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 * The maximum value of 49 allows for an array up to length 178 * Integer.MAX_VALUE-4, if array is filled by the worst case stack size 179 * increasing scenario. More explanations are given in section 4 of: 180 * http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf 181 */ 182 int stackLen = (len < 120 ? 5 : 183 len < 1542 ? 10 : 184 len < 119151 ? 24 : 49); 185 runBase = new int[stackLen]; 186 runLen = new int[stackLen]; 187 } 188 189 /* 190 * The next method (package private and static) constitutes the 191 * entire API of this class. 192 */ 193 194 /** 195 * Sorts the given range, using the given workspace array slice 196 * for temp storage when possible. This method is designed to be 197 * invoked from public methods (in class Arrays) after performing 198 * any necessary array bounds checks and expanding parameters into 199 * the required forms. 200 * 201 * @param a the array to be sorted 202 * @param lo the index of the first element, inclusive, to be sorted 203 * @param hi the index of the last element, exclusive, to be sorted 204 * @param c the comparator to use 205 * @param work a workspace array (slice) 206 * @param workBase origin of usable space in work array 207 * @param workLen usable size of work array 208 * @since 1.8 209 */ sort(T[] a, int lo, int hi, Comparator<? super T> c, T[] work, int workBase, int workLen)210 static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c, 211 T[] work, int workBase, int workLen) { 212 assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length; 213 214 int nRemaining = hi - lo; 215 if (nRemaining < 2) 216 return; // Arrays of size 0 and 1 are always sorted 217 218 // If array is small, do a "mini-TimSort" with no merges 219 if (nRemaining < MIN_MERGE) { 220 int initRunLen = countRunAndMakeAscending(a, lo, hi, c); 221 binarySort(a, lo, hi, lo + initRunLen, c); 222 return; 223 } 224 225 /** 226 * March over the array once, left to right, finding natural runs, 227 * extending short natural runs to minRun elements, and merging runs 228 * to maintain stack invariant. 229 */ 230 TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen); 231 int minRun = minRunLength(nRemaining); 232 do { 233 // Identify next run 234 int runLen = countRunAndMakeAscending(a, lo, hi, c); 235 236 // If run is short, extend to min(minRun, nRemaining) 237 if (runLen < minRun) { 238 int force = nRemaining <= minRun ? nRemaining : minRun; 239 binarySort(a, lo, lo + force, lo + runLen, c); 240 runLen = force; 241 } 242 243 // Push run onto pending-run stack, and maybe merge 244 ts.pushRun(lo, runLen); 245 ts.mergeCollapse(); 246 247 // Advance to find next run 248 lo += runLen; 249 nRemaining -= runLen; 250 } while (nRemaining != 0); 251 252 // Merge all remaining runs to complete sort 253 assert lo == hi; 254 ts.mergeForceCollapse(); 255 assert ts.stackSize == 1; 256 } 257 258 /** 259 * Sorts the specified portion of the specified array using a binary 260 * insertion sort. This is the best method for sorting small numbers 261 * of elements. It requires O(n log n) compares, but O(n^2) data 262 * movement (worst case). 263 * 264 * If the initial part of the specified range is already sorted, 265 * this method can take advantage of it: the method assumes that the 266 * elements from index {@code lo}, inclusive, to {@code start}, 267 * exclusive are already sorted. 268 * 269 * @param a the array in which a range is to be sorted 270 * @param lo the index of the first element in the range to be sorted 271 * @param hi the index after the last element in the range to be sorted 272 * @param start the index of the first element in the range that is 273 * not already known to be sorted ({@code lo <= start <= hi}) 274 * @param c comparator to used for the sort 275 */ 276 @SuppressWarnings("fallthrough") binarySort(T[] a, int lo, int hi, int start, Comparator<? super T> c)277 private static <T> void binarySort(T[] a, int lo, int hi, int start, 278 Comparator<? super T> c) { 279 assert lo <= start && start <= hi; 280 if (start == lo) 281 start++; 282 for ( ; start < hi; start++) { 283 T pivot = a[start]; 284 285 // Set left (and right) to the index where a[start] (pivot) belongs 286 int left = lo; 287 int right = start; 288 assert left <= right; 289 /* 290 * Invariants: 291 * pivot >= all in [lo, left). 292 * pivot < all in [right, start). 293 */ 294 while (left < right) { 295 int mid = (left + right) >>> 1; 296 if (c.compare(pivot, a[mid]) < 0) 297 right = mid; 298 else 299 left = mid + 1; 300 } 301 assert left == right; 302 303 /* 304 * The invariants still hold: pivot >= all in [lo, left) and 305 * pivot < all in [left, start), so pivot belongs at left. Note 306 * that if there are elements equal to pivot, left points to the 307 * first slot after them -- that's why this sort is stable. 308 * Slide elements over to make room for pivot. 309 */ 310 int n = start - left; // The number of elements to move 311 // Switch is just an optimization for arraycopy in default case 312 switch (n) { 313 case 2: a[left + 2] = a[left + 1]; 314 case 1: a[left + 1] = a[left]; 315 break; 316 default: System.arraycopy(a, left, a, left + 1, n); 317 } 318 a[left] = pivot; 319 } 320 } 321 322 /** 323 * Returns the length of the run beginning at the specified position in 324 * the specified array and reverses the run if it is descending (ensuring 325 * that the run will always be ascending when the method returns). 326 * 327 * A run is the longest ascending sequence with: 328 * 329 * a[lo] <= a[lo + 1] <= a[lo + 2] <= ... 330 * 331 * or the longest descending sequence with: 332 * 333 * a[lo] > a[lo + 1] > a[lo + 2] > ... 334 * 335 * For its intended use in a stable mergesort, the strictness of the 336 * definition of "descending" is needed so that the call can safely 337 * reverse a descending sequence without violating stability. 338 * 339 * @param a the array in which a run is to be counted and possibly reversed 340 * @param lo index of the first element in the run 341 * @param hi index after the last element that may be contained in the run. 342 It is required that {@code lo < hi}. 343 * @param c the comparator to used for the sort 344 * @return the length of the run beginning at the specified position in 345 * the specified array 346 */ countRunAndMakeAscending(T[] a, int lo, int hi, Comparator<? super T> c)347 private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi, 348 Comparator<? super T> c) { 349 assert lo < hi; 350 int runHi = lo + 1; 351 if (runHi == hi) 352 return 1; 353 354 // Find end of run, and reverse range if descending 355 if (c.compare(a[runHi++], a[lo]) < 0) { // Descending 356 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0) 357 runHi++; 358 reverseRange(a, lo, runHi); 359 } else { // Ascending 360 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0) 361 runHi++; 362 } 363 364 return runHi - lo; 365 } 366 367 /** 368 * Reverse the specified range of the specified array. 369 * 370 * @param a the array in which a range is to be reversed 371 * @param lo the index of the first element in the range to be reversed 372 * @param hi the index after the last element in the range to be reversed 373 */ 374 private static void reverseRange(Object[] a, int lo, int hi) { 375 hi--; 376 while (lo < hi) { 377 Object t = a[lo]; 378 a[lo++] = a[hi]; 379 a[hi--] = t; 380 } 381 } 382 383 /** 384 * Returns the minimum acceptable run length for an array of the specified 385 * length. Natural runs shorter than this will be extended with 386 * {@link #binarySort}. 387 * 388 * Roughly speaking, the computation is: 389 * 390 * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff). 391 * Else if n is an exact power of 2, return MIN_MERGE/2. 392 * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k 393 * is close to, but strictly less than, an exact power of 2. 394 * 395 * For the rationale, see listsort.txt. 396 * 397 * @param n the length of the array to be sorted 398 * @return the length of the minimum run to be merged 399 */ 400 private static int minRunLength(int n) { 401 assert n >= 0; 402 int r = 0; // Becomes 1 if any 1 bits are shifted off 403 while (n >= MIN_MERGE) { 404 r |= (n & 1); 405 n >>= 1; 406 } 407 return n + r; 408 } 409 410 /** 411 * Pushes the specified run onto the pending-run stack. 412 * 413 * @param runBase index of the first element in the run 414 * @param runLen the number of elements in the run 415 */ 416 private void pushRun(int runBase, int runLen) { 417 this.runBase[stackSize] = runBase; 418 this.runLen[stackSize] = runLen; 419 stackSize++; 420 } 421 422 /** 423 * Examines the stack of runs waiting to be merged and merges adjacent runs 424 * until the stack invariants are reestablished: 425 * 426 * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] 427 * 2. runLen[i - 2] > runLen[i - 1] 428 * 429 * This method is called each time a new run is pushed onto the stack, 430 * so the invariants are guaranteed to hold for i < stackSize upon 431 * entry to the method. 432 */ 433 private void mergeCollapse() { 434 while (stackSize > 1) { 435 int n = stackSize - 2; 436 if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) { 437 if (runLen[n - 1] < runLen[n + 1]) 438 n--; 439 mergeAt(n); 440 } else if (runLen[n] <= runLen[n + 1]) { 441 mergeAt(n); 442 } else { 443 break; // Invariant is established 444 } 445 } 446 } 447 448 /** 449 * Merges all runs on the stack until only one remains. This method is 450 * called once, to complete the sort. 451 */ mergeForceCollapse()452 private void mergeForceCollapse() { 453 while (stackSize > 1) { 454 int n = stackSize - 2; 455 if (n > 0 && runLen[n - 1] < runLen[n + 1]) 456 n--; 457 mergeAt(n); 458 } 459 } 460 461 /** 462 * Merges the two runs at stack indices i and i+1. Run i must be 463 * the penultimate or antepenultimate run on the stack. In other words, 464 * i must be equal to stackSize-2 or stackSize-3. 465 * 466 * @param i stack index of the first of the two runs to merge 467 */ mergeAt(int i)468 private void mergeAt(int i) { 469 assert stackSize >= 2; 470 assert i >= 0; 471 assert i == stackSize - 2 || i == stackSize - 3; 472 473 int base1 = runBase[i]; 474 int len1 = runLen[i]; 475 int base2 = runBase[i + 1]; 476 int len2 = runLen[i + 1]; 477 assert len1 > 0 && len2 > 0; 478 assert base1 + len1 == base2; 479 480 /* 481 * Record the length of the combined runs; if i is the 3rd-last 482 * run now, also slide over the last run (which isn't involved 483 * in this merge). The current run (i+1) goes away in any case. 484 */ 485 runLen[i] = len1 + len2; 486 if (i == stackSize - 3) { 487 runBase[i + 1] = runBase[i + 2]; 488 runLen[i + 1] = runLen[i + 2]; 489 } 490 stackSize--; 491 492 /* 493 * Find where the first element of run2 goes in run1. Prior elements 494 * in run1 can be ignored (because they're already in place). 495 */ 496 int k = gallopRight(a[base2], a, base1, len1, 0, c); 497 assert k >= 0; 498 base1 += k; 499 len1 -= k; 500 if (len1 == 0) 501 return; 502 503 /* 504 * Find where the last element of run1 goes in run2. Subsequent elements 505 * in run2 can be ignored (because they're already in place). 506 */ 507 len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c); 508 assert len2 >= 0; 509 if (len2 == 0) 510 return; 511 512 // Merge remaining runs, using tmp array with min(len1, len2) elements 513 if (len1 <= len2) 514 mergeLo(base1, len1, base2, len2); 515 else 516 mergeHi(base1, len1, base2, len2); 517 } 518 519 /** 520 * Locates the position at which to insert the specified key into the 521 * specified sorted range; if the range contains an element equal to key, 522 * returns the index of the leftmost equal element. 523 * 524 * @param key the key whose insertion point to search for 525 * @param a the array in which to search 526 * @param base the index of the first element in the range 527 * @param len the length of the range; must be > 0 528 * @param hint the index at which to begin the search, 0 <= hint < n. 529 * The closer hint is to the result, the faster this method will run. 530 * @param c the comparator used to order the range, and to search 531 * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k], 532 * pretending that a[b - 1] is minus infinity and a[b + n] is infinity. 533 * In other words, key belongs at index b + k; or in other words, 534 * the first k elements of a should precede key, and the last n - k 535 * should follow it. 536 */ gallopLeft(T key, T[] a, int base, int len, int hint, Comparator<? super T> c)537 private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint, 538 Comparator<? super T> c) { 539 assert len > 0 && hint >= 0 && hint < len; 540 int lastOfs = 0; 541 int ofs = 1; 542 if (c.compare(key, a[base + hint]) > 0) { 543 // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs] 544 int maxOfs = len - hint; 545 while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) { 546 lastOfs = ofs; 547 ofs = (ofs << 1) + 1; 548 if (ofs <= 0) // int overflow 549 ofs = maxOfs; 550 } 551 if (ofs > maxOfs) 552 ofs = maxOfs; 553 554 // Make offsets relative to base 555 lastOfs += hint; 556 ofs += hint; 557 } else { // key <= a[base + hint] 558 // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs] 559 final int maxOfs = hint + 1; 560 while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) { 561 lastOfs = ofs; 562 ofs = (ofs << 1) + 1; 563 if (ofs <= 0) // int overflow 564 ofs = maxOfs; 565 } 566 if (ofs > maxOfs) 567 ofs = maxOfs; 568 569 // Make offsets relative to base 570 int tmp = lastOfs; 571 lastOfs = hint - ofs; 572 ofs = hint - tmp; 573 } 574 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; 575 576 /* 577 * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere 578 * to the right of lastOfs but no farther right than ofs. Do a binary 579 * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs]. 580 */ 581 lastOfs++; 582 while (lastOfs < ofs) { 583 int m = lastOfs + ((ofs - lastOfs) >>> 1); 584 585 if (c.compare(key, a[base + m]) > 0) 586 lastOfs = m + 1; // a[base + m] < key 587 else 588 ofs = m; // key <= a[base + m] 589 } 590 assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs] 591 return ofs; 592 } 593 594 /** 595 * Like gallopLeft, except that if the range contains an element equal to 596 * key, gallopRight returns the index after the rightmost equal element. 597 * 598 * @param key the key whose insertion point to search for 599 * @param a the array in which to search 600 * @param base the index of the first element in the range 601 * @param len the length of the range; must be > 0 602 * @param hint the index at which to begin the search, 0 <= hint < n. 603 * The closer hint is to the result, the faster this method will run. 604 * @param c the comparator used to order the range, and to search 605 * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k] 606 */ gallopRight(T key, T[] a, int base, int len, int hint, Comparator<? super T> c)607 private static <T> int gallopRight(T key, T[] a, int base, int len, 608 int hint, Comparator<? super T> c) { 609 assert len > 0 && hint >= 0 && hint < len; 610 611 int ofs = 1; 612 int lastOfs = 0; 613 if (c.compare(key, a[base + hint]) < 0) { 614 // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs] 615 int maxOfs = hint + 1; 616 while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) { 617 lastOfs = ofs; 618 ofs = (ofs << 1) + 1; 619 if (ofs <= 0) // int overflow 620 ofs = maxOfs; 621 } 622 if (ofs > maxOfs) 623 ofs = maxOfs; 624 625 // Make offsets relative to b 626 int tmp = lastOfs; 627 lastOfs = hint - ofs; 628 ofs = hint - tmp; 629 } else { // a[b + hint] <= key 630 // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs] 631 int maxOfs = len - hint; 632 while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) { 633 lastOfs = ofs; 634 ofs = (ofs << 1) + 1; 635 if (ofs <= 0) // int overflow 636 ofs = maxOfs; 637 } 638 if (ofs > maxOfs) 639 ofs = maxOfs; 640 641 // Make offsets relative to b 642 lastOfs += hint; 643 ofs += hint; 644 } 645 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; 646 647 /* 648 * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to 649 * the right of lastOfs but no farther right than ofs. Do a binary 650 * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs]. 651 */ 652 lastOfs++; 653 while (lastOfs < ofs) { 654 int m = lastOfs + ((ofs - lastOfs) >>> 1); 655 656 if (c.compare(key, a[base + m]) < 0) 657 ofs = m; // key < a[b + m] 658 else 659 lastOfs = m + 1; // a[b + m] <= key 660 } 661 assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs] 662 return ofs; 663 } 664 665 /** 666 * Merges two adjacent runs in place, in a stable fashion. The first 667 * element of the first run must be greater than the first element of the 668 * second run (a[base1] > a[base2]), and the last element of the first run 669 * (a[base1 + len1-1]) must be greater than all elements of the second run. 670 * 671 * For performance, this method should be called only when len1 <= len2; 672 * its twin, mergeHi should be called if len1 >= len2. (Either method 673 * may be called if len1 == len2.) 674 * 675 * @param base1 index of first element in first run to be merged 676 * @param len1 length of first run to be merged (must be > 0) 677 * @param base2 index of first element in second run to be merged 678 * (must be aBase + aLen) 679 * @param len2 length of second run to be merged (must be > 0) 680 */ 681 private void mergeLo(int base1, int len1, int base2, int len2) { 682 assert len1 > 0 && len2 > 0 && base1 + len1 == base2; 683 684 // Copy first run into temp array 685 T[] a = this.a; // For performance 686 T[] tmp = ensureCapacity(len1); 687 int cursor1 = tmpBase; // Indexes into tmp array 688 int cursor2 = base2; // Indexes int a 689 int dest = base1; // Indexes int a 690 System.arraycopy(a, base1, tmp, cursor1, len1); 691 692 // Move first element of second run and deal with degenerate cases 693 a[dest++] = a[cursor2++]; 694 if (--len2 == 0) { 695 System.arraycopy(tmp, cursor1, a, dest, len1); 696 return; 697 } 698 if (len1 == 1) { 699 System.arraycopy(a, cursor2, a, dest, len2); 700 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge 701 return; 702 } 703 704 Comparator<? super T> c = this.c; // Use local variable for performance 705 int minGallop = this.minGallop; // " " " " " 706 outer: 707 while (true) { 708 int count1 = 0; // Number of times in a row that first run won 709 int count2 = 0; // Number of times in a row that second run won 710 711 /* 712 * Do the straightforward thing until (if ever) one run starts 713 * winning consistently. 714 */ 715 do { 716 assert len1 > 1 && len2 > 0; 717 if (c.compare(a[cursor2], tmp[cursor1]) < 0) { 718 a[dest++] = a[cursor2++]; 719 count2++; 720 count1 = 0; 721 if (--len2 == 0) 722 break outer; 723 } else { 724 a[dest++] = tmp[cursor1++]; 725 count1++; 726 count2 = 0; 727 if (--len1 == 1) 728 break outer; 729 } 730 } while ((count1 | count2) < minGallop); 731 732 /* 733 * One run is winning so consistently that galloping may be a 734 * huge win. So try that, and continue galloping until (if ever) 735 * neither run appears to be winning consistently anymore. 736 */ 737 do { 738 assert len1 > 1 && len2 > 0; 739 count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c); 740 if (count1 != 0) { 741 System.arraycopy(tmp, cursor1, a, dest, count1); 742 dest += count1; 743 cursor1 += count1; 744 len1 -= count1; 745 if (len1 <= 1) // len1 == 1 || len1 == 0 746 break outer; 747 } 748 a[dest++] = a[cursor2++]; 749 if (--len2 == 0) 750 break outer; 751 752 count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c); 753 if (count2 != 0) { 754 System.arraycopy(a, cursor2, a, dest, count2); 755 dest += count2; 756 cursor2 += count2; 757 len2 -= count2; 758 if (len2 == 0) 759 break outer; 760 } 761 a[dest++] = tmp[cursor1++]; 762 if (--len1 == 1) 763 break outer; 764 minGallop--; 765 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); 766 if (minGallop < 0) 767 minGallop = 0; 768 minGallop += 2; // Penalize for leaving gallop mode 769 } // End of "outer" loop 770 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field 771 772 if (len1 == 1) { 773 assert len2 > 0; 774 System.arraycopy(a, cursor2, a, dest, len2); 775 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge 776 } else if (len1 == 0) { 777 throw new IllegalArgumentException( 778 "Comparison method violates its general contract!"); 779 } else { 780 assert len2 == 0; 781 assert len1 > 1; 782 System.arraycopy(tmp, cursor1, a, dest, len1); 783 } 784 } 785 786 /** 787 * Like mergeLo, except that this method should be called only if 788 * len1 >= len2; mergeLo should be called if len1 <= len2. (Either method 789 * may be called if len1 == len2.) 790 * 791 * @param base1 index of first element in first run to be merged 792 * @param len1 length of first run to be merged (must be > 0) 793 * @param base2 index of first element in second run to be merged 794 * (must be aBase + aLen) 795 * @param len2 length of second run to be merged (must be > 0) 796 */ 797 private void mergeHi(int base1, int len1, int base2, int len2) { 798 assert len1 > 0 && len2 > 0 && base1 + len1 == base2; 799 800 // Copy second run into temp array 801 T[] a = this.a; // For performance 802 T[] tmp = ensureCapacity(len2); 803 int tmpBase = this.tmpBase; 804 System.arraycopy(a, base2, tmp, tmpBase, len2); 805 806 int cursor1 = base1 + len1 - 1; // Indexes into a 807 int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array 808 int dest = base2 + len2 - 1; // Indexes into a 809 810 // Move last element of first run and deal with degenerate cases 811 a[dest--] = a[cursor1--]; 812 if (--len1 == 0) { 813 System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); 814 return; 815 } 816 if (len2 == 1) { 817 dest -= len1; 818 cursor1 -= len1; 819 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); 820 a[dest] = tmp[cursor2]; 821 return; 822 } 823 824 Comparator<? super T> c = this.c; // Use local variable for performance 825 int minGallop = this.minGallop; // " " " " " 826 outer: 827 while (true) { 828 int count1 = 0; // Number of times in a row that first run won 829 int count2 = 0; // Number of times in a row that second run won 830 831 /* 832 * Do the straightforward thing until (if ever) one run 833 * appears to win consistently. 834 */ 835 do { 836 assert len1 > 0 && len2 > 1; 837 if (c.compare(tmp[cursor2], a[cursor1]) < 0) { 838 a[dest--] = a[cursor1--]; 839 count1++; 840 count2 = 0; 841 if (--len1 == 0) 842 break outer; 843 } else { 844 a[dest--] = tmp[cursor2--]; 845 count2++; 846 count1 = 0; 847 if (--len2 == 1) 848 break outer; 849 } 850 } while ((count1 | count2) < minGallop); 851 852 /* 853 * One run is winning so consistently that galloping may be a 854 * huge win. So try that, and continue galloping until (if ever) 855 * neither run appears to be winning consistently anymore. 856 */ 857 do { 858 assert len1 > 0 && len2 > 1; 859 count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c); 860 if (count1 != 0) { 861 dest -= count1; 862 cursor1 -= count1; 863 len1 -= count1; 864 System.arraycopy(a, cursor1 + 1, a, dest + 1, count1); 865 if (len1 == 0) 866 break outer; 867 } 868 a[dest--] = tmp[cursor2--]; 869 if (--len2 == 1) 870 break outer; 871 872 count2 = len2 - gallopLeft(a[cursor1], tmp, tmpBase, len2, len2 - 1, c); 873 if (count2 != 0) { 874 dest -= count2; 875 cursor2 -= count2; 876 len2 -= count2; 877 System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2); 878 if (len2 <= 1) // len2 == 1 || len2 == 0 879 break outer; 880 } 881 a[dest--] = a[cursor1--]; 882 if (--len1 == 0) 883 break outer; 884 minGallop--; 885 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); 886 if (minGallop < 0) 887 minGallop = 0; 888 minGallop += 2; // Penalize for leaving gallop mode 889 } // End of "outer" loop 890 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field 891 892 if (len2 == 1) { 893 assert len1 > 0; 894 dest -= len1; 895 cursor1 -= len1; System.arraycopy(a, cursor1 + 1, a, dest + 1, len1)896 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); 897 a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge 898 } else if (len2 == 0) { 899 throw new IllegalArgumentException( 900 "Comparison method violates its general contract!"); 901 } else { 902 assert len1 == 0; 903 assert len2 > 0; 904 System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); 905 } 906 } 907 908 /** 909 * Ensures that the external array tmp has at least the specified 910 * number of elements, increasing its size if necessary. The size 911 * increases exponentially to ensure amortized linear time complexity. 912 * 913 * @param minCapacity the minimum required capacity of the tmp array 914 * @return tmp, whether or not it grew 915 */ 916 private T[] ensureCapacity(int minCapacity) { 917 if (tmpLen < minCapacity) { 918 // Compute smallest power of 2 > minCapacity 919 int newSize = minCapacity; 920 newSize |= newSize >> 1; 921 newSize |= newSize >> 2; 922 newSize |= newSize >> 4; 923 newSize |= newSize >> 8; 924 newSize |= newSize >> 16; 925 newSize++; 926 927 if (newSize < 0) // Not bloody likely! 928 newSize = minCapacity; 929 else 930 newSize = Math.min(newSize, a.length >>> 1); 931 932 @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) 933 T[] newArray = (T[])java.lang.reflect.Array.newInstance 934 (a.getClass().getComponentType(), newSize); 935 tmp = newArray; 936 tmpLen = newSize; 937 tmpBase = 0; 938 } 939 return tmp; 940 } 941 } 942