1 /* 2 * Copyright 2009 Google Inc. All Rights Reserved. 3 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. 4 * 5 * This code is free software; you can redistribute it and/or modify it 6 * under the terms of the GNU General Public License version 2 only, as 7 * published by the Free Software Foundation. Oracle designates this 8 * particular file as subject to the "Classpath" exception as provided 9 * by Oracle in the LICENSE file that accompanied this code. 10 * 11 * This code is distributed in the hope that it will be useful, but WITHOUT 12 * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or 13 * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License 14 * version 2 for more details (a copy is included in the LICENSE file that 15 * accompanied this code). 16 * 17 * You should have received a copy of the GNU General Public License version 18 * 2 along with this work; if not, write to the Free Software Foundation, 19 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. 20 * 21 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA 22 * or visit www.oracle.com if you need additional information or have any 23 * questions. 24 */ 25 26 package java.util; 27 28 /** 29 * A stable, adaptive, iterative mergesort that requires far fewer than 30 * n lg(n) comparisons when running on partially sorted arrays, while 31 * offering performance comparable to a traditional mergesort when run 32 * on random arrays. Like all proper mergesorts, this sort is stable and 33 * runs O(n log n) time (worst case). In the worst case, this sort requires 34 * temporary storage space for n/2 object references; in the best case, 35 * it requires only a small constant amount of space. 36 * 37 * This implementation was adapted from Tim Peters's list sort for 38 * Python, which is described in detail here: 39 * 40 * http://svn.python.org/projects/python/trunk/Objects/listsort.txt 41 * 42 * Tim's C code may be found here: 43 * 44 * http://svn.python.org/projects/python/trunk/Objects/listobject.c 45 * 46 * The underlying techniques are described in this paper (and may have 47 * even earlier origins): 48 * 49 * "Optimistic Sorting and Information Theoretic Complexity" 50 * Peter McIlroy 51 * SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms), 52 * pp 467-474, Austin, Texas, 25-27 January 1993. 53 * 54 * While the API to this class consists solely of static methods, it is 55 * (privately) instantiable; a TimSort instance holds the state of an ongoing 56 * sort, assuming the input array is large enough to warrant the full-blown 57 * TimSort. Small arrays are sorted in place, using a binary insertion sort. 58 * 59 * @author Josh Bloch 60 */ 61 class TimSort<T> { 62 /** 63 * This is the minimum sized sequence that will be merged. Shorter 64 * sequences will be lengthened by calling binarySort. If the entire 65 * array is less than this length, no merges will be performed. 66 * 67 * This constant should be a power of two. It was 64 in Tim Peter's C 68 * implementation, but 32 was empirically determined to work better in 69 * this implementation. In the unlikely event that you set this constant 70 * to be a number that's not a power of two, you'll need to change the 71 * {@link #minRunLength} computation. 72 * 73 * If you decrease this constant, you must change the stackLen 74 * computation in the TimSort constructor, or you risk an 75 * ArrayOutOfBounds exception. See listsort.txt for a discussion 76 * of the minimum stack length required as a function of the length 77 * of the array being sorted and the minimum merge sequence length. 78 */ 79 private static final int MIN_MERGE = 32; 80 81 /** 82 * The array being sorted. 83 */ 84 private final T[] a; 85 86 /** 87 * The comparator for this sort. 88 */ 89 private final Comparator<? super T> c; 90 91 /** 92 * When we get into galloping mode, we stay there until both runs win less 93 * often than MIN_GALLOP consecutive times. 94 */ 95 private static final int MIN_GALLOP = 7; 96 97 /** 98 * This controls when we get *into* galloping mode. It is initialized 99 * to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for 100 * random data, and lower for highly structured data. 101 */ 102 private int minGallop = MIN_GALLOP; 103 104 /** 105 * Maximum initial size of tmp array, which is used for merging. The array 106 * can grow to accommodate demand. 107 * 108 * Unlike Tim's original C version, we do not allocate this much storage 109 * when sorting smaller arrays. This change was required for performance. 110 */ 111 private static final int INITIAL_TMP_STORAGE_LENGTH = 256; 112 113 /** 114 * Temp storage for merges. A workspace array may optionally be 115 * provided in constructor, and if so will be used as long as it 116 * is big enough. 117 */ 118 private T[] tmp; 119 private int tmpBase; // base of tmp array slice 120 private int tmpLen; // length of tmp array slice 121 122 /** 123 * A stack of pending runs yet to be merged. Run i starts at 124 * address base[i] and extends for len[i] elements. It's always 125 * true (so long as the indices are in bounds) that: 126 * 127 * runBase[i] + runLen[i] == runBase[i + 1] 128 * 129 * so we could cut the storage for this, but it's a minor amount, 130 * and keeping all the info explicit simplifies the code. 131 */ 132 private int stackSize = 0; // Number of pending runs on stack 133 private final int[] runBase; 134 private final int[] runLen; 135 136 /** 137 * Creates a TimSort instance to maintain the state of an ongoing sort. 138 * 139 * @param a the array to be sorted 140 * @param c the comparator to determine the order of the sort 141 * @param work a workspace array (slice) 142 * @param workBase origin of usable space in work array 143 * @param workLen usable size of work array 144 */ 145 private TimSort(T[] a, Comparator<? super T> c, T[] work, int workBase, int workLen) { 146 this.a = a; 147 this.c = c; 148 149 // Allocate temp storage (which may be increased later if necessary) 150 int len = a.length; 151 int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ? 152 len >>> 1 : INITIAL_TMP_STORAGE_LENGTH; 153 if (work == null || workLen < tlen || workBase + tlen > work.length) { 154 @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) 155 T[] newArray = (T[])java.lang.reflect.Array.newInstance 156 (a.getClass().getComponentType(), tlen); 157 tmp = newArray; 158 tmpBase = 0; 159 tmpLen = tlen; 160 } 161 else { 162 tmp = work; 163 tmpBase = workBase; 164 tmpLen = workLen; 165 } 166 167 /* 168 * Allocate runs-to-be-merged stack (which cannot be expanded). The 169 * stack length requirements are described in listsort.txt. The C 170 * version always uses the same stack length (85), but this was 171 * measured to be too expensive when sorting "mid-sized" arrays (e.g., 172 * 100 elements) in Java. Therefore, we use smaller (but sufficiently 173 * large) stack lengths for smaller arrays. The "magic numbers" in the 174 * computation below must be changed if MIN_MERGE is decreased. See 175 * the MIN_MERGE declaration above for more information. 176 */ 177 int stackLen = (len < 120 ? 5 : 178 len < 1542 ? 10 : 179 len < 119151 ? 19 : 40); 180 runBase = new int[stackLen]; 181 runLen = new int[stackLen]; 182 } 183 184 /* 185 * The next method (package private and static) constitutes the 186 * entire API of this class. 187 */ 188 189 /** 190 * Sorts the given range, using the given workspace array slice 191 * for temp storage when possible. This method is designed to be 192 * invoked from public methods (in class Arrays) after performing 193 * any necessary array bounds checks and expanding parameters into 194 * the required forms. 195 * 196 * @param a the array to be sorted 197 * @param lo the index of the first element, inclusive, to be sorted 198 * @param hi the index of the last element, exclusive, to be sorted 199 * @param c the comparator to use 200 * @param work a workspace array (slice) 201 * @param workBase origin of usable space in work array 202 * @param workLen usable size of work array 203 * @since 1.8 204 */ 205 static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c, 206 T[] work, int workBase, int workLen) { 207 assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length; 208 209 int nRemaining = hi - lo; 210 if (nRemaining < 2) 211 return; // Arrays of size 0 and 1 are always sorted 212 213 // If array is small, do a "mini-TimSort" with no merges 214 if (nRemaining < MIN_MERGE) { 215 int initRunLen = countRunAndMakeAscending(a, lo, hi, c); 216 binarySort(a, lo, hi, lo + initRunLen, c); 217 return; 218 } 219 220 /** 221 * March over the array once, left to right, finding natural runs, 222 * extending short natural runs to minRun elements, and merging runs 223 * to maintain stack invariant. 224 */ 225 TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen); 226 int minRun = minRunLength(nRemaining); 227 do { 228 // Identify next run 229 int runLen = countRunAndMakeAscending(a, lo, hi, c); 230 231 // If run is short, extend to min(minRun, nRemaining) 232 if (runLen < minRun) { 233 int force = nRemaining <= minRun ? nRemaining : minRun; 234 binarySort(a, lo, lo + force, lo + runLen, c); 235 runLen = force; 236 } 237 238 // Push run onto pending-run stack, and maybe merge 239 ts.pushRun(lo, runLen); 240 ts.mergeCollapse(); 241 242 // Advance to find next run 243 lo += runLen; 244 nRemaining -= runLen; 245 } while (nRemaining != 0); 246 247 // Merge all remaining runs to complete sort 248 assert lo == hi; 249 ts.mergeForceCollapse(); 250 assert ts.stackSize == 1; 251 } 252 253 /** 254 * Sorts the specified portion of the specified array using a binary 255 * insertion sort. This is the best method for sorting small numbers 256 * of elements. It requires O(n log n) compares, but O(n^2) data 257 * movement (worst case). 258 * 259 * If the initial part of the specified range is already sorted, 260 * this method can take advantage of it: the method assumes that the 261 * elements from index {@code lo}, inclusive, to {@code start}, 262 * exclusive are already sorted. 263 * 264 * @param a the array in which a range is to be sorted 265 * @param lo the index of the first element in the range to be sorted 266 * @param hi the index after the last element in the range to be sorted 267 * @param start the index of the first element in the range that is 268 * not already known to be sorted ({@code lo <= start <= hi}) 269 * @param c comparator to used for the sort 270 */ 271 @SuppressWarnings("fallthrough") 272 private static <T> void binarySort(T[] a, int lo, int hi, int start, 273 Comparator<? super T> c) { 274 assert lo <= start && start <= hi; 275 if (start == lo) 276 start++; 277 for ( ; start < hi; start++) { 278 T pivot = a[start]; 279 280 // Set left (and right) to the index where a[start] (pivot) belongs 281 int left = lo; 282 int right = start; 283 assert left <= right; 284 /* 285 * Invariants: 286 * pivot >= all in [lo, left). 287 * pivot < all in [right, start). 288 */ 289 while (left < right) { 290 int mid = (left + right) >>> 1; 291 if (c.compare(pivot, a[mid]) < 0) 292 right = mid; 293 else 294 left = mid + 1; 295 } 296 assert left == right; 297 298 /* 299 * The invariants still hold: pivot >= all in [lo, left) and 300 * pivot < all in [left, start), so pivot belongs at left. Note 301 * that if there are elements equal to pivot, left points to the 302 * first slot after them -- that's why this sort is stable. 303 * Slide elements over to make room for pivot. 304 */ 305 int n = start - left; // The number of elements to move 306 // Switch is just an optimization for arraycopy in default case 307 switch (n) { 308 case 2: a[left + 2] = a[left + 1]; 309 case 1: a[left + 1] = a[left]; 310 break; 311 default: System.arraycopy(a, left, a, left + 1, n); 312 } 313 a[left] = pivot; 314 } 315 } 316 317 /** 318 * Returns the length of the run beginning at the specified position in 319 * the specified array and reverses the run if it is descending (ensuring 320 * that the run will always be ascending when the method returns). 321 * 322 * A run is the longest ascending sequence with: 323 * 324 * a[lo] <= a[lo + 1] <= a[lo + 2] <= ... 325 * 326 * or the longest descending sequence with: 327 * 328 * a[lo] > a[lo + 1] > a[lo + 2] > ... 329 * 330 * For its intended use in a stable mergesort, the strictness of the 331 * definition of "descending" is needed so that the call can safely 332 * reverse a descending sequence without violating stability. 333 * 334 * @param a the array in which a run is to be counted and possibly reversed 335 * @param lo index of the first element in the run 336 * @param hi index after the last element that may be contained in the run. 337 It is required that {@code lo < hi}. 338 * @param c the comparator to used for the sort 339 * @return the length of the run beginning at the specified position in 340 * the specified array 341 */ 342 private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi, 343 Comparator<? super T> c) { 344 assert lo < hi; 345 int runHi = lo + 1; 346 if (runHi == hi) 347 return 1; 348 349 // Find end of run, and reverse range if descending 350 if (c.compare(a[runHi++], a[lo]) < 0) { // Descending 351 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0) 352 runHi++; 353 reverseRange(a, lo, runHi); 354 } else { // Ascending 355 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0) 356 runHi++; 357 } 358 359 return runHi - lo; 360 } 361 362 /** 363 * Reverse the specified range of the specified array. 364 * 365 * @param a the array in which a range is to be reversed 366 * @param lo the index of the first element in the range to be reversed 367 * @param hi the index after the last element in the range to be reversed 368 */ 369 private static void reverseRange(Object[] a, int lo, int hi) { 370 hi--; 371 while (lo < hi) { 372 Object t = a[lo]; 373 a[lo++] = a[hi]; 374 a[hi--] = t; 375 } 376 } 377 378 /** 379 * Returns the minimum acceptable run length for an array of the specified 380 * length. Natural runs shorter than this will be extended with 381 * {@link #binarySort}. 382 * 383 * Roughly speaking, the computation is: 384 * 385 * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff). 386 * Else if n is an exact power of 2, return MIN_MERGE/2. 387 * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k 388 * is close to, but strictly less than, an exact power of 2. 389 * 390 * For the rationale, see listsort.txt. 391 * 392 * @param n the length of the array to be sorted 393 * @return the length of the minimum run to be merged 394 */ 395 private static int minRunLength(int n) { 396 assert n >= 0; 397 int r = 0; // Becomes 1 if any 1 bits are shifted off 398 while (n >= MIN_MERGE) { 399 r |= (n & 1); 400 n >>= 1; 401 } 402 return n + r; 403 } 404 405 /** 406 * Pushes the specified run onto the pending-run stack. 407 * 408 * @param runBase index of the first element in the run 409 * @param runLen the number of elements in the run 410 */ 411 private void pushRun(int runBase, int runLen) { 412 this.runBase[stackSize] = runBase; 413 this.runLen[stackSize] = runLen; 414 stackSize++; 415 } 416 417 /** 418 * Examines the stack of runs waiting to be merged and merges adjacent runs 419 * until the stack invariants are reestablished: 420 * 421 * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] 422 * 2. runLen[i - 2] > runLen[i - 1] 423 * 424 * This method is called each time a new run is pushed onto the stack, 425 * so the invariants are guaranteed to hold for i < stackSize upon 426 * entry to the method. 427 */ 428 private void mergeCollapse() { 429 while (stackSize > 1) { 430 int n = stackSize - 2; 431 if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) { 432 if (runLen[n - 1] < runLen[n + 1]) 433 n--; 434 mergeAt(n); 435 } else if (runLen[n] <= runLen[n + 1]) { 436 mergeAt(n); 437 } else { 438 break; // Invariant is established 439 } 440 } 441 } 442 443 /** 444 * Merges all runs on the stack until only one remains. This method is 445 * called once, to complete the sort. 446 */ 447 private void mergeForceCollapse() { 448 while (stackSize > 1) { 449 int n = stackSize - 2; 450 if (n > 0 && runLen[n - 1] < runLen[n + 1]) 451 n--; 452 mergeAt(n); 453 } 454 } 455 456 /** 457 * Merges the two runs at stack indices i and i+1. Run i must be 458 * the penultimate or antepenultimate run on the stack. In other words, 459 * i must be equal to stackSize-2 or stackSize-3. 460 * 461 * @param i stack index of the first of the two runs to merge 462 */ 463 private void mergeAt(int i) { 464 assert stackSize >= 2; 465 assert i >= 0; 466 assert i == stackSize - 2 || i == stackSize - 3; 467 468 int base1 = runBase[i]; 469 int len1 = runLen[i]; 470 int base2 = runBase[i + 1]; 471 int len2 = runLen[i + 1]; 472 assert len1 > 0 && len2 > 0; 473 assert base1 + len1 == base2; 474 475 /* 476 * Record the length of the combined runs; if i is the 3rd-last 477 * run now, also slide over the last run (which isn't involved 478 * in this merge). The current run (i+1) goes away in any case. 479 */ 480 runLen[i] = len1 + len2; 481 if (i == stackSize - 3) { 482 runBase[i + 1] = runBase[i + 2]; 483 runLen[i + 1] = runLen[i + 2]; 484 } 485 stackSize--; 486 487 /* 488 * Find where the first element of run2 goes in run1. Prior elements 489 * in run1 can be ignored (because they're already in place). 490 */ 491 int k = gallopRight(a[base2], a, base1, len1, 0, c); 492 assert k >= 0; 493 base1 += k; 494 len1 -= k; 495 if (len1 == 0) 496 return; 497 498 /* 499 * Find where the last element of run1 goes in run2. Subsequent elements 500 * in run2 can be ignored (because they're already in place). 501 */ 502 len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c); 503 assert len2 >= 0; 504 if (len2 == 0) 505 return; 506 507 // Merge remaining runs, using tmp array with min(len1, len2) elements 508 if (len1 <= len2) 509 mergeLo(base1, len1, base2, len2); 510 else 511 mergeHi(base1, len1, base2, len2); 512 } 513 514 /** 515 * Locates the position at which to insert the specified key into the 516 * specified sorted range; if the range contains an element equal to key, 517 * returns the index of the leftmost equal element. 518 * 519 * @param key the key whose insertion point to search for 520 * @param a the array in which to search 521 * @param base the index of the first element in the range 522 * @param len the length of the range; must be > 0 523 * @param hint the index at which to begin the search, 0 <= hint < n. 524 * The closer hint is to the result, the faster this method will run. 525 * @param c the comparator used to order the range, and to search 526 * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k], 527 * pretending that a[b - 1] is minus infinity and a[b + n] is infinity. 528 * In other words, key belongs at index b + k; or in other words, 529 * the first k elements of a should precede key, and the last n - k 530 * should follow it. 531 */ 532 private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint, 533 Comparator<? super T> c) { 534 assert len > 0 && hint >= 0 && hint < len; 535 int lastOfs = 0; 536 int ofs = 1; 537 if (c.compare(key, a[base + hint]) > 0) { 538 // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs] 539 int maxOfs = len - hint; 540 while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) { 541 lastOfs = ofs; 542 ofs = (ofs << 1) + 1; 543 if (ofs <= 0) // int overflow 544 ofs = maxOfs; 545 } 546 if (ofs > maxOfs) 547 ofs = maxOfs; 548 549 // Make offsets relative to base 550 lastOfs += hint; 551 ofs += hint; 552 } else { // key <= a[base + hint] 553 // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs] 554 final int maxOfs = hint + 1; 555 while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) { 556 lastOfs = ofs; 557 ofs = (ofs << 1) + 1; 558 if (ofs <= 0) // int overflow 559 ofs = maxOfs; 560 } 561 if (ofs > maxOfs) 562 ofs = maxOfs; 563 564 // Make offsets relative to base 565 int tmp = lastOfs; 566 lastOfs = hint - ofs; 567 ofs = hint - tmp; 568 } 569 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; 570 571 /* 572 * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere 573 * to the right of lastOfs but no farther right than ofs. Do a binary 574 * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs]. 575 */ 576 lastOfs++; 577 while (lastOfs < ofs) { 578 int m = lastOfs + ((ofs - lastOfs) >>> 1); 579 580 if (c.compare(key, a[base + m]) > 0) 581 lastOfs = m + 1; // a[base + m] < key 582 else 583 ofs = m; // key <= a[base + m] 584 } 585 assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs] 586 return ofs; 587 } 588 589 /** 590 * Like gallopLeft, except that if the range contains an element equal to 591 * key, gallopRight returns the index after the rightmost equal element. 592 * 593 * @param key the key whose insertion point to search for 594 * @param a the array in which to search 595 * @param base the index of the first element in the range 596 * @param len the length of the range; must be > 0 597 * @param hint the index at which to begin the search, 0 <= hint < n. 598 * The closer hint is to the result, the faster this method will run. 599 * @param c the comparator used to order the range, and to search 600 * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k] 601 */ 602 private static <T> int gallopRight(T key, T[] a, int base, int len, 603 int hint, Comparator<? super T> c) { 604 assert len > 0 && hint >= 0 && hint < len; 605 606 int ofs = 1; 607 int lastOfs = 0; 608 if (c.compare(key, a[base + hint]) < 0) { 609 // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs] 610 int maxOfs = hint + 1; 611 while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) { 612 lastOfs = ofs; 613 ofs = (ofs << 1) + 1; 614 if (ofs <= 0) // int overflow 615 ofs = maxOfs; 616 } 617 if (ofs > maxOfs) 618 ofs = maxOfs; 619 620 // Make offsets relative to b 621 int tmp = lastOfs; 622 lastOfs = hint - ofs; 623 ofs = hint - tmp; 624 } else { // a[b + hint] <= key 625 // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs] 626 int maxOfs = len - hint; 627 while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) { 628 lastOfs = ofs; 629 ofs = (ofs << 1) + 1; 630 if (ofs <= 0) // int overflow 631 ofs = maxOfs; 632 } 633 if (ofs > maxOfs) 634 ofs = maxOfs; 635 636 // Make offsets relative to b 637 lastOfs += hint; 638 ofs += hint; 639 } 640 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; 641 642 /* 643 * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to 644 * the right of lastOfs but no farther right than ofs. Do a binary 645 * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs]. 646 */ 647 lastOfs++; 648 while (lastOfs < ofs) { 649 int m = lastOfs + ((ofs - lastOfs) >>> 1); 650 651 if (c.compare(key, a[base + m]) < 0) 652 ofs = m; // key < a[b + m] 653 else 654 lastOfs = m + 1; // a[b + m] <= key 655 } 656 assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs] 657 return ofs; 658 } 659 660 /** 661 * Merges two adjacent runs in place, in a stable fashion. The first 662 * element of the first run must be greater than the first element of the 663 * second run (a[base1] > a[base2]), and the last element of the first run 664 * (a[base1 + len1-1]) must be greater than all elements of the second run. 665 * 666 * For performance, this method should be called only when len1 <= len2; 667 * its twin, mergeHi should be called if len1 >= len2. (Either method 668 * may be called if len1 == len2.) 669 * 670 * @param base1 index of first element in first run to be merged 671 * @param len1 length of first run to be merged (must be > 0) 672 * @param base2 index of first element in second run to be merged 673 * (must be aBase + aLen) 674 * @param len2 length of second run to be merged (must be > 0) 675 */ 676 private void mergeLo(int base1, int len1, int base2, int len2) { 677 assert len1 > 0 && len2 > 0 && base1 + len1 == base2; 678 679 // Copy first run into temp array 680 T[] a = this.a; // For performance 681 T[] tmp = ensureCapacity(len1); 682 int cursor1 = tmpBase; // Indexes into tmp array 683 int cursor2 = base2; // Indexes int a 684 int dest = base1; // Indexes int a 685 System.arraycopy(a, base1, tmp, cursor1, len1); 686 687 // Move first element of second run and deal with degenerate cases 688 a[dest++] = a[cursor2++]; 689 if (--len2 == 0) { 690 System.arraycopy(tmp, cursor1, a, dest, len1); 691 return; 692 } 693 if (len1 == 1) { 694 System.arraycopy(a, cursor2, a, dest, len2); 695 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge 696 return; 697 } 698 699 Comparator<? super T> c = this.c; // Use local variable for performance 700 int minGallop = this.minGallop; // " " " " " 701 outer: 702 while (true) { 703 int count1 = 0; // Number of times in a row that first run won 704 int count2 = 0; // Number of times in a row that second run won 705 706 /* 707 * Do the straightforward thing until (if ever) one run starts 708 * winning consistently. 709 */ 710 do { 711 assert len1 > 1 && len2 > 0; 712 if (c.compare(a[cursor2], tmp[cursor1]) < 0) { 713 a[dest++] = a[cursor2++]; 714 count2++; 715 count1 = 0; 716 if (--len2 == 0) 717 break outer; 718 } else { 719 a[dest++] = tmp[cursor1++]; 720 count1++; 721 count2 = 0; 722 if (--len1 == 1) 723 break outer; 724 } 725 } while ((count1 | count2) < minGallop); 726 727 /* 728 * One run is winning so consistently that galloping may be a 729 * huge win. So try that, and continue galloping until (if ever) 730 * neither run appears to be winning consistently anymore. 731 */ 732 do { 733 assert len1 > 1 && len2 > 0; 734 count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c); 735 if (count1 != 0) { 736 System.arraycopy(tmp, cursor1, a, dest, count1); 737 dest += count1; 738 cursor1 += count1; 739 len1 -= count1; 740 if (len1 <= 1) // len1 == 1 || len1 == 0 741 break outer; 742 } 743 a[dest++] = a[cursor2++]; 744 if (--len2 == 0) 745 break outer; 746 747 count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c); 748 if (count2 != 0) { 749 System.arraycopy(a, cursor2, a, dest, count2); 750 dest += count2; 751 cursor2 += count2; 752 len2 -= count2; 753 if (len2 == 0) 754 break outer; 755 } 756 a[dest++] = tmp[cursor1++]; 757 if (--len1 == 1) 758 break outer; 759 minGallop--; 760 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); 761 if (minGallop < 0) 762 minGallop = 0; 763 minGallop += 2; // Penalize for leaving gallop mode 764 } // End of "outer" loop 765 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field 766 767 if (len1 == 1) { 768 assert len2 > 0; 769 System.arraycopy(a, cursor2, a, dest, len2); 770 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge 771 } else if (len1 == 0) { 772 throw new IllegalArgumentException( 773 "Comparison method violates its general contract!"); 774 } else { 775 assert len2 == 0; 776 assert len1 > 1; 777 System.arraycopy(tmp, cursor1, a, dest, len1); 778 } 779 } 780 781 /** 782 * Like mergeLo, except that this method should be called only if 783 * len1 >= len2; mergeLo should be called if len1 <= len2. (Either method 784 * may be called if len1 == len2.) 785 * 786 * @param base1 index of first element in first run to be merged 787 * @param len1 length of first run to be merged (must be > 0) 788 * @param base2 index of first element in second run to be merged 789 * (must be aBase + aLen) 790 * @param len2 length of second run to be merged (must be > 0) 791 */ 792 private void mergeHi(int base1, int len1, int base2, int len2) { 793 assert len1 > 0 && len2 > 0 && base1 + len1 == base2; 794 795 // Copy second run into temp array 796 T[] a = this.a; // For performance 797 T[] tmp = ensureCapacity(len2); 798 int tmpBase = this.tmpBase; 799 System.arraycopy(a, base2, tmp, tmpBase, len2); 800 801 int cursor1 = base1 + len1 - 1; // Indexes into a 802 int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array 803 int dest = base2 + len2 - 1; // Indexes into a 804 805 // Move last element of first run and deal with degenerate cases 806 a[dest--] = a[cursor1--]; 807 if (--len1 == 0) { 808 System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); 809 return; 810 } 811 if (len2 == 1) { 812 dest -= len1; 813 cursor1 -= len1; 814 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); 815 a[dest] = tmp[cursor2]; 816 return; 817 } 818 819 Comparator<? super T> c = this.c; // Use local variable for performance 820 int minGallop = this.minGallop; // " " " " " 821 outer: 822 while (true) { 823 int count1 = 0; // Number of times in a row that first run won 824 int count2 = 0; // Number of times in a row that second run won 825 826 /* 827 * Do the straightforward thing until (if ever) one run 828 * appears to win consistently. 829 */ 830 do { 831 assert len1 > 0 && len2 > 1; 832 if (c.compare(tmp[cursor2], a[cursor1]) < 0) { 833 a[dest--] = a[cursor1--]; 834 count1++; 835 count2 = 0; 836 if (--len1 == 0) 837 break outer; 838 } else { 839 a[dest--] = tmp[cursor2--]; 840 count2++; 841 count1 = 0; 842 if (--len2 == 1) 843 break outer; 844 } 845 } while ((count1 | count2) < minGallop); 846 847 /* 848 * One run is winning so consistently that galloping may be a 849 * huge win. So try that, and continue galloping until (if ever) 850 * neither run appears to be winning consistently anymore. 851 */ 852 do { 853 assert len1 > 0 && len2 > 1; 854 count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c); 855 if (count1 != 0) { 856 dest -= count1; 857 cursor1 -= count1; 858 len1 -= count1; 859 System.arraycopy(a, cursor1 + 1, a, dest + 1, count1); 860 if (len1 == 0) 861 break outer; 862 } 863 a[dest--] = tmp[cursor2--]; 864 if (--len2 == 1) 865 break outer; 866 867 count2 = len2 - gallopLeft(a[cursor1], tmp, tmpBase, len2, len2 - 1, c); 868 if (count2 != 0) { 869 dest -= count2; 870 cursor2 -= count2; 871 len2 -= count2; 872 System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2); 873 if (len2 <= 1) // len2 == 1 || len2 == 0 874 break outer; 875 } 876 a[dest--] = a[cursor1--]; 877 if (--len1 == 0) 878 break outer; 879 minGallop--; 880 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); 881 if (minGallop < 0) 882 minGallop = 0; 883 minGallop += 2; // Penalize for leaving gallop mode 884 } // End of "outer" loop 885 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field 886 887 if (len2 == 1) { 888 assert len1 > 0; 889 dest -= len1; 890 cursor1 -= len1; 891 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); 892 a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge 893 } else if (len2 == 0) { 894 throw new IllegalArgumentException( 895 "Comparison method violates its general contract!"); 896 } else { 897 assert len1 == 0; 898 assert len2 > 0; 899 System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); 900 } 901 } 902 903 /** 904 * Ensures that the external array tmp has at least the specified 905 * number of elements, increasing its size if necessary. The size 906 * increases exponentially to ensure amortized linear time complexity. 907 * 908 * @param minCapacity the minimum required capacity of the tmp array 909 * @return tmp, whether or not it grew 910 */ 911 private T[] ensureCapacity(int minCapacity) { 912 if (tmpLen < minCapacity) { 913 // Compute smallest power of 2 > minCapacity 914 int newSize = minCapacity; 915 newSize |= newSize >> 1; 916 newSize |= newSize >> 2; 917 newSize |= newSize >> 4; 918 newSize |= newSize >> 8; 919 newSize |= newSize >> 16; 920 newSize++; 921 922 if (newSize < 0) // Not bloody likely! 923 newSize = minCapacity; 924 else 925 newSize = Math.min(newSize, a.length >>> 1); 926 927 @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) 928 T[] newArray = (T[])java.lang.reflect.Array.newInstance 929 (a.getClass().getComponentType(), newSize); 930 tmp = newArray; 931 tmpLen = newSize; 932 tmpBase = 0; 933 } 934 return tmp; 935 } 936 }