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