1 /*JDK 1.8 2 */ 3 4 package java.util; 5 6 /** 7 * A stable, adaptive, iterative mergesort that requires far fewer than 8 * n lg(n) comparisons when running on partially sorted arrays, while 9 * offering performance comparable to a traditional mergesort when run 10 * on random arrays. Like all proper mergesorts, this sort is stable and 11 * runs O(n log n) time (worst case). In the worst case, this sort requires 12 * temporary storage space for n/2 object references; in the best case, 13 * it requires only a small constant amount of space. 14 * 15 * This implementation was adapted from Tim Peters's list sort for 16 * Python, which is described in detail here: 17 * 18 * http://svn.python.org/projects/python/trunk/Objects/listsort.txt 19 * 20 * Tim's C code may be found here: 21 * 22 * http://svn.python.org/projects/python/trunk/Objects/listobject.c 23 * 24 * The underlying techniques are described in this paper (and may have 25 * even earlier origins): 26 * 27 * "Optimistic Sorting and Information Theoretic Complexity" 28 * Peter McIlroy 29 * SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms), 30 * pp 467-474, Austin, Texas, 25-27 January 1993. 31 * 32 * While the API to this class consists solely of static methods, it is 33 * (privately) instantiable; a TimSort instance holds the state of an ongoing 34 * sort, assuming the input array is large enough to warrant the full-blown 35 * TimSort. Small arrays are sorted in place, using a binary insertion sort. 36 * 37 * @author Josh Bloch 38 */ 39 class TimSort<T> { 40 /** 41 * This is the minimum sized sequence that will be merged. Shorter 42 * sequences will be lengthened by calling binarySort. If the entire 43 * array is less than this length, no merges will be performed. 44 * 45 * This constant should be a power of two. It was 64 in Tim Peter's C 46 * implementation, but 32 was empirically determined to work better in 47 * this implementation. In the unlikely event that you set this constant 48 * to be a number that's not a power of two, you'll need to change the 49 * {@link #minRunLength} computation. 50 * 51 * If you decrease this constant, you must change the stackLen 52 * computation in the TimSort constructor, or you risk an 53 * ArrayOutOfBounds exception. See listsort.txt for a discussion 54 * of the minimum stack length required as a function of the length 55 * of the array being sorted and the minimum merge sequence length. 56 */ 57 private static final int MIN_MERGE = 32; 58 59 /** 60 * The array being sorted. 61 */ 62 private final T[] a; 63 64 /** 65 * The comparator for this sort. 66 */ 67 private final Comparator<? super T> c; 68 69 /** 70 * When we get into galloping mode, we stay there until both runs win less 71 * often than MIN_GALLOP consecutive times. 72 */ 73 private static final int MIN_GALLOP = 7; 74 75 /** 76 * This controls when we get *into* galloping mode. It is initialized 77 * to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for 78 * random data, and lower for highly structured data. 79 */ 80 private int minGallop = MIN_GALLOP; 81 82 /** 83 * Maximum initial size of tmp array, which is used for merging. The array 84 * can grow to accommodate demand. 85 * 86 * Unlike Tim's original C version, we do not allocate this much storage 87 * when sorting smaller arrays. This change was required for performance. 88 */ 89 private static final int INITIAL_TMP_STORAGE_LENGTH = 256; 90 91 /** 92 * Temp storage for merges. A workspace array may optionally be 93 * provided in constructor, and if so will be used as long as it 94 * is big enough. 95 */ 96 private T[] tmp; 97 private int tmpBase; // base of tmp array slice 98 private int tmpLen; // length of tmp array slice 99 100 /** 101 * A stack of pending runs yet to be merged. Run i starts at 102 * address base[i] and extends for len[i] elements. It's always 103 * true (so long as the indices are in bounds) that: 104 * 105 * runBase[i] + runLen[i] == runBase[i + 1] 106 * 107 * so we could cut the storage for this, but it's a minor amount, 108 * and keeping all the info explicit simplifies the code. 109 */ 110 private int stackSize = 0; // Number of pending runs on stack 111 private final int[] runBase; 112 private final int[] runLen; 113 114 /** 115 * Creates a TimSort instance to maintain the state of an ongoing sort. 116 * 117 * @param a the array to be sorted 118 * @param c the comparator to determine the order of the sort 119 * @param work a workspace array (slice) 120 * @param workBase origin of usable space in work array 121 * @param workLen usable size of work array 122 */ 123 private TimSort(T[] a, Comparator<? super T> c, T[] work, int workBase, int workLen) { 124 this.a = a; 125 this.c = c; 126 127 // Allocate temp storage (which may be increased later if necessary) 128 int len = a.length; 129 // 肯定臨時數組的長度， 若是低於默認值256的2倍， 則空間大小爲原始數組a的長度乘以2， 不然爲默認長度 130 int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ? 131 len >>> 1 : INITIAL_TMP_STORAGE_LENGTH; 132 if (work == null || workLen < tlen || workBase + tlen > work.length) { 133 @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) 134 T[] newArray = (T[])java.lang.reflect.Array.newInstance 135 (a.getClass().getComponentType(), tlen); 136 tmp = newArray; 137 tmpBase = 0; 138 tmpLen = tlen; 139 } 140 else { 141 // 當指定的work數組不爲空， 且workLen大於計算出的tlen的長度， 而且work數組的有效長度大於tlen的長度時， 使用指定的臨時數組 142 tmp = work; 143 tmpBase = workBase; 144 tmpLen = workLen; 145 } 146 147 /* 148 * Allocate runs-to-be-merged stack (which cannot be expanded). The 149 * stack length requirements are described in listsort.txt. The C 150 * version always uses the same stack length (85), but this was 151 * measured to be too expensive when sorting "mid-sized" arrays (e.g., 152 * 100 elements) in Java. Therefore, we use smaller (but sufficiently 153 * large) stack lengths for smaller arrays. The "magic numbers" in the 154 * computation below must be changed if MIN_MERGE is decreased. See 155 * the MIN_MERGE declaration above for more information. 156 * The maximum value of 49 allows for an array up to length 157 * Integer.MAX_VALUE-4, if array is filled by the worst case stack size 158 * increasing scenario. More explanations are given in section 4 of: 159 * http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf 160 */ 161 int stackLen = (len < 120 ? 5 : 162 len < 1542 ? 10 : 163 len < 119151 ? 24 : 49); 164 runBase = new int[stackLen]; 165 runLen = new int[stackLen]; 166 } 167 168 /* 169 * The next method (package private and static) constitutes the 170 * entire API of this class. 171 */ 172 173 /** 174 * Sorts the given range, using the given workspace array slice 175 * for temp storage when possible. This method is designed to be 176 * invoked from public methods (in class Arrays) after performing 177 * any necessary array bounds checks and expanding parameters into 178 * the required forms. 179 * 180 * @param a the array to be sorted 181 * @param lo the index of the first element, inclusive, to be sorted 182 * @param hi the index of the last element, exclusive, to be sorted 183 * @param c the comparator to use 184 * @param work a workspace array (slice) 185 * @param workBase origin of usable space in work array 186 * @param workLen usable size of work array 187 * @since 1.8 188 */ 189 static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c, 190 T[] work, int workBase, int workLen) { 191 assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length; 192 193 int nRemaining = hi - lo; //總共的待排序的元素個數 194 if (nRemaining < 2) 195 return; // Arrays of size 0 and 1 are always sorted 196 197 // If array is small, do a "mini-TimSort" with no merges 198 // 當元素個數小於7時， 使用折半插入排序， 由於插入排序對於元素個數少的數組更快。 199 if (nRemaining < MIN_MERGE) { 200 // initRunLen是初始的有序的元素的個數，而從索引位置lo + initRunLen開始， 後面爲亂序的元素。 一種優化方法， lo + initRunLen以前的不用排序了 201 int initRunLen = countRunAndMakeAscending(a, lo, hi, c); 202 // 折半插入排序 203 binarySort(a, lo, hi, lo + initRunLen, c); 204 return; 205 } 206 207 /** 208 * March over the array once, left to right, finding natural runs, 209 * extending short natural runs to minRun elements, and merging runs 210 * to maintain stack invariant. 211 */ 212 // 新建一個TimSort實例， 存儲運行時的狀態， 好比臨時的run(一個run即一個有序的數組) 213 TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen); 214 // 查找一個minRun值， 小於minRun時使用折半插入排序，MIN_MERGE/2 <= minRun <= MIN_MERGE 215 int minRun = minRunLength(nRemaining); 216 do { 217 // Identify next run 218 // 找到下一個有序的數組的長度 219 int runLen = countRunAndMakeAscending(a, lo, hi, c); 220 221 // If run is short, extend to min(minRun, nRemaining) 222 if (runLen < minRun) { 223 // 使用折半插入排序擴展數組， 使之達到minRun， 由於元素個數小於minRun， 折半插入排序更快速 224 int force = nRemaining <= minRun ? nRemaining : minRun; 225 binarySort(a, lo, lo + force, lo + runLen, c); 226 runLen = force; 227 } 228 229 // Push run onto pending-run stack, and maybe merge 230 // 把這個有序數組壓入棧中 231 ts.pushRun(lo, runLen); 232 /** 233 * 判斷當 234 * runLen[i - 3] <= runLen[i - 2] + runLen[i - 1] 235 * 且 runLen[i-3] < runLen[i-1]時 236 * 或 237 * runLen[i - 2] <= runLen[i - 1] 238 * 合併較小的兩個有序數組， 以達到最大的平衡（即每一個數組大小基本相同） 239 */ 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 //合併剩餘數組 250 ts.mergeForceCollapse(); 251 assert ts.stackSize == 1; 252 } 253 254 /** 255 * Sorts the specified portion of the specified array using a binary 256 * insertion sort. This is the best method for sorting small numbers 257 * of elements. It requires O(n log n) compares, but O(n^2) data 258 * movement (worst case). 259 * 260 * If the initial part of the specified range is already sorted, 261 * this method can take advantage of it: the method assumes that the 262 * elements from index {@code lo}, inclusive, to {@code start}, 263 * exclusive are already sorted. 264 * 265 * @param a the array in which a range is to be sorted 266 * @param lo the index of the first element in the range to be sorted 267 * @param hi the index after the last element in the range to be sorted 268 * @param start the index of the first element in the range that is 269 * not already known to be sorted ({@code lo <= start <= hi}) 270 * @param c comparator to used for the sort 271 */ 272 @SuppressWarnings("fallthrough") 273 private static <T> void binarySort(T[] a, int lo, int hi, int start, 274 Comparator<? super T> c) { 275 assert lo <= start && start <= hi; 276 // start以前的有序元素直接略過 277 if (start == lo) 278 start++; 279 // 從start到hi， 使用折半插入排序進行數組排序 280 for ( ; start < hi; start++) { 281 //待插入的元素 282 T pivot = a[start]; 283 284 // Set left (and right) to the index where a[start] (pivot) belongs 285 // 從left到right， 找到插入位置 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 // left即爲最終的插入位置， 由於start>=lo && start <=hi， 因此最終必定會找到一個位置使得pivot>=a[mid]， 所以最終必定是pivot >= right， 所以最終爲left的位置， 即mid+1 302 assert left == right; 303 304 /* 305 * The invariants still hold: pivot >= all in [lo, left) and 306 * pivot < all in [left, start), so pivot belongs at left. Note 307 * that if there are elements equal to pivot, left points to the 308 * first slot after them -- that's why this sort is stable. 309 * Slide elements over to make room for pivot. 310 */ 311 int n = start - left; // The number of elements to move 312 // Switch is just an optimization for arraycopy in default case 313 switch (n) { 314 case 2: a[left + 2] = a[left + 1]; // 若是待移動元素個數小於等於2則直接移動 315 case 1: a[left + 1] = a[left]; 316 break; 317 default: System.arraycopy(a, left, a, left + 1, n); // 從left開始日後移， 而後把start位置的元素插入到原來的left的位置 318 } 319 a[left] = pivot; 320 } 321 } 322 323 /** 324 * Returns the length of the run beginning at the specified position in 325 * the specified array and reverses the run if it is descending (ensuring 326 * that the run will always be ascending when the method returns). 327 * 328 * A run is the longest ascending sequence with: 329 * 330 * a[lo] <= a[lo + 1] <= a[lo + 2] <= ... 331 * 332 * or the longest descending sequence with: 333 * 334 * a[lo] > a[lo + 1] > a[lo + 2] > ... 335 * 336 * For its intended use in a stable mergesort, the strictness of the 337 * definition of "descending" is needed so that the call can safely 338 * reverse a descending sequence without violating stability. 339 * 340 * @param a the array in which a run is to be counted and possibly reversed 341 * @param lo index of the first element in the run 342 * @param hi index after the last element that may be contained in the run. 343 It is required that {@code lo < hi}. 344 * @param c the comparator to used for the sort 345 * @return the length of the run beginning at the specified position in 346 * the specified array 347 */ 348 private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi, 349 Comparator<? super T> c) { 350 assert lo < hi; 351 int runHi = lo + 1; 352 if (runHi == hi) 353 return 1; // lo < hi， 且lo + 1 = hi， 所以是有序且升序的， 直接返回 354 355 // Find end of run, and reverse range if descending 356 if (c.compare(a[runHi++], a[lo]) < 0) { // Descending 357 // 若是是降序的， 找到最長的有序降序序列的長度， 而且把序列倒置， 使之升序 358 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0) 359 runHi++; 360 reverseRange(a, lo, runHi); 361 } else { // Ascending 362 // 若是是升序的， 一樣找到最長的有序序列的長度 363 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0) 364 runHi++; 365 } 366 367 // 返回有序序列的長度 368 return runHi - lo; 369 } 370 371 /** 372 * Reverse the specified range of the specified array. 373 * 374 * @param a the array in which a range is to be reversed 375 * @param lo the index of the first element in the range to be reversed 376 * @param hi the index after the last element in the range to be reversed 377 */ 378 private static void reverseRange(Object[] a, int lo, int hi) { 379 hi--; 380 // 首尾倒置 381 while (lo < hi) { 382 Object t = a[lo]; 383 a[lo++] = a[hi]; 384 a[hi--] = t; 385 } 386 } 387 388 /** 389 * Returns the minimum acceptable run length for an array of the specified 390 * length. Natural runs shorter than this will be extended with 391 * {@link #binarySort}. 392 * 393 * Roughly speaking, the computation is: 394 * 395 * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff). 396 * Else if n is an exact power of 2, return MIN_MERGE/2. 397 * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k 398 * is close to, but strictly less than, an exact power of 2. 399 * 400 * For the rationale, see listsort.txt. 401 * 402 * @param n the length of the array to be sorted 403 * @return the length of the minimum run to be merged 404 */ 405 private static int minRunLength(int n) { 406 assert n >= 0; 407 int r = 0; // Becomes 1 if any 1 bits are shifted off 408 while (n >= MIN_MERGE) { 409 // n&1是判斷n是可否被2整除， 若是不能被2整除， 最後一個Bit位必定是1,則1&1爲1， r = r | 1 爲1 410 r |= (n & 1); 411 n >>= 1; 412 } 413 return n + r; 414 } 415 416 /** 417 * Pushes the specified run onto the pending-run stack. 418 * 419 * @param runBase index of the first element in the run 420 * @param runLen the number of elements in the run 421 */ 422 // 把有序序列起始位置和長度放入棧中 423 private void pushRun(int runBase, int runLen) { 424 this.runBase[stackSize] = runBase; 425 this.runLen[stackSize] = runLen; 426 stackSize++; 427 } 428 429 /** 430 * Examines the stack of runs waiting to be merged and merges adjacent runs 431 * until the stack invariants are reestablished: 432 * 433 * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] 434 * 2. runLen[i - 2] > runLen[i - 1] 435 * 436 * This method is called each time a new run is pushed onto the stack, 437 * so the invariants are guaranteed to hold for i < stackSize upon 438 * entry to the method. 439 */ 440 // 判斷合併棧頂的三個元素中較小的兩個， 或若是第二個元素比第一個小， 則合併， 使棧中全部的序列大小達到近似相等 441 private void mergeCollapse() { 442 while (stackSize > 1) { 443 int n = stackSize - 2; 444 if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) { 445 if (runLen[n - 1] < runLen[n + 1]) 446 n--; 447 mergeAt(n); 448 } else if (runLen[n] <= runLen[n + 1]) { 449 mergeAt(n); 450 } else { 451 break; // Invariant is established 452 } 453 } 454 } 455 456 /** 457 * Merges all runs on the stack until only one remains. This method is 458 * called once, to complete the sort. 459 */ 460 // 最後合併棧中全部的序列， 直到最後只剩一個有序序列 461 private void mergeForceCollapse() { 462 while (stackSize > 1) { 463 int n = stackSize - 2; 464 // 若是runLen[n-1] < runLen[n+1]， 則合併較小的較小的runBase[n-1]和runBase[n]， 不然合併runBase[n]和runBase[n+1] 465 if (n > 0 && runLen[n - 1] < runLen[n + 1]) 466 n--; 467 mergeAt(n); 468 } 469 } 470 471 /** 472 * Merges the two runs at stack indices i and i+1. Run i must be 473 * the penultimate or antepenultimate run on the stack. In other words, 474 * i must be equal to stackSize-2 or stackSize-3. 475 * 476 * @param i stack index of the first of the two runs to merge 477 */ 478 private void mergeAt(int i) { 479 assert stackSize >= 2; 480 assert i >= 0; 481 assert i == stackSize - 2 || i == stackSize - 3; 482 483 int base1 = runBase[i]; 484 int len1 = runLen[i]; 485 int base2 = runBase[i + 1]; 486 int len2 = runLen[i + 1]; 487 assert len1 > 0 && len2 > 0; 488 assert base1 + len1 == base2; 489 490 /* 491 * Record the length of the combined runs; if i is the 3rd-last 492 * run now, also slide over the last run (which isn't involved 493 * in this merge). The current run (i+1) goes away in any case. 494 */ 495 runLen[i] = len1 + len2; 496 // 若是i是棧頂倒數第三個元素， 則最後i+1必定會合進i數組， 所以i+1的位置替換成i+2 497 if (i == stackSize - 3) { 498 runBase[i + 1] = runBase[i + 2]; 499 runLen[i + 1] = runLen[i + 2]; 500 } 501 stackSize--; 502 503 /* 504 * Find where the first element of run2 goes in run1. Prior elements 505 * in run1 can be ignored (because they're already in place). 506 */ 507 // 找到run2的首元素在run1中的位置 508 int k = gallopRight(a[base2], a, base1, len1, 0, c); 509 assert k >= 0; 510 // 忽略k以前的序列， 由於已經有序， 減小比較次數 511 base1 += k; 512 len1 -= k; 513 if (len1 == 0) 514 return; 515 516 /* 517 * Find where the last element of run1 goes in run2. Subsequent elements 518 * in run2 can be ignored (because they're already in place). 519 */ 520 // 找打run1的尾元素在run2中的位置 521 len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c); 522 assert len2 >= 0; 523 // len2 == 0， 說明run1和run2已是一個總體有序的序列了， 直接返回。 524 if (len2 == 0) 525 return; 526 527 // Merge remaining runs, using tmp array with min(len1, len2) elements 528 if (len1 <= len2) 529 mergeLo(base1, len1, base2, len2); 530 else 531 mergeHi(base1, len1, base2, len2); 532 } 533 534 /** 535 * Locates the position at which to insert the specified key into the 536 * specified sorted range; if the range contains an element equal to key, 537 * returns the index of the leftmost equal element. 538 * 539 * @param key the key whose insertion point to search for 540 * @param a the array in which to search 541 * @param base the index of the first element in the range 542 * @param len the length of the range; must be > 0 543 * @param hint the index at which to begin the search, 0 <= hint < n. 544 * The closer hint is to the result, the faster this method will run. 545 * @param c the comparator used to order the range, and to search 546 * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k], 547 * pretending that a[b - 1] is minus infinity and a[b + n] is infinity. 548 * In other words, key belongs at index b + k; or in other words, 549 * the first k elements of a should precede key, and the last n - k 550 * should follow it. 551 */ 552 private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint, 553 Comparator<? super T> c) { 554 assert len > 0 && hint >= 0 && hint < len; 555 int lastOfs = 0; 556 int ofs = 1; 557 if (c.compare(key, a[base + hint]) > 0) { 558 // 查找區間[base + hint, len], 查找使得a[base+hint+lastOfs] < key <= a[base+hint+ofs]成立的ofs值。 559 int maxOfs = len - hint; 560 // 向右查找， 最大可能Ofs爲maxOfs-1， 即len - hint - 1， 即a[base + len - 1] 561 while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) { 562 // 記錄上一次ofs的值， 存到lastOfs中 563 lastOfs = ofs; 564 // ofs乘以2再加1 565 ofs = (ofs << 1) + 1; 566 // 整數溢出 567 if (ofs <= 0) 568 ofs = maxOfs; 569 } 570 571 // ofs最大值爲maxOfs 572 if (ofs > maxOfs) 573 ofs = maxOfs; 574 575 // Make offsets relative to base 576 // 以前的ofs和lastOfs都是相對於hint位置的， 如今把它重置爲相對於base的位置 577 lastOfs += hint; 578 ofs += hint; 579 } else { // key <= a[base + hint] 580 // 從base+hint向前搜索，查找區間[base, base + hint]， 直到找到ofs值使得a[base+hint-ofs] < key <= a[base+hint-lastOfs] 581 // maxOfs爲hint+1， 而ofs < maxOfs, 所以當ofs = maxOfs -1 時， 比較到的最左邊的元素爲a[base + hint - hint] == a[base] 582 final int maxOfs = hint + 1; 583 while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) { 584 lastOfs = ofs; 585 // ofs乘以2再加1 586 ofs = (ofs << 1) + 1; 587 // 正整數溢出 588 if (ofs <= 0) 589 ofs = maxOfs; 590 } 591 // 最大爲maxOfs 592 if (ofs > maxOfs) 593 ofs = maxOfs; 594 595 // 重置ofs和lastOfs爲相對於base的位置索引 596 int tmp = lastOfs; 597 lastOfs = hint - ofs; 598 ofs = hint - tmp; 599 } 600 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; 601 602 /* 603 * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere 604 * to the right of lastOfs but no farther right than ofs. Do a binary 605 * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs]. 606 */ 607 lastOfs++; 608 // 查找準確位置 609 while (lastOfs < ofs) { 610 // 中間位置 611 int m = lastOfs + ((ofs - lastOfs) >>> 1); 612 613 if (c.compare(key, a[base + m]) > 0) 614 lastOfs = m + 1; // a[base + m] < key 615 else 616 ofs = m; // key <= a[base + m] 617 } 618 assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs] 619 return ofs; 620 } 621 622 /** 623 * Like gallopLeft, except that if the range contains an element equal to 624 * key, gallopRight returns the index after the rightmost equal element. 625 * 626 * @param key the key whose insertion point to search for 627 * @param a the array in which to search 628 * @param base the index of the first element in the range 629 * @param len the length of the range; must be > 0 630 * @param hint the index at which to begin the search, 0 <= hint < n. 631 * The closer hint is to the result, the faster this method will run. 632 * @param c the comparator used to order the range, and to search 633 * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k] 634 */ 635 private static <T> int gallopRight(T key, T[] a, int base, int len, 636 int hint, Comparator<? super T> c) { 637 assert len > 0 && hint >= 0 && hint < len; 638 639 int ofs = 1; 640 int lastOfs = 0; 641 if (c.compare(key, a[base + hint]) < 0) { 642 // 從base + hint位置向前搜索區間[base, base + hint]， 使得a[b+hint - ofs] <= key < a[b+hint - lastOfs] 643 int maxOfs = hint + 1; 644 while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) { 645 // 記錄上次查找位置 646 lastOfs = ofs; 647 // 乘以2 加1 648 ofs = (ofs << 1) + 1; 649 // 正整數溢出 650 if (ofs <= 0) 651 ofs = maxOfs; 652 } 653 // 最大爲maxOfs 654 if (ofs > maxOfs) 655 ofs = maxOfs; 656 657 // 重置ofs和lastOfs爲相對於base的位置索引 658 int tmp = lastOfs; 659 lastOfs = hint - ofs; 660 ofs = hint - tmp; 661 } else { // a[b + hint] <= key 662 // 搜索區間[base + hint, base + len -1]使得a[b+hint + lastOfs] <= key < a[b+hint + ofs] 663 int maxOfs = len - hint; 664 while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) { 665 lastOfs = ofs; 666 ofs = (ofs << 1) + 1; 667 // 正整數溢出 668 if (ofs <= 0) 669 ofs = maxOfs; 670 } 671 if (ofs > maxOfs) 672 ofs = maxOfs; 673 674 // 重置ofs和lastOfs爲相對於base的位置索引 675 lastOfs += hint; 676 ofs += hint; 677 } 678 assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; 679 680 lastOfs++; 681 // 查找key的準確位置 682 while (lastOfs < ofs) { 683 int m = lastOfs + ((ofs - lastOfs) >>> 1); 684 685 if (c.compare(key, a[base + m]) < 0) 686 ofs = m; // key < a[b + m] 687 else 688 lastOfs = m + 1; // a[b + m] <= key 689 } 690 // 最終會找到m使得k >= a[base + m]， 而此時lastOfs == ofs且lastOfs = m +1， 則ofs = m +1， 所以a[base + ofs] > k >= a[base + ofs -1]， ofs即m+1， ofs - 1即爲m， 所以ofs位置的值大於key 691 assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs] 692 return ofs; 693 } 694 695 /** 696 *基於以上gallopRight方法最後查找key的索引的解釋， 所以run1的第一個元素必定大於run2的第一個元素， *而run2中run1最後一個元素所在索引位置以後的值也被忽略掉， 所以run1的最後一個元素大於run2中的全部元素的值。*/ 697 698 699 /** 700 * Merges two adjacent runs in place, in a stable fashion. The first 701 * element of the first run must be greater than the first element of the 702 * second run (a[base1] > a[base2]), and the last element of the first run 703 * (a[base1 + len1-1]) must be greater than all elements of the second run. 704 * 705 * For performance, this method should be called only when len1 <= len2; 706 * its twin, mergeHi should be called if len1 >= len2. (Either method 707 * may be called if len1 == len2.) 708 * 709 * @param base1 index of first element in first run to be merged 710 * @param len1 length of first run to be merged (must be > 0) 711 * @param base2 index of first element in second run to be merged 712 * (must be aBase + aLen) 713 * @param len2 length of second run to be merged (must be > 0) 714 */ 715 private void mergeLo(int base1, int len1, int base2, int len2) { 716 assert len1 > 0 && len2 > 0 && base1 + len1 == base2; 717 718 // Copy first run into temp array 719 T[] a = this.a; // For performance 720 T[] tmp = ensureCapacity(len1); 721 int cursor1 = tmpBase; // Indexes into tmp array 722 int cursor2 = base2; // Indexes int a 723 // 目標位置從base1的索引開始， 由於base1在base2以前， 下面會將base1的內容放入臨時數組， 這樣run1中的內容就能夠覆蓋了 724 int dest = base1; // Indexes int a 725 // 把第一個序列的內容放入臨時數組tmp中 726 System.arraycopy(a, base1, tmp, cursor1, len1); 727 728 // Move first element of second run and deal with degenerate cases 729 // 由於run1的第一個元素大於run2的第一個元素， 所以將run2的第一個元素先放進 730 a[dest++] = a[cursor2++]; 731 // 若是run2只有一個元素， 則將run1中剩餘的元素放入正確位置後返回 732 if (--len2 == 0) { 733 System.arraycopy(tmp, cursor1, a, dest, len1); 734 return; 735 } 736 // 若是run1只有一個元素， 由於run1的最後一個元素大於run2中的全部元素， 所以先將run2中的元素放入正確位置， 而後將run1的惟一的一個元素放入最後一個位置， 而後返回 737 if (len1 == 1) { 738 System.arraycopy(a, cursor2, a, dest, len2); 739 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge 740 return; 741 } 742 743 Comparator<? super T> c = this.c; // Use local variable for performance 744 int minGallop = this.minGallop; // " " " " " 745 outer: 746 while (true) { 747 int count1 = 0; // Number of times in a row that first run won 748 int count2 = 0; // Number of times in a row that second run won 749 750 /* 751 * Do the straightforward thing until (if ever) one run starts 752 * winning consistently. 753 */ 754 do { 755 assert len1 > 1 && len2 > 0; 756 if (c.compare(a[cursor2], tmp[cursor1]) < 0) { 757 a[dest++] = a[cursor2++]; 758 count2++; 759 count1 = 0; 760 if (--len2 == 0) 761 break outer; 762 } else { 763 a[dest++] = tmp[cursor1++]; 764 count1++; 765 count2 = 0; 766 if (--len1 == 1) 767 break outer; 768 } 769 // 當每一個序列中的連續放入目標位置的元素個數小於minGallop時， 這樣分別拷貝就能夠了 770 } while ((count1 | count2) < minGallop); 771 772 /* 773 * One run is winning so consistently that galloping may be a 774 * huge win. So try that, and continue galloping until (if ever) 775 * neither run appears to be winning consistently anymore. 776 */ 777 // 由於兩個run序列的大小是近似相等的， 若是一個序列連續超過minGallop個數的元素被放入目標位置， 則另外一個有接近大小的連續序列等待被放入正確位置，切換成Gallopping模式 778 do { 779 assert len1 > 1 && len2 > 0; 780 //查找run1第一個大於run2中第一個元素的元素的位置索引 781 count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c); 782 if (count1 != 0) { 783 // run1中count1以前的元素所有放入目標序列 784 System.arraycopy(tmp, cursor1, a, dest, count1); 785 // 移動索引位置 786 dest += count1; 787 cursor1 += count1; 788 len1 -= count1; 789 if (len1 <= 1) // len1 == 1 || len1 == 0 790 break outer; 791 } 792 // 移動run2的第一個元素到目標序列中 793 a[dest++] = a[cursor2++]; 794 // 若是run2中沒有其餘元素則跳出 795 if (--len2 == 0) 796 break outer; 797 798 // 查找run2中第一個小於等於run1當前元素的元素的位置索引 799 count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c); 800 if (count2 != 0) { 801 // 拷貝count2以後的元素到目標序列 802 System.arraycopy(a, cursor2, a, dest, count2); 803 dest += count2; 804 cursor2 += count2; 805 len2 -= count2; 806 // run2中沒有其餘元素則跳出 807 if (len2 == 0) 808 break outer; 809 } 810 811 // 此時run2中的第一個元素大於等於run1中的第一個元素， 拷貝run1中的第一個元素到目標序列 812 a[dest++] = tmp[cursor1++]; 813 // 若是run1中只有一個元素則跳出 814 if (--len1 == 1) 815 break outer; 816 // 動態調整minGallop的值 817 minGallop--; 818 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); 819 if (minGallop < 0) 820 minGallop = 0; 821 // 調整minGallop的值， 使得在有序序列很少的狀況下不用Gallopping模式 822 minGallop += 2; // Penalize for leaving gallop mode 823 } // End of "outer" loop 824 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field 825 826 // run1中只有一個元素 827 if (len1 == 1) { 828 assert len2 > 0; 829 System.arraycopy(a, cursor2, a, dest, len2); 830 a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge 831 } else if (len1 == 0) { 832 // 由於run1中的最後一個元素大於run2中的全部元素， 所以這種狀況不存在 833 throw new IllegalArgumentException( 834 "Comparison method violates its general contract!"); 835 } else { 836 // run2中已經沒有元素 837 assert len2 == 0; 838 assert len1 > 1; 839 System.arraycopy(tmp, cursor1, a, dest, len1); 840 } 841 } 842 843 /** 844 * Like mergeLo, except that this method should be called only if 845 * len1 >= len2; mergeLo should be called if len1 <= len2. (Either method 846 * may be called if len1 == len2.) 847 * 848 * @param base1 index of first element in first run to be merged 849 * @param len1 length of first run to be merged (must be > 0) 850 * @param base2 index of first element in second run to be merged 851 * (must be aBase + aLen) 852 * @param len2 length of second run to be merged (must be > 0) 853 */ 854 private void mergeHi(int base1, int len1, int base2, int len2) { 855 assert len1 > 0 && len2 > 0 && base1 + len1 == base2; 856 857 // Copy second run into temp array 858 T[] a = this.a; // For performance 859 T[] tmp = ensureCapacity(len2); 860 int tmpBase = this.tmpBase; 861 // 將run2中的全部元素放入臨時數組tmp中 862 System.arraycopy(a, base2, tmp, tmpBase, len2); 863 864 int cursor1 = base1 + len1 - 1; // Indexes into a 865 int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array 866 // 從後往前插入元素 867 int dest = base2 + len2 - 1; // Indexes into a 868 869 // Move last element of first run and deal with degenerate cases 870 // run1的最後一個元素導入目標位置 871 a[dest--] = a[cursor1--]; 872 // 若是run1中只有一個元素， 將run2中的剩餘元素放入目標位置（從後往前） 873 if (--len1 == 0) { 874 System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); 875 return; 876 } 877 if (len2 == 1) { 878 // run2中只有一個元素， 由於run1的第一個元素大於run2的第一個元素， 所以， run2中惟一的一個元素小於run1中全部的元素， 所以將run1中的元素所有放入目標位置， 最後將惟一的run2中的一個元素放入第一個位置 879 dest -= len1; 880 cursor1 -= len1; 881 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); 882 a[dest] = tmp[cursor2]; 883 return; 884 } 885 886 Comparator<? super T> c = this.c; // Use local variable for performance 887 int minGallop = this.minGallop; // " " " " " 888 outer: 889 while (true) { 890 int count1 = 0; // Number of times in a row that first run won 891 int count2 = 0; // Number of times in a row that second run won 892 893 /* 894 * Do the straightforward thing until (if ever) one run 895 * appears to win consistently. 896 */ 897 do { 898 assert len1 > 0 && len2 > 1; 899 if (c.compare(tmp[cursor2], a[cursor1]) < 0) { 900 // 從後往前放入目標位置 901 a[dest--] = a[cursor1--]; 902 count1++; 903 count2 = 0; 904 // run1中沒有了元素 905 if (--len1 == 0) 906 break outer; 907 } else { 908 a[dest--] = tmp[cursor2--]; 909 count2++; 910 count1 = 0; 911 // run2中只有一個剩餘元素 912 if (--len2 == 1) 913 break outer; 914 } 915 } while ((count1 | count2) < minGallop); 916 917 /* 918 * One run is winning so consistently that galloping may be a 919 * huge win. So try that, and continue galloping until (if ever) 920 * neither run appears to be winning consistently anymore. 921 */ 922 do { 923 assert len1 > 0 && len2 > 1; 924 // 找到大於run2當前位置的元素的run1中元素， 由於是從後往前查找， 所以找到的位置好比k1，k1以後的全部元素大於run1, run2中的剩餘全部元素 925 count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c); 926 if (count1 != 0) { 927 dest -= count1; 928 cursor1 -= count1; 929 len1 -= count1; 930 // 拷貝run1中的大的元素 931 System.arraycopy(a, cursor1 + 1, a, dest + 1, count1); 932 // run1中沒有元素， 跳出 933 if (len1 == 0) 934 break outer; 935 } 936 // run2的當前元素拷貝到目標位置 937 a[dest--] = tmp[cursor2--]; 938 if (--len2 == 1) 939 break outer; 940 941 // 找到run2中大於等於run1當前元素的元素的位置索引好比K2， 則k2以後的全部元素大於run1, run2中的剩餘元素 942 count2 = len2 - gallopLeft(a[cursor1], tmp, tmpBase, len2, len2 - 1, c); 943 if (count2 != 0) { 944 dest -= count2; 945 cursor2 -= count2; 946 len2 -= count2; 947 System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2); 948 if (len2 <= 1) // len2 == 1 || len2 == 0 949 break outer; 950 } 951 // 拷貝run1的當前元素到目標位置， 由於a[cursior1]大於等於run2中的剩餘元素 952 a[dest--] = a[cursor1--]; 953 if (--len1 == 0) 954 break outer; 955 minGallop--; 956 } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); 957 if (minGallop < 0) 958 minGallop = 0; 959 minGallop += 2; // Penalize for leaving gallop mode 960 } // End of "outer" loop 961 this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field 962 963 // run2中只有一個元素 964 if (len2 == 1) { 965 assert len1 > 0; 966 dest -= len1; 967 cursor1 -= len1; 968 // 拷貝run1中的全部元素到目標位置 969 System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); 970 // run2的最後一個元素放入第一個位置 971 a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge 972 } else if (len2 == 0) { 973 // 由於run2的第一個元素小於run1， run2中的全部元素， 所以run2不可能最後爲空 974 throw new IllegalArgumentException( 975 "Comparison method violates its general contract!"); 976 } else { 977 assert len1 == 0; 978 assert len2 > 0; 979 // 拷貝run2中的剩餘元素 980 System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2); 981 } 982 } 983 984 /** 985 * Ensures that the external array tmp has at least the specified 986 * number of elements, increasing its size if necessary. The size 987 * increases exponentially to ensure amortized linear time complexity. 988 * 989 * @param minCapacity the minimum required capacity of the tmp array 990 * @return tmp, whether or not it grew 991 */ 992 private T[] ensureCapacity(int minCapacity) { 993 if (tmpLen < minCapacity) { 994 // Compute smallest power of 2 > minCapacity 995 int newSize = minCapacity; 996 newSize |= newSize >> 1; 997 newSize |= newSize >> 2; 998 newSize |= newSize >> 4; 999 newSize |= newSize >> 8; 1000 newSize |= newSize >> 16; 1001 newSize++; 1002 1003 if (newSize < 0) // Not bloody likely! 1004 newSize = minCapacity; 1005 else 1006 newSize = Math.min(newSize, a.length >>> 1); 1007 1008 @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) 1009 T[] newArray = (T[])java.lang.reflect.Array.newInstance 1010 (a.getClass().getComponentType(), newSize); 1011 tmp = newArray; 1012 tmpLen = newSize; 1013 tmpBase = 0; 1014 } 1015 return tmp; 1016 } 1017 }
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