index of the first of the two runs to merge 463 */ mergeAt(int i)464 private void mergeAt(int i) { 465 assert stackSize >= 2; 466 assert i >= 0; 467 assert i == stackSize - 2 || i == stackSize - 3; 468 469 int base1 = runBase[i]; 470 int len1 = runLen[i]; 471 int base2 = runBase[i + 1]; 472 int len2 = runLen[i + 1]; 473 assert len1 > 0 && len2 > 0; 474 assert base1 + len1 == base2; 475 476 /* 477 * Record the length of the combined runs; if i is the 3rd-last 478 * run now, also slide over the last run (which isn't involved 479 * in this merge). The current run (i+1) goes away in any case. 480 */ 481 runLen[i] = len1 + len2; 482 if (i == stackSize - 3) { 483 runBase[i + 1] = runBase[i + 2]; 484 runLen[i + 1] = runLen[i + 2]; 485 } 486 stackSize--; 487 488 /* 489 * Find where the first element of run2 goes in run1. Prior elements 490 * in run1 can be ignored (because they're already in place). 491 */ 492 int k = gallopRight(a[base2], a, base1, len1, 0, c); 493 assert k >= 0; 494 base1 += k; 495 len1 -= k; 496 if (len1 == 0) 497 return; 498 499 /* 500 * Find where the last element of run1 goes in run2. Subsequent elements 501 * in run2 can be ignored (because they're already in place). 502 */ 503 len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c); 504 assert len2 >= 0; 505 if (len2 == 0) 506 return; 507 508 // Merge remaining runs, using tmp array with min(len1, len2) elements 509 if (len1 <= len2) 510 mergeLo(base1, len1, base2, len2); 511 else 512 mergeHi(base1, len1, base2, len2); 513 } 514 515 /** 516 * Locates the position at which to insert the specified key into the 517 * specified sorted range; if the range contains an element equal to key, 518 * returns the index of the leftmost equal element. 519 * 520 * @param key the key whose insertion point to search for 521 * @param a the array in which to search 522 * @param base the index of the first element in the range 523 * @param len the length of the range; must be > 0 524 * @param hint the index at which to begin the search, 0 <= hint < n. 525 * The closer hint is to the result, the faster this method will run. 526 * @param c the comparator used to order the range, and to search 527 * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k], 528 * pretending that a[b - 1] is minus infinity and a[b + n] is infinity. 529 * In other words, key belongs at index b + k; or in other words, 530 * the first k elements of a should precede key, and the last n - k 531 * should follow it. 532 */ gallopLeft(T key, T[] a, int base, int len, int hint, Comparator<? super T> c)533 private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint, 534 Comparator<? super T> c) { 535 assert len > 0 && hint >= 0 && hint < len; 536 int lastOfs = 0; 537 int ofs = 1; 538 if (c.compare(key, a[base + hint]) > 0) { 539 // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs] 540 int maxOfs = len - hint; 541 while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) { 542 lastOfs = ofs; 543 ofs = (ofs << 1) + 1; 544 if (ofs <= 0) // int overflow 545 ofs = maxOfs; 546 } 547 if (ofs > maxOfs) 548 ofs = maxOfs; 549 550 // Make offsets relative to base 551 lastOfs += hint; 552 ofs += hint; 553 } else { // key <= a[base + hint] 554 // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs] 555 final int maxOfs = hint + 1; 556 while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) { 557 lastOfs = ofs; 558 ofs = (ofs << 1) + 1; 559 if (ofs <= 0) // int overflow 560 ofs = maxOfs; 561 } 562 if (ofs > maxOfs) 563 ofs = maxOfs; 564 565 // Make offsets relative to base 566 int tmp = lastOfs; 567 lastOfs = hint - ofs; 568 ofs = hint - tmp; 569 } 570 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; 571 572 /* 573 * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere 574 * to the right of lastOfs but no farther right than ofs. Do a binary 575 * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs]. 576 */ 577 lastOfs++; 578 while (lastOfs < ofs) { 579 int m = lastOfs + ((ofs - lastOfs) >>> 1); 580 581 if (c.compare(key, a[base + m]) > 0) 582 lastOfs = m + 1; // a[base + m] < key 583 else 584 ofs = m; // key <= a[base + m] 585 } 586 assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs] 587 return ofs; 588 } 589 590 /** 591 * Like gallopLeft, except that if the range contains an element equal to 592 * key, gallopRight returns the index after the rightmost equal element. 593 * 594 * @param key the key whose insertion point to search for 595 * @param a the array in which to search 596 * @param base the index of the first element in the range 597 * @param len the length of the range; must be > 0 598 * @param hint the index at which to begin the search, 0 <= hint < n. 599 * The closer hint is to the result, the faster this method will run. 600 * @param c the comparator used to order the range, and to search 601 * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k] 602 */ gallopRight(T key, T[] a, int base, int len, int hint, Comparator<? super T> c)603 private static <T> int gallopRight(T key, T[] a, int base, int len, 604 int hint, Comparator<? super T> c) { 605 assert len > 0 && hint >= 0 && hint < len; 606 607 int ofs = 1; 608 int lastOfs = 0; 609 if (c.compare(key, a[base + hint]) < 0) { 610 // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs] 611 int maxOfs = hint + 1; 612 while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) { 613 lastOfs = ofs; 614 ofs = (ofs << 1) + 1; 615 if (ofs <= 0) // int overflow 616 ofs = maxOfs; 617 } 618 if (ofs > maxOfs) 619 ofs = maxOfs; 620 621 // Make offsets relative to b 622 int tmp = lastOfs; 623 lastOfs = hint - ofs; 624 ofs = hint - tmp; 625 } else { // a[b + hint] <= key 626 // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs] 627 int maxOfs = len - hint; 628 while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) { 629 lastOfs = ofs; 630 ofs = (ofs << 1) + 1; 631 if (ofs <= 0) // int overflow 632 ofs = maxOfs; 633 } 634 if (ofs > maxOfs) 635 ofs = maxOfs; 636 637 // Make offsets relative to b 638 lastOfs += hint; 639 ofs += hint; 640 } 641 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; 642 643 /* 644 * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to 645 * the right of lastOfs but no farther right than ofs. Do a binary 646 * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs]. 647 */ 648 lastOfs++; 649 while (lastOfs < ofs) { 650 int m = lastOfs + ((ofs - lastOfs) >>> 1); 651 652 if (c.compare(key, a[base + m]) < 0) 653 ofs = m; // key < a[b + m] 654 else 655 lastOfs = m + 1; // a[b + m] <= key 656 } 657 assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs] 658 return ofs; 659 } 660 661 /** 662 * Merges two adjacent runs in place, in a stable fashion. The first 663 * element of the first run must be greater than the first element of the 664 * second run (a[base1] > a[base2]), and the last element of the first run 665 * (a[base1 + len1-1]) must be greater than all elements of the second run. 666 * 667 * For performance, this method should be called only when len1 <= len2; 668 * its twin, mergeHi should be called if len1 >= len2. (Either method 669 * may be called if len1 == len2.) 670 * 671 * @param base1 index of first element in first run to be merged 672 * @param len1 length of first run to be merged (must be > 0) 673 * @param base2 index of first element in second run to be merged 674 * (must be aBase + aLen) 675 * @param len2 length of second run to be merged (must be > 0) 676 */ 677 private void mergeLo(int base1, int len1, int base2, int len2) { 678 assert len1 > 0 && len2 > 0 && base1 + len1 == base2; 679 680 // Copy first run into temp array 681 T[] a = this.a; // For performance 682 T[] tmp = ensureCapacity(len1); 683 int cursor1 = tmpBase; // Indexes into tmp array 684 int cursor2 = base2; // Indexes int a 685 int dest = base1; // Indexes int a 686 System.arraycopy(a, base1, tmp, cursor1, len1); 687 688 // Move first element of second run and deal with degenerate cases 689 a[dest++] = a[cursor2++]; 690 if (--len2 == 0) { 691 System.arraycopy(tmp, cursor1, a, dest, len1); 692 return; 693 } 694 if (len1 == 1) { 695 System.arraycopy(a, cursor2, a, dest, len2); 696 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge 697 return; 698 } 699 700 Comparator<? super T> c = this.c; // Use local variable for performance 701 int minGallop = this.minGallop; // " " " " " 702 outer: 703 while (true) { 704 int count1 = 0; // Number of times in a row that first run won 705 int count2 = 0; // Number of times in a row that second run won 706 707 /* 708 * Do the straightforward thing until (if ever) one run starts 709 * winning consistently. 710 */ 711 do { 712 assert len1 > 1 && len2 > 0; 713 if (c.compare(a[cursor2], tmp[cursor1]) < 0) { 714 a[dest++] = a[cursor2++]; 715 count2++; 716 count1 = 0; 717 if (--len2 == 0) 718 break outer; 719 } else { 720 a[dest++] = tmp[cursor1++]; 721 count1++; 722 count2 = 0; 723 if (--len1 == 1) 724 break outer; 725 } 726 } while ((count1 | count2) < minGallop); 727 728 /* 729 * One run is winning so consistently that galloping may be a 730 * huge win. So try that, and continue galloping until (if ever) 731 * neither run appears to be winning consistently anymore. 732 */ 733 do { 734 assert len1 > 1 && len2 > 0; 735 count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c); 736 if (count1 != 0) { 737 System.arraycopy(tmp, cursor1, a, dest, count1); 738 dest += count1; 739 cursor1 += count1; 740 len1 -= count1; 741 if (len1 <= 1) // len1 == 1 || len1 == 0 742 break outer; 743 } 744 a[dest++] = a[cursor2++]; 745 if (--len2 == 0) 746 break outer; 747 748 count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c); 749 if (count2 != 0) { 750 System.arraycopy(a, cursor2, a, dest, count2); 751 dest += count2; 752 cursor2 += count2; 753 len2 -= count2; 754 if (len2 == 0) 755 break outer; 756 } 757 a[dest++] = tmp[cursor1++]; 758 if (--len1 == 1) 759 break outer; 760 minGallop--; 761 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); 762 if (minGallop < 0) 763 minGallop = 0; 764 minGallop += 2; // Penalize for leaving gallop mode 765 } // End of "outer" loop 766 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field 767 768 if (len1 == 1) { 769 assert len2 > 0; 770 System.arraycopy(a, cursor2, a, dest, len2); 771 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge 772 } else if (len1 == 0) { 773 throw new IllegalArgumentException( 774 "Comparison method violates its general contract!"); 775 } else { 776 assert len2 == 0; 777 assert len1 > 1; 778 System.arraycopy(tmp, cursor1, a, dest, len1); 779 } 780 } 781 782 /** 783 * Like mergeLo, except that this method should be called only if 784 * len1 >= len2; mergeLo should be called if len1 <= len2. (Either method 785 * may be called if len1 == len2.) 786 * 787 * @param base1 index of first element in first run to be merged 788 * @param len1 length of first run to be merged (must be > 0) 789 * @param base2 index of first element in second run to be merged 790 * (must be aBase + aLen) 791 * @param len2 length of second run to be merged (must be > 0) 792 */ 793 private void mergeHi(int base1, int len1, int base2, int len2) { 794 assert len1 > 0 && len2 > 0 && base1 + len1 == base2; 795 796 // Copy second run into temp array 797 T[] a = this.a; // For performance 798 T[] tmp = ensureCapacity(len2); 799 int tmpBase = this.tmpBase; 800 System.arraycopy(a, base2, tmp, tmpBase, len2); 801 802 int cursor1 = base1 + len1 - 1; // Indexes into a 803 int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array 804 int dest = base2 + len2 - 1; // Indexes into a 805 806 // Move last element of first run and deal with degenerate cases 807 a[dest--] = a[cursor1--]; 808 if (--len1 == 0) { 809 System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); 810 return; 811 } 812 if (len2 == 1) { 813 dest -= len1; 814 cursor1 -= len1; 815 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); 816 a[dest] = tmp[cursor2]; 817 return; 818 } 819 820 Comparator<? super T> c = this.c; // Use local variable for performance 821 int minGallop = this.minGallop; // " " " " " 822 outer: 823 while (true) { 824 int count1 = 0; // Number of times in a row that first run won 825 int count2 = 0; // Number of times in a row that second run won 826 827 /* 828 * Do the straightforward thing until (if ever) one run 829 * appears to win consistently. 830 */ 831 do { 832 assert len1 > 0 && len2 > 1; 833 if (c.compare(tmp[cursor2], a[cursor1]) < 0) { 834 a[dest--] = a[cursor1--]; 835 count1++; 836 count2 = 0; 837 if (--len1 == 0) 838 break outer; 839 } else { 840 a[dest--] = tmp[cursor2--]; 841 count2++; 842 count1 = 0; 843 if (--len2 == 1) 844 break outer; 845 } 846 } while ((count1 | count2) < minGallop); 847 848 /* 849 * One run is winning so consistently that galloping may be a 850 * huge win. So try that, and continue galloping until (if ever) 851 * neither run appears to be winning consistently anymore. 852 */ 853 do { 854 assert len1 > 0 && len2 > 1; 855 count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c); 856 if (count1 != 0) { 857 dest -= count1; 858 cursor1 -= count1; 859 len1 -= count1; 860 System.arraycopy(a, cursor1 + 1, a, dest + 1, count1); 861 if (len1 == 0) 862 break outer; 863 } 864 a[dest--] = tmp[cursor2--]; 865 if (--len2 == 1) 866 break outer; 867 868 count2 = len2 - gallopLeft(a[cursor1], tmp, tmpBase, len2, len2 - 1, c); 869 if (count2 != 0) { 870 dest -= count2; 871 cursor2 -= count2; 872 len2 -= count2; 873 System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2); 874 if (len2 <= 1) // len2 == 1 || len2 == 0 875 break outer; 876 } 877 a[dest--] = a[cursor1--]; 878 if (--len1 == 0) 879 break outer; 880 minGallop--; 881 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); 882 if (minGallop < 0) 883 minGallop = 0; 884 minGallop += 2; // Penalize for leaving gallop mode 885 } // End of "outer" loop 886 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field 887 888 if (len2 == 1) { 889 assert len1 > 0; 890 dest -= len1; 891 cursor1 -= len1; System.arraycopy(a, cursor1 + 1, a, dest + 1, len1)892 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); 893 a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge 894 } else if (len2 == 0) { 895 throw new IllegalArgumentException( 896 "Comparison method violates its general contract!"); 897 } else { 898 assert len1 == 0; 899 assert len2 > 0; 900 System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); 901 } 902 } 903 904 /** 905 * Ensures that the external array tmp has at least the specified 906 * number of elements, increasing its size if necessary. The size 907 * increases exponentially to ensure amortized linear time complexity. 908 * 909 * @param minCapacity the minimum required capacity of the tmp array 910 * @return tmp, whether or not it grew 911 */ 912 private T[] ensureCapacity(int minCapacity) { 913 if (tmpLen < minCapacity) { 914 // Compute smallest power of 2 > minCapacity 915 int newSize = minCapacity; 916 newSize |= newSize >> 1; 917 newSize |= newSize >> 2; 918 newSize |= newSize >> 4; 919 newSize |= newSize >> 8; 920 newSize |= newSize >> 16; 921 newSize++; 922 923 if (newSize < 0) // Not bloody likely! 924 newSize = minCapacity; 925 else 926 newSize = Math.min(newSize, a.length >>> 1); 927 928 @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) 929 T[] newArray = (T[])java.lang.reflect.Array.newInstance 930 (a.getClass().getComponentType(), newSize); 931 tmp = newArray; 932 tmpLen = newSize; 933 tmpBase = 0; 934 } 935 return tmp; 936 } 937 } 938