1.1 --- a/emul/compact/src/main/java/java/util/TimSort.java Fri Mar 22 16:59:47 2013 +0100
1.2 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000
1.3 @@ -1,929 +0,0 @@
1.4 -/*
1.5 - * Copyright 2009 Google Inc. All Rights Reserved.
1.6 - * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
1.7 - *
1.8 - * This code is free software; you can redistribute it and/or modify it
1.9 - * under the terms of the GNU General Public License version 2 only, as
1.10 - * published by the Free Software Foundation. Oracle designates this
1.11 - * particular file as subject to the "Classpath" exception as provided
1.12 - * by Oracle in the LICENSE file that accompanied this code.
1.13 - *
1.14 - * This code is distributed in the hope that it will be useful, but WITHOUT
1.15 - * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
1.16 - * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
1.17 - * version 2 for more details (a copy is included in the LICENSE file that
1.18 - * accompanied this code).
1.19 - *
1.20 - * You should have received a copy of the GNU General Public License version
1.21 - * 2 along with this work; if not, write to the Free Software Foundation,
1.22 - * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
1.23 - *
1.24 - * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
1.25 - * or visit www.oracle.com if you need additional information or have any
1.26 - * questions.
1.27 - */
1.28 -
1.29 -package java.util;
1.30 -
1.31 -
1.32 -/**
1.33 - * A stable, adaptive, iterative mergesort that requires far fewer than
1.34 - * n lg(n) comparisons when running on partially sorted arrays, while
1.35 - * offering performance comparable to a traditional mergesort when run
1.36 - * on random arrays. Like all proper mergesorts, this sort is stable and
1.37 - * runs O(n log n) time (worst case). In the worst case, this sort requires
1.38 - * temporary storage space for n/2 object references; in the best case,
1.39 - * it requires only a small constant amount of space.
1.40 - *
1.41 - * This implementation was adapted from Tim Peters's list sort for
1.42 - * Python, which is described in detail here:
1.43 - *
1.44 - * http://svn.python.org/projects/python/trunk/Objects/listsort.txt
1.45 - *
1.46 - * Tim's C code may be found here:
1.47 - *
1.48 - * http://svn.python.org/projects/python/trunk/Objects/listobject.c
1.49 - *
1.50 - * The underlying techniques are described in this paper (and may have
1.51 - * even earlier origins):
1.52 - *
1.53 - * "Optimistic Sorting and Information Theoretic Complexity"
1.54 - * Peter McIlroy
1.55 - * SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),
1.56 - * pp 467-474, Austin, Texas, 25-27 January 1993.
1.57 - *
1.58 - * While the API to this class consists solely of static methods, it is
1.59 - * (privately) instantiable; a TimSort instance holds the state of an ongoing
1.60 - * sort, assuming the input array is large enough to warrant the full-blown
1.61 - * TimSort. Small arrays are sorted in place, using a binary insertion sort.
1.62 - *
1.63 - * @author Josh Bloch
1.64 - */
1.65 -class TimSort<T> {
1.66 - /**
1.67 - * This is the minimum sized sequence that will be merged. Shorter
1.68 - * sequences will be lengthened by calling binarySort. If the entire
1.69 - * array is less than this length, no merges will be performed.
1.70 - *
1.71 - * This constant should be a power of two. It was 64 in Tim Peter's C
1.72 - * implementation, but 32 was empirically determined to work better in
1.73 - * this implementation. In the unlikely event that you set this constant
1.74 - * to be a number that's not a power of two, you'll need to change the
1.75 - * {@link #minRunLength} computation.
1.76 - *
1.77 - * If you decrease this constant, you must change the stackLen
1.78 - * computation in the TimSort constructor, or you risk an
1.79 - * ArrayOutOfBounds exception. See listsort.txt for a discussion
1.80 - * of the minimum stack length required as a function of the length
1.81 - * of the array being sorted and the minimum merge sequence length.
1.82 - */
1.83 - private static final int MIN_MERGE = 32;
1.84 -
1.85 - /**
1.86 - * The array being sorted.
1.87 - */
1.88 - private final T[] a;
1.89 -
1.90 - /**
1.91 - * The comparator for this sort.
1.92 - */
1.93 - private final Comparator<? super T> c;
1.94 -
1.95 - /**
1.96 - * When we get into galloping mode, we stay there until both runs win less
1.97 - * often than MIN_GALLOP consecutive times.
1.98 - */
1.99 - private static final int MIN_GALLOP = 7;
1.100 -
1.101 - /**
1.102 - * This controls when we get *into* galloping mode. It is initialized
1.103 - * to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
1.104 - * random data, and lower for highly structured data.
1.105 - */
1.106 - private int minGallop = MIN_GALLOP;
1.107 -
1.108 - /**
1.109 - * Maximum initial size of tmp array, which is used for merging. The array
1.110 - * can grow to accommodate demand.
1.111 - *
1.112 - * Unlike Tim's original C version, we do not allocate this much storage
1.113 - * when sorting smaller arrays. This change was required for performance.
1.114 - */
1.115 - private static final int INITIAL_TMP_STORAGE_LENGTH = 256;
1.116 -
1.117 - /**
1.118 - * Temp storage for merges.
1.119 - */
1.120 - private T[] tmp; // Actual runtime type will be Object[], regardless of T
1.121 -
1.122 - /**
1.123 - * A stack of pending runs yet to be merged. Run i starts at
1.124 - * address base[i] and extends for len[i] elements. It's always
1.125 - * true (so long as the indices are in bounds) that:
1.126 - *
1.127 - * runBase[i] + runLen[i] == runBase[i + 1]
1.128 - *
1.129 - * so we could cut the storage for this, but it's a minor amount,
1.130 - * and keeping all the info explicit simplifies the code.
1.131 - */
1.132 - private int stackSize = 0; // Number of pending runs on stack
1.133 - private final int[] runBase;
1.134 - private final int[] runLen;
1.135 -
1.136 - /**
1.137 - * Creates a TimSort instance to maintain the state of an ongoing sort.
1.138 - *
1.139 - * @param a the array to be sorted
1.140 - * @param c the comparator to determine the order of the sort
1.141 - */
1.142 - private TimSort(T[] a, Comparator<? super T> c) {
1.143 - this.a = a;
1.144 - this.c = c;
1.145 -
1.146 - // Allocate temp storage (which may be increased later if necessary)
1.147 - int len = a.length;
1.148 - @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
1.149 - T[] newArray = (T[]) new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ?
1.150 - len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];
1.151 - tmp = newArray;
1.152 -
1.153 - /*
1.154 - * Allocate runs-to-be-merged stack (which cannot be expanded). The
1.155 - * stack length requirements are described in listsort.txt. The C
1.156 - * version always uses the same stack length (85), but this was
1.157 - * measured to be too expensive when sorting "mid-sized" arrays (e.g.,
1.158 - * 100 elements) in Java. Therefore, we use smaller (but sufficiently
1.159 - * large) stack lengths for smaller arrays. The "magic numbers" in the
1.160 - * computation below must be changed if MIN_MERGE is decreased. See
1.161 - * the MIN_MERGE declaration above for more information.
1.162 - */
1.163 - int stackLen = (len < 120 ? 5 :
1.164 - len < 1542 ? 10 :
1.165 - len < 119151 ? 19 : 40);
1.166 - runBase = new int[stackLen];
1.167 - runLen = new int[stackLen];
1.168 - }
1.169 -
1.170 - /*
1.171 - * The next two methods (which are package private and static) constitute
1.172 - * the entire API of this class. Each of these methods obeys the contract
1.173 - * of the public method with the same signature in java.util.Arrays.
1.174 - */
1.175 -
1.176 - static <T> void sort(T[] a, Comparator<? super T> c) {
1.177 - sort(a, 0, a.length, c);
1.178 - }
1.179 -
1.180 - static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c) {
1.181 - if (c == null) {
1.182 - Arrays.sort(a, lo, hi);
1.183 - return;
1.184 - }
1.185 -
1.186 - rangeCheck(a.length, lo, hi);
1.187 - int nRemaining = hi - lo;
1.188 - if (nRemaining < 2)
1.189 - return; // Arrays of size 0 and 1 are always sorted
1.190 -
1.191 - // If array is small, do a "mini-TimSort" with no merges
1.192 - if (nRemaining < MIN_MERGE) {
1.193 - int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
1.194 - binarySort(a, lo, hi, lo + initRunLen, c);
1.195 - return;
1.196 - }
1.197 -
1.198 - /**
1.199 - * March over the array once, left to right, finding natural runs,
1.200 - * extending short natural runs to minRun elements, and merging runs
1.201 - * to maintain stack invariant.
1.202 - */
1.203 - TimSort<T> ts = new TimSort<>(a, c);
1.204 - int minRun = minRunLength(nRemaining);
1.205 - do {
1.206 - // Identify next run
1.207 - int runLen = countRunAndMakeAscending(a, lo, hi, c);
1.208 -
1.209 - // If run is short, extend to min(minRun, nRemaining)
1.210 - if (runLen < minRun) {
1.211 - int force = nRemaining <= minRun ? nRemaining : minRun;
1.212 - binarySort(a, lo, lo + force, lo + runLen, c);
1.213 - runLen = force;
1.214 - }
1.215 -
1.216 - // Push run onto pending-run stack, and maybe merge
1.217 - ts.pushRun(lo, runLen);
1.218 - ts.mergeCollapse();
1.219 -
1.220 - // Advance to find next run
1.221 - lo += runLen;
1.222 - nRemaining -= runLen;
1.223 - } while (nRemaining != 0);
1.224 -
1.225 - // Merge all remaining runs to complete sort
1.226 - assert lo == hi;
1.227 - ts.mergeForceCollapse();
1.228 - assert ts.stackSize == 1;
1.229 - }
1.230 -
1.231 - /**
1.232 - * Sorts the specified portion of the specified array using a binary
1.233 - * insertion sort. This is the best method for sorting small numbers
1.234 - * of elements. It requires O(n log n) compares, but O(n^2) data
1.235 - * movement (worst case).
1.236 - *
1.237 - * If the initial part of the specified range is already sorted,
1.238 - * this method can take advantage of it: the method assumes that the
1.239 - * elements from index {@code lo}, inclusive, to {@code start},
1.240 - * exclusive are already sorted.
1.241 - *
1.242 - * @param a the array in which a range is to be sorted
1.243 - * @param lo the index of the first element in the range to be sorted
1.244 - * @param hi the index after the last element in the range to be sorted
1.245 - * @param start the index of the first element in the range that is
1.246 - * not already known to be sorted ({@code lo <= start <= hi})
1.247 - * @param c comparator to used for the sort
1.248 - */
1.249 - @SuppressWarnings("fallthrough")
1.250 - private static <T> void binarySort(T[] a, int lo, int hi, int start,
1.251 - Comparator<? super T> c) {
1.252 - assert lo <= start && start <= hi;
1.253 - if (start == lo)
1.254 - start++;
1.255 - for ( ; start < hi; start++) {
1.256 - T pivot = a[start];
1.257 -
1.258 - // Set left (and right) to the index where a[start] (pivot) belongs
1.259 - int left = lo;
1.260 - int right = start;
1.261 - assert left <= right;
1.262 - /*
1.263 - * Invariants:
1.264 - * pivot >= all in [lo, left).
1.265 - * pivot < all in [right, start).
1.266 - */
1.267 - while (left < right) {
1.268 - int mid = (left + right) >>> 1;
1.269 - if (c.compare(pivot, a[mid]) < 0)
1.270 - right = mid;
1.271 - else
1.272 - left = mid + 1;
1.273 - }
1.274 - assert left == right;
1.275 -
1.276 - /*
1.277 - * The invariants still hold: pivot >= all in [lo, left) and
1.278 - * pivot < all in [left, start), so pivot belongs at left. Note
1.279 - * that if there are elements equal to pivot, left points to the
1.280 - * first slot after them -- that's why this sort is stable.
1.281 - * Slide elements over to make room for pivot.
1.282 - */
1.283 - int n = start - left; // The number of elements to move
1.284 - // Switch is just an optimization for arraycopy in default case
1.285 - switch (n) {
1.286 - case 2: a[left + 2] = a[left + 1];
1.287 - case 1: a[left + 1] = a[left];
1.288 - break;
1.289 - default: System.arraycopy(a, left, a, left + 1, n);
1.290 - }
1.291 - a[left] = pivot;
1.292 - }
1.293 - }
1.294 -
1.295 - /**
1.296 - * Returns the length of the run beginning at the specified position in
1.297 - * the specified array and reverses the run if it is descending (ensuring
1.298 - * that the run will always be ascending when the method returns).
1.299 - *
1.300 - * A run is the longest ascending sequence with:
1.301 - *
1.302 - * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
1.303 - *
1.304 - * or the longest descending sequence with:
1.305 - *
1.306 - * a[lo] > a[lo + 1] > a[lo + 2] > ...
1.307 - *
1.308 - * For its intended use in a stable mergesort, the strictness of the
1.309 - * definition of "descending" is needed so that the call can safely
1.310 - * reverse a descending sequence without violating stability.
1.311 - *
1.312 - * @param a the array in which a run is to be counted and possibly reversed
1.313 - * @param lo index of the first element in the run
1.314 - * @param hi index after the last element that may be contained in the run.
1.315 - It is required that {@code lo < hi}.
1.316 - * @param c the comparator to used for the sort
1.317 - * @return the length of the run beginning at the specified position in
1.318 - * the specified array
1.319 - */
1.320 - private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,
1.321 - Comparator<? super T> c) {
1.322 - assert lo < hi;
1.323 - int runHi = lo + 1;
1.324 - if (runHi == hi)
1.325 - return 1;
1.326 -
1.327 - // Find end of run, and reverse range if descending
1.328 - if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
1.329 - while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
1.330 - runHi++;
1.331 - reverseRange(a, lo, runHi);
1.332 - } else { // Ascending
1.333 - while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
1.334 - runHi++;
1.335 - }
1.336 -
1.337 - return runHi - lo;
1.338 - }
1.339 -
1.340 - /**
1.341 - * Reverse the specified range of the specified array.
1.342 - *
1.343 - * @param a the array in which a range is to be reversed
1.344 - * @param lo the index of the first element in the range to be reversed
1.345 - * @param hi the index after the last element in the range to be reversed
1.346 - */
1.347 - private static void reverseRange(Object[] a, int lo, int hi) {
1.348 - hi--;
1.349 - while (lo < hi) {
1.350 - Object t = a[lo];
1.351 - a[lo++] = a[hi];
1.352 - a[hi--] = t;
1.353 - }
1.354 - }
1.355 -
1.356 - /**
1.357 - * Returns the minimum acceptable run length for an array of the specified
1.358 - * length. Natural runs shorter than this will be extended with
1.359 - * {@link #binarySort}.
1.360 - *
1.361 - * Roughly speaking, the computation is:
1.362 - *
1.363 - * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
1.364 - * Else if n is an exact power of 2, return MIN_MERGE/2.
1.365 - * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
1.366 - * is close to, but strictly less than, an exact power of 2.
1.367 - *
1.368 - * For the rationale, see listsort.txt.
1.369 - *
1.370 - * @param n the length of the array to be sorted
1.371 - * @return the length of the minimum run to be merged
1.372 - */
1.373 - private static int minRunLength(int n) {
1.374 - assert n >= 0;
1.375 - int r = 0; // Becomes 1 if any 1 bits are shifted off
1.376 - while (n >= MIN_MERGE) {
1.377 - r |= (n & 1);
1.378 - n >>= 1;
1.379 - }
1.380 - return n + r;
1.381 - }
1.382 -
1.383 - /**
1.384 - * Pushes the specified run onto the pending-run stack.
1.385 - *
1.386 - * @param runBase index of the first element in the run
1.387 - * @param runLen the number of elements in the run
1.388 - */
1.389 - private void pushRun(int runBase, int runLen) {
1.390 - this.runBase[stackSize] = runBase;
1.391 - this.runLen[stackSize] = runLen;
1.392 - stackSize++;
1.393 - }
1.394 -
1.395 - /**
1.396 - * Examines the stack of runs waiting to be merged and merges adjacent runs
1.397 - * until the stack invariants are reestablished:
1.398 - *
1.399 - * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
1.400 - * 2. runLen[i - 2] > runLen[i - 1]
1.401 - *
1.402 - * This method is called each time a new run is pushed onto the stack,
1.403 - * so the invariants are guaranteed to hold for i < stackSize upon
1.404 - * entry to the method.
1.405 - */
1.406 - private void mergeCollapse() {
1.407 - while (stackSize > 1) {
1.408 - int n = stackSize - 2;
1.409 - if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
1.410 - if (runLen[n - 1] < runLen[n + 1])
1.411 - n--;
1.412 - mergeAt(n);
1.413 - } else if (runLen[n] <= runLen[n + 1]) {
1.414 - mergeAt(n);
1.415 - } else {
1.416 - break; // Invariant is established
1.417 - }
1.418 - }
1.419 - }
1.420 -
1.421 - /**
1.422 - * Merges all runs on the stack until only one remains. This method is
1.423 - * called once, to complete the sort.
1.424 - */
1.425 - private void mergeForceCollapse() {
1.426 - while (stackSize > 1) {
1.427 - int n = stackSize - 2;
1.428 - if (n > 0 && runLen[n - 1] < runLen[n + 1])
1.429 - n--;
1.430 - mergeAt(n);
1.431 - }
1.432 - }
1.433 -
1.434 - /**
1.435 - * Merges the two runs at stack indices i and i+1. Run i must be
1.436 - * the penultimate or antepenultimate run on the stack. In other words,
1.437 - * i must be equal to stackSize-2 or stackSize-3.
1.438 - *
1.439 - * @param i stack index of the first of the two runs to merge
1.440 - */
1.441 - private void mergeAt(int i) {
1.442 - assert stackSize >= 2;
1.443 - assert i >= 0;
1.444 - assert i == stackSize - 2 || i == stackSize - 3;
1.445 -
1.446 - int base1 = runBase[i];
1.447 - int len1 = runLen[i];
1.448 - int base2 = runBase[i + 1];
1.449 - int len2 = runLen[i + 1];
1.450 - assert len1 > 0 && len2 > 0;
1.451 - assert base1 + len1 == base2;
1.452 -
1.453 - /*
1.454 - * Record the length of the combined runs; if i is the 3rd-last
1.455 - * run now, also slide over the last run (which isn't involved
1.456 - * in this merge). The current run (i+1) goes away in any case.
1.457 - */
1.458 - runLen[i] = len1 + len2;
1.459 - if (i == stackSize - 3) {
1.460 - runBase[i + 1] = runBase[i + 2];
1.461 - runLen[i + 1] = runLen[i + 2];
1.462 - }
1.463 - stackSize--;
1.464 -
1.465 - /*
1.466 - * Find where the first element of run2 goes in run1. Prior elements
1.467 - * in run1 can be ignored (because they're already in place).
1.468 - */
1.469 - int k = gallopRight(a[base2], a, base1, len1, 0, c);
1.470 - assert k >= 0;
1.471 - base1 += k;
1.472 - len1 -= k;
1.473 - if (len1 == 0)
1.474 - return;
1.475 -
1.476 - /*
1.477 - * Find where the last element of run1 goes in run2. Subsequent elements
1.478 - * in run2 can be ignored (because they're already in place).
1.479 - */
1.480 - len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
1.481 - assert len2 >= 0;
1.482 - if (len2 == 0)
1.483 - return;
1.484 -
1.485 - // Merge remaining runs, using tmp array with min(len1, len2) elements
1.486 - if (len1 <= len2)
1.487 - mergeLo(base1, len1, base2, len2);
1.488 - else
1.489 - mergeHi(base1, len1, base2, len2);
1.490 - }
1.491 -
1.492 - /**
1.493 - * Locates the position at which to insert the specified key into the
1.494 - * specified sorted range; if the range contains an element equal to key,
1.495 - * returns the index of the leftmost equal element.
1.496 - *
1.497 - * @param key the key whose insertion point to search for
1.498 - * @param a the array in which to search
1.499 - * @param base the index of the first element in the range
1.500 - * @param len the length of the range; must be > 0
1.501 - * @param hint the index at which to begin the search, 0 <= hint < n.
1.502 - * The closer hint is to the result, the faster this method will run.
1.503 - * @param c the comparator used to order the range, and to search
1.504 - * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
1.505 - * pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
1.506 - * In other words, key belongs at index b + k; or in other words,
1.507 - * the first k elements of a should precede key, and the last n - k
1.508 - * should follow it.
1.509 - */
1.510 - private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint,
1.511 - Comparator<? super T> c) {
1.512 - assert len > 0 && hint >= 0 && hint < len;
1.513 - int lastOfs = 0;
1.514 - int ofs = 1;
1.515 - if (c.compare(key, a[base + hint]) > 0) {
1.516 - // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
1.517 - int maxOfs = len - hint;
1.518 - while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) {
1.519 - lastOfs = ofs;
1.520 - ofs = (ofs << 1) + 1;
1.521 - if (ofs <= 0) // int overflow
1.522 - ofs = maxOfs;
1.523 - }
1.524 - if (ofs > maxOfs)
1.525 - ofs = maxOfs;
1.526 -
1.527 - // Make offsets relative to base
1.528 - lastOfs += hint;
1.529 - ofs += hint;
1.530 - } else { // key <= a[base + hint]
1.531 - // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
1.532 - final int maxOfs = hint + 1;
1.533 - while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) {
1.534 - lastOfs = ofs;
1.535 - ofs = (ofs << 1) + 1;
1.536 - if (ofs <= 0) // int overflow
1.537 - ofs = maxOfs;
1.538 - }
1.539 - if (ofs > maxOfs)
1.540 - ofs = maxOfs;
1.541 -
1.542 - // Make offsets relative to base
1.543 - int tmp = lastOfs;
1.544 - lastOfs = hint - ofs;
1.545 - ofs = hint - tmp;
1.546 - }
1.547 - assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
1.548 -
1.549 - /*
1.550 - * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
1.551 - * to the right of lastOfs but no farther right than ofs. Do a binary
1.552 - * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
1.553 - */
1.554 - lastOfs++;
1.555 - while (lastOfs < ofs) {
1.556 - int m = lastOfs + ((ofs - lastOfs) >>> 1);
1.557 -
1.558 - if (c.compare(key, a[base + m]) > 0)
1.559 - lastOfs = m + 1; // a[base + m] < key
1.560 - else
1.561 - ofs = m; // key <= a[base + m]
1.562 - }
1.563 - assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
1.564 - return ofs;
1.565 - }
1.566 -
1.567 - /**
1.568 - * Like gallopLeft, except that if the range contains an element equal to
1.569 - * key, gallopRight returns the index after the rightmost equal element.
1.570 - *
1.571 - * @param key the key whose insertion point to search for
1.572 - * @param a the array in which to search
1.573 - * @param base the index of the first element in the range
1.574 - * @param len the length of the range; must be > 0
1.575 - * @param hint the index at which to begin the search, 0 <= hint < n.
1.576 - * The closer hint is to the result, the faster this method will run.
1.577 - * @param c the comparator used to order the range, and to search
1.578 - * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
1.579 - */
1.580 - private static <T> int gallopRight(T key, T[] a, int base, int len,
1.581 - int hint, Comparator<? super T> c) {
1.582 - assert len > 0 && hint >= 0 && hint < len;
1.583 -
1.584 - int ofs = 1;
1.585 - int lastOfs = 0;
1.586 - if (c.compare(key, a[base + hint]) < 0) {
1.587 - // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
1.588 - int maxOfs = hint + 1;
1.589 - while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) {
1.590 - lastOfs = ofs;
1.591 - ofs = (ofs << 1) + 1;
1.592 - if (ofs <= 0) // int overflow
1.593 - ofs = maxOfs;
1.594 - }
1.595 - if (ofs > maxOfs)
1.596 - ofs = maxOfs;
1.597 -
1.598 - // Make offsets relative to b
1.599 - int tmp = lastOfs;
1.600 - lastOfs = hint - ofs;
1.601 - ofs = hint - tmp;
1.602 - } else { // a[b + hint] <= key
1.603 - // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
1.604 - int maxOfs = len - hint;
1.605 - while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) {
1.606 - lastOfs = ofs;
1.607 - ofs = (ofs << 1) + 1;
1.608 - if (ofs <= 0) // int overflow
1.609 - ofs = maxOfs;
1.610 - }
1.611 - if (ofs > maxOfs)
1.612 - ofs = maxOfs;
1.613 -
1.614 - // Make offsets relative to b
1.615 - lastOfs += hint;
1.616 - ofs += hint;
1.617 - }
1.618 - assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
1.619 -
1.620 - /*
1.621 - * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
1.622 - * the right of lastOfs but no farther right than ofs. Do a binary
1.623 - * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
1.624 - */
1.625 - lastOfs++;
1.626 - while (lastOfs < ofs) {
1.627 - int m = lastOfs + ((ofs - lastOfs) >>> 1);
1.628 -
1.629 - if (c.compare(key, a[base + m]) < 0)
1.630 - ofs = m; // key < a[b + m]
1.631 - else
1.632 - lastOfs = m + 1; // a[b + m] <= key
1.633 - }
1.634 - assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
1.635 - return ofs;
1.636 - }
1.637 -
1.638 - /**
1.639 - * Merges two adjacent runs in place, in a stable fashion. The first
1.640 - * element of the first run must be greater than the first element of the
1.641 - * second run (a[base1] > a[base2]), and the last element of the first run
1.642 - * (a[base1 + len1-1]) must be greater than all elements of the second run.
1.643 - *
1.644 - * For performance, this method should be called only when len1 <= len2;
1.645 - * its twin, mergeHi should be called if len1 >= len2. (Either method
1.646 - * may be called if len1 == len2.)
1.647 - *
1.648 - * @param base1 index of first element in first run to be merged
1.649 - * @param len1 length of first run to be merged (must be > 0)
1.650 - * @param base2 index of first element in second run to be merged
1.651 - * (must be aBase + aLen)
1.652 - * @param len2 length of second run to be merged (must be > 0)
1.653 - */
1.654 - private void mergeLo(int base1, int len1, int base2, int len2) {
1.655 - assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
1.656 -
1.657 - // Copy first run into temp array
1.658 - T[] a = this.a; // For performance
1.659 - T[] tmp = ensureCapacity(len1);
1.660 - System.arraycopy(a, base1, tmp, 0, len1);
1.661 -
1.662 - int cursor1 = 0; // Indexes into tmp array
1.663 - int cursor2 = base2; // Indexes int a
1.664 - int dest = base1; // Indexes int a
1.665 -
1.666 - // Move first element of second run and deal with degenerate cases
1.667 - a[dest++] = a[cursor2++];
1.668 - if (--len2 == 0) {
1.669 - System.arraycopy(tmp, cursor1, a, dest, len1);
1.670 - return;
1.671 - }
1.672 - if (len1 == 1) {
1.673 - System.arraycopy(a, cursor2, a, dest, len2);
1.674 - a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
1.675 - return;
1.676 - }
1.677 -
1.678 - Comparator<? super T> c = this.c; // Use local variable for performance
1.679 - int minGallop = this.minGallop; // " " " " "
1.680 - outer:
1.681 - while (true) {
1.682 - int count1 = 0; // Number of times in a row that first run won
1.683 - int count2 = 0; // Number of times in a row that second run won
1.684 -
1.685 - /*
1.686 - * Do the straightforward thing until (if ever) one run starts
1.687 - * winning consistently.
1.688 - */
1.689 - do {
1.690 - assert len1 > 1 && len2 > 0;
1.691 - if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
1.692 - a[dest++] = a[cursor2++];
1.693 - count2++;
1.694 - count1 = 0;
1.695 - if (--len2 == 0)
1.696 - break outer;
1.697 - } else {
1.698 - a[dest++] = tmp[cursor1++];
1.699 - count1++;
1.700 - count2 = 0;
1.701 - if (--len1 == 1)
1.702 - break outer;
1.703 - }
1.704 - } while ((count1 | count2) < minGallop);
1.705 -
1.706 - /*
1.707 - * One run is winning so consistently that galloping may be a
1.708 - * huge win. So try that, and continue galloping until (if ever)
1.709 - * neither run appears to be winning consistently anymore.
1.710 - */
1.711 - do {
1.712 - assert len1 > 1 && len2 > 0;
1.713 - count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
1.714 - if (count1 != 0) {
1.715 - System.arraycopy(tmp, cursor1, a, dest, count1);
1.716 - dest += count1;
1.717 - cursor1 += count1;
1.718 - len1 -= count1;
1.719 - if (len1 <= 1) // len1 == 1 || len1 == 0
1.720 - break outer;
1.721 - }
1.722 - a[dest++] = a[cursor2++];
1.723 - if (--len2 == 0)
1.724 - break outer;
1.725 -
1.726 - count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
1.727 - if (count2 != 0) {
1.728 - System.arraycopy(a, cursor2, a, dest, count2);
1.729 - dest += count2;
1.730 - cursor2 += count2;
1.731 - len2 -= count2;
1.732 - if (len2 == 0)
1.733 - break outer;
1.734 - }
1.735 - a[dest++] = tmp[cursor1++];
1.736 - if (--len1 == 1)
1.737 - break outer;
1.738 - minGallop--;
1.739 - } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
1.740 - if (minGallop < 0)
1.741 - minGallop = 0;
1.742 - minGallop += 2; // Penalize for leaving gallop mode
1.743 - } // End of "outer" loop
1.744 - this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
1.745 -
1.746 - if (len1 == 1) {
1.747 - assert len2 > 0;
1.748 - System.arraycopy(a, cursor2, a, dest, len2);
1.749 - a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
1.750 - } else if (len1 == 0) {
1.751 - throw new IllegalArgumentException(
1.752 - "Comparison method violates its general contract!");
1.753 - } else {
1.754 - assert len2 == 0;
1.755 - assert len1 > 1;
1.756 - System.arraycopy(tmp, cursor1, a, dest, len1);
1.757 - }
1.758 - }
1.759 -
1.760 - /**
1.761 - * Like mergeLo, except that this method should be called only if
1.762 - * len1 >= len2; mergeLo should be called if len1 <= len2. (Either method
1.763 - * may be called if len1 == len2.)
1.764 - *
1.765 - * @param base1 index of first element in first run to be merged
1.766 - * @param len1 length of first run to be merged (must be > 0)
1.767 - * @param base2 index of first element in second run to be merged
1.768 - * (must be aBase + aLen)
1.769 - * @param len2 length of second run to be merged (must be > 0)
1.770 - */
1.771 - private void mergeHi(int base1, int len1, int base2, int len2) {
1.772 - assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
1.773 -
1.774 - // Copy second run into temp array
1.775 - T[] a = this.a; // For performance
1.776 - T[] tmp = ensureCapacity(len2);
1.777 - System.arraycopy(a, base2, tmp, 0, len2);
1.778 -
1.779 - int cursor1 = base1 + len1 - 1; // Indexes into a
1.780 - int cursor2 = len2 - 1; // Indexes into tmp array
1.781 - int dest = base2 + len2 - 1; // Indexes into a
1.782 -
1.783 - // Move last element of first run and deal with degenerate cases
1.784 - a[dest--] = a[cursor1--];
1.785 - if (--len1 == 0) {
1.786 - System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
1.787 - return;
1.788 - }
1.789 - if (len2 == 1) {
1.790 - dest -= len1;
1.791 - cursor1 -= len1;
1.792 - System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
1.793 - a[dest] = tmp[cursor2];
1.794 - return;
1.795 - }
1.796 -
1.797 - Comparator<? super T> c = this.c; // Use local variable for performance
1.798 - int minGallop = this.minGallop; // " " " " "
1.799 - outer:
1.800 - while (true) {
1.801 - int count1 = 0; // Number of times in a row that first run won
1.802 - int count2 = 0; // Number of times in a row that second run won
1.803 -
1.804 - /*
1.805 - * Do the straightforward thing until (if ever) one run
1.806 - * appears to win consistently.
1.807 - */
1.808 - do {
1.809 - assert len1 > 0 && len2 > 1;
1.810 - if (c.compare(tmp[cursor2], a[cursor1]) < 0) {
1.811 - a[dest--] = a[cursor1--];
1.812 - count1++;
1.813 - count2 = 0;
1.814 - if (--len1 == 0)
1.815 - break outer;
1.816 - } else {
1.817 - a[dest--] = tmp[cursor2--];
1.818 - count2++;
1.819 - count1 = 0;
1.820 - if (--len2 == 1)
1.821 - break outer;
1.822 - }
1.823 - } while ((count1 | count2) < minGallop);
1.824 -
1.825 - /*
1.826 - * One run is winning so consistently that galloping may be a
1.827 - * huge win. So try that, and continue galloping until (if ever)
1.828 - * neither run appears to be winning consistently anymore.
1.829 - */
1.830 - do {
1.831 - assert len1 > 0 && len2 > 1;
1.832 - count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
1.833 - if (count1 != 0) {
1.834 - dest -= count1;
1.835 - cursor1 -= count1;
1.836 - len1 -= count1;
1.837 - System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
1.838 - if (len1 == 0)
1.839 - break outer;
1.840 - }
1.841 - a[dest--] = tmp[cursor2--];
1.842 - if (--len2 == 1)
1.843 - break outer;
1.844 -
1.845 - count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);
1.846 - if (count2 != 0) {
1.847 - dest -= count2;
1.848 - cursor2 -= count2;
1.849 - len2 -= count2;
1.850 - System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
1.851 - if (len2 <= 1) // len2 == 1 || len2 == 0
1.852 - break outer;
1.853 - }
1.854 - a[dest--] = a[cursor1--];
1.855 - if (--len1 == 0)
1.856 - break outer;
1.857 - minGallop--;
1.858 - } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
1.859 - if (minGallop < 0)
1.860 - minGallop = 0;
1.861 - minGallop += 2; // Penalize for leaving gallop mode
1.862 - } // End of "outer" loop
1.863 - this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
1.864 -
1.865 - if (len2 == 1) {
1.866 - assert len1 > 0;
1.867 - dest -= len1;
1.868 - cursor1 -= len1;
1.869 - System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
1.870 - a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
1.871 - } else if (len2 == 0) {
1.872 - throw new IllegalArgumentException(
1.873 - "Comparison method violates its general contract!");
1.874 - } else {
1.875 - assert len1 == 0;
1.876 - assert len2 > 0;
1.877 - System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
1.878 - }
1.879 - }
1.880 -
1.881 - /**
1.882 - * Ensures that the external array tmp has at least the specified
1.883 - * number of elements, increasing its size if necessary. The size
1.884 - * increases exponentially to ensure amortized linear time complexity.
1.885 - *
1.886 - * @param minCapacity the minimum required capacity of the tmp array
1.887 - * @return tmp, whether or not it grew
1.888 - */
1.889 - private T[] ensureCapacity(int minCapacity) {
1.890 - if (tmp.length < minCapacity) {
1.891 - // Compute smallest power of 2 > minCapacity
1.892 - int newSize = minCapacity;
1.893 - newSize |= newSize >> 1;
1.894 - newSize |= newSize >> 2;
1.895 - newSize |= newSize >> 4;
1.896 - newSize |= newSize >> 8;
1.897 - newSize |= newSize >> 16;
1.898 - newSize++;
1.899 -
1.900 - if (newSize < 0) // Not bloody likely!
1.901 - newSize = minCapacity;
1.902 - else
1.903 - newSize = Math.min(newSize, a.length >>> 1);
1.904 -
1.905 - @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
1.906 - T[] newArray = (T[]) new Object[newSize];
1.907 - tmp = newArray;
1.908 - }
1.909 - return tmp;
1.910 - }
1.911 -
1.912 - /**
1.913 - * Checks that fromIndex and toIndex are in range, and throws an
1.914 - * appropriate exception if they aren't.
1.915 - *
1.916 - * @param arrayLen the length of the array
1.917 - * @param fromIndex the index of the first element of the range
1.918 - * @param toIndex the index after the last element of the range
1.919 - * @throws IllegalArgumentException if fromIndex > toIndex
1.920 - * @throws ArrayIndexOutOfBoundsException if fromIndex < 0
1.921 - * or toIndex > arrayLen
1.922 - */
1.923 - private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) {
1.924 - if (fromIndex > toIndex)
1.925 - throw new IllegalArgumentException("fromIndex(" + fromIndex +
1.926 - ") > toIndex(" + toIndex+")");
1.927 - if (fromIndex < 0)
1.928 - throw new ArrayIndexOutOfBoundsException(fromIndex);
1.929 - if (toIndex > arrayLen)
1.930 - throw new ArrayIndexOutOfBoundsException(toIndex);
1.931 - }
1.932 -}