1.1 --- a/emul/compact/src/main/java/java/util/ComparableTimSort.java Mon Feb 25 19:00:08 2013 +0100
1.2 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000
1.3 @@ -1,896 +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 - * This is a near duplicate of {@link TimSort}, modified for use with
1.34 - * arrays of objects that implement {@link Comparable}, instead of using
1.35 - * explicit comparators.
1.36 - *
1.37 - * <p>If you are using an optimizing VM, you may find that ComparableTimSort
1.38 - * offers no performance benefit over TimSort in conjunction with a
1.39 - * comparator that simply returns {@code ((Comparable)first).compareTo(Second)}.
1.40 - * If this is the case, you are better off deleting ComparableTimSort to
1.41 - * eliminate the code duplication. (See Arrays.java for details.)
1.42 - *
1.43 - * @author Josh Bloch
1.44 - */
1.45 -class ComparableTimSort {
1.46 - /**
1.47 - * This is the minimum sized sequence that will be merged. Shorter
1.48 - * sequences will be lengthened by calling binarySort. If the entire
1.49 - * array is less than this length, no merges will be performed.
1.50 - *
1.51 - * This constant should be a power of two. It was 64 in Tim Peter's C
1.52 - * implementation, but 32 was empirically determined to work better in
1.53 - * this implementation. In the unlikely event that you set this constant
1.54 - * to be a number that's not a power of two, you'll need to change the
1.55 - * {@link #minRunLength} computation.
1.56 - *
1.57 - * If you decrease this constant, you must change the stackLen
1.58 - * computation in the TimSort constructor, or you risk an
1.59 - * ArrayOutOfBounds exception. See listsort.txt for a discussion
1.60 - * of the minimum stack length required as a function of the length
1.61 - * of the array being sorted and the minimum merge sequence length.
1.62 - */
1.63 - private static final int MIN_MERGE = 32;
1.64 -
1.65 - /**
1.66 - * The array being sorted.
1.67 - */
1.68 - private final Object[] a;
1.69 -
1.70 - /**
1.71 - * When we get into galloping mode, we stay there until both runs win less
1.72 - * often than MIN_GALLOP consecutive times.
1.73 - */
1.74 - private static final int MIN_GALLOP = 7;
1.75 -
1.76 - /**
1.77 - * This controls when we get *into* galloping mode. It is initialized
1.78 - * to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
1.79 - * random data, and lower for highly structured data.
1.80 - */
1.81 - private int minGallop = MIN_GALLOP;
1.82 -
1.83 - /**
1.84 - * Maximum initial size of tmp array, which is used for merging. The array
1.85 - * can grow to accommodate demand.
1.86 - *
1.87 - * Unlike Tim's original C version, we do not allocate this much storage
1.88 - * when sorting smaller arrays. This change was required for performance.
1.89 - */
1.90 - private static final int INITIAL_TMP_STORAGE_LENGTH = 256;
1.91 -
1.92 - /**
1.93 - * Temp storage for merges.
1.94 - */
1.95 - private Object[] tmp;
1.96 -
1.97 - /**
1.98 - * A stack of pending runs yet to be merged. Run i starts at
1.99 - * address base[i] and extends for len[i] elements. It's always
1.100 - * true (so long as the indices are in bounds) that:
1.101 - *
1.102 - * runBase[i] + runLen[i] == runBase[i + 1]
1.103 - *
1.104 - * so we could cut the storage for this, but it's a minor amount,
1.105 - * and keeping all the info explicit simplifies the code.
1.106 - */
1.107 - private int stackSize = 0; // Number of pending runs on stack
1.108 - private final int[] runBase;
1.109 - private final int[] runLen;
1.110 -
1.111 - /**
1.112 - * Creates a TimSort instance to maintain the state of an ongoing sort.
1.113 - *
1.114 - * @param a the array to be sorted
1.115 - */
1.116 - private ComparableTimSort(Object[] a) {
1.117 - this.a = a;
1.118 -
1.119 - // Allocate temp storage (which may be increased later if necessary)
1.120 - int len = a.length;
1.121 - @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
1.122 - Object[] newArray = new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ?
1.123 - len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];
1.124 - tmp = newArray;
1.125 -
1.126 - /*
1.127 - * Allocate runs-to-be-merged stack (which cannot be expanded). The
1.128 - * stack length requirements are described in listsort.txt. The C
1.129 - * version always uses the same stack length (85), but this was
1.130 - * measured to be too expensive when sorting "mid-sized" arrays (e.g.,
1.131 - * 100 elements) in Java. Therefore, we use smaller (but sufficiently
1.132 - * large) stack lengths for smaller arrays. The "magic numbers" in the
1.133 - * computation below must be changed if MIN_MERGE is decreased. See
1.134 - * the MIN_MERGE declaration above for more information.
1.135 - */
1.136 - int stackLen = (len < 120 ? 5 :
1.137 - len < 1542 ? 10 :
1.138 - len < 119151 ? 19 : 40);
1.139 - runBase = new int[stackLen];
1.140 - runLen = new int[stackLen];
1.141 - }
1.142 -
1.143 - /*
1.144 - * The next two methods (which are package private and static) constitute
1.145 - * the entire API of this class. Each of these methods obeys the contract
1.146 - * of the public method with the same signature in java.util.Arrays.
1.147 - */
1.148 -
1.149 - static void sort(Object[] a) {
1.150 - sort(a, 0, a.length);
1.151 - }
1.152 -
1.153 - static void sort(Object[] a, int lo, int hi) {
1.154 - rangeCheck(a.length, lo, hi);
1.155 - int nRemaining = hi - lo;
1.156 - if (nRemaining < 2)
1.157 - return; // Arrays of size 0 and 1 are always sorted
1.158 -
1.159 - // If array is small, do a "mini-TimSort" with no merges
1.160 - if (nRemaining < MIN_MERGE) {
1.161 - int initRunLen = countRunAndMakeAscending(a, lo, hi);
1.162 - binarySort(a, lo, hi, lo + initRunLen);
1.163 - return;
1.164 - }
1.165 -
1.166 - /**
1.167 - * March over the array once, left to right, finding natural runs,
1.168 - * extending short natural runs to minRun elements, and merging runs
1.169 - * to maintain stack invariant.
1.170 - */
1.171 - ComparableTimSort ts = new ComparableTimSort(a);
1.172 - int minRun = minRunLength(nRemaining);
1.173 - do {
1.174 - // Identify next run
1.175 - int runLen = countRunAndMakeAscending(a, lo, hi);
1.176 -
1.177 - // If run is short, extend to min(minRun, nRemaining)
1.178 - if (runLen < minRun) {
1.179 - int force = nRemaining <= minRun ? nRemaining : minRun;
1.180 - binarySort(a, lo, lo + force, lo + runLen);
1.181 - runLen = force;
1.182 - }
1.183 -
1.184 - // Push run onto pending-run stack, and maybe merge
1.185 - ts.pushRun(lo, runLen);
1.186 - ts.mergeCollapse();
1.187 -
1.188 - // Advance to find next run
1.189 - lo += runLen;
1.190 - nRemaining -= runLen;
1.191 - } while (nRemaining != 0);
1.192 -
1.193 - // Merge all remaining runs to complete sort
1.194 - assert lo == hi;
1.195 - ts.mergeForceCollapse();
1.196 - assert ts.stackSize == 1;
1.197 - }
1.198 -
1.199 - /**
1.200 - * Sorts the specified portion of the specified array using a binary
1.201 - * insertion sort. This is the best method for sorting small numbers
1.202 - * of elements. It requires O(n log n) compares, but O(n^2) data
1.203 - * movement (worst case).
1.204 - *
1.205 - * If the initial part of the specified range is already sorted,
1.206 - * this method can take advantage of it: the method assumes that the
1.207 - * elements from index {@code lo}, inclusive, to {@code start},
1.208 - * exclusive are already sorted.
1.209 - *
1.210 - * @param a the array in which a range is to be sorted
1.211 - * @param lo the index of the first element in the range to be sorted
1.212 - * @param hi the index after the last element in the range to be sorted
1.213 - * @param start the index of the first element in the range that is
1.214 - * not already known to be sorted ({@code lo <= start <= hi})
1.215 - */
1.216 - @SuppressWarnings("fallthrough")
1.217 - private static void binarySort(Object[] a, int lo, int hi, int start) {
1.218 - assert lo <= start && start <= hi;
1.219 - if (start == lo)
1.220 - start++;
1.221 - for ( ; start < hi; start++) {
1.222 - @SuppressWarnings("unchecked")
1.223 - Comparable<Object> pivot = (Comparable) a[start];
1.224 -
1.225 - // Set left (and right) to the index where a[start] (pivot) belongs
1.226 - int left = lo;
1.227 - int right = start;
1.228 - assert left <= right;
1.229 - /*
1.230 - * Invariants:
1.231 - * pivot >= all in [lo, left).
1.232 - * pivot < all in [right, start).
1.233 - */
1.234 - while (left < right) {
1.235 - int mid = (left + right) >>> 1;
1.236 - if (pivot.compareTo(a[mid]) < 0)
1.237 - right = mid;
1.238 - else
1.239 - left = mid + 1;
1.240 - }
1.241 - assert left == right;
1.242 -
1.243 - /*
1.244 - * The invariants still hold: pivot >= all in [lo, left) and
1.245 - * pivot < all in [left, start), so pivot belongs at left. Note
1.246 - * that if there are elements equal to pivot, left points to the
1.247 - * first slot after them -- that's why this sort is stable.
1.248 - * Slide elements over to make room for pivot.
1.249 - */
1.250 - int n = start - left; // The number of elements to move
1.251 - // Switch is just an optimization for arraycopy in default case
1.252 - switch (n) {
1.253 - case 2: a[left + 2] = a[left + 1];
1.254 - case 1: a[left + 1] = a[left];
1.255 - break;
1.256 - default: System.arraycopy(a, left, a, left + 1, n);
1.257 - }
1.258 - a[left] = pivot;
1.259 - }
1.260 - }
1.261 -
1.262 - /**
1.263 - * Returns the length of the run beginning at the specified position in
1.264 - * the specified array and reverses the run if it is descending (ensuring
1.265 - * that the run will always be ascending when the method returns).
1.266 - *
1.267 - * A run is the longest ascending sequence with:
1.268 - *
1.269 - * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
1.270 - *
1.271 - * or the longest descending sequence with:
1.272 - *
1.273 - * a[lo] > a[lo + 1] > a[lo + 2] > ...
1.274 - *
1.275 - * For its intended use in a stable mergesort, the strictness of the
1.276 - * definition of "descending" is needed so that the call can safely
1.277 - * reverse a descending sequence without violating stability.
1.278 - *
1.279 - * @param a the array in which a run is to be counted and possibly reversed
1.280 - * @param lo index of the first element in the run
1.281 - * @param hi index after the last element that may be contained in the run.
1.282 - It is required that {@code lo < hi}.
1.283 - * @return the length of the run beginning at the specified position in
1.284 - * the specified array
1.285 - */
1.286 - @SuppressWarnings("unchecked")
1.287 - private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
1.288 - assert lo < hi;
1.289 - int runHi = lo + 1;
1.290 - if (runHi == hi)
1.291 - return 1;
1.292 -
1.293 - // Find end of run, and reverse range if descending
1.294 - if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending
1.295 - while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0)
1.296 - runHi++;
1.297 - reverseRange(a, lo, runHi);
1.298 - } else { // Ascending
1.299 - while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0)
1.300 - runHi++;
1.301 - }
1.302 -
1.303 - return runHi - lo;
1.304 - }
1.305 -
1.306 - /**
1.307 - * Reverse the specified range of the specified array.
1.308 - *
1.309 - * @param a the array in which a range is to be reversed
1.310 - * @param lo the index of the first element in the range to be reversed
1.311 - * @param hi the index after the last element in the range to be reversed
1.312 - */
1.313 - private static void reverseRange(Object[] a, int lo, int hi) {
1.314 - hi--;
1.315 - while (lo < hi) {
1.316 - Object t = a[lo];
1.317 - a[lo++] = a[hi];
1.318 - a[hi--] = t;
1.319 - }
1.320 - }
1.321 -
1.322 - /**
1.323 - * Returns the minimum acceptable run length for an array of the specified
1.324 - * length. Natural runs shorter than this will be extended with
1.325 - * {@link #binarySort}.
1.326 - *
1.327 - * Roughly speaking, the computation is:
1.328 - *
1.329 - * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
1.330 - * Else if n is an exact power of 2, return MIN_MERGE/2.
1.331 - * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
1.332 - * is close to, but strictly less than, an exact power of 2.
1.333 - *
1.334 - * For the rationale, see listsort.txt.
1.335 - *
1.336 - * @param n the length of the array to be sorted
1.337 - * @return the length of the minimum run to be merged
1.338 - */
1.339 - private static int minRunLength(int n) {
1.340 - assert n >= 0;
1.341 - int r = 0; // Becomes 1 if any 1 bits are shifted off
1.342 - while (n >= MIN_MERGE) {
1.343 - r |= (n & 1);
1.344 - n >>= 1;
1.345 - }
1.346 - return n + r;
1.347 - }
1.348 -
1.349 - /**
1.350 - * Pushes the specified run onto the pending-run stack.
1.351 - *
1.352 - * @param runBase index of the first element in the run
1.353 - * @param runLen the number of elements in the run
1.354 - */
1.355 - private void pushRun(int runBase, int runLen) {
1.356 - this.runBase[stackSize] = runBase;
1.357 - this.runLen[stackSize] = runLen;
1.358 - stackSize++;
1.359 - }
1.360 -
1.361 - /**
1.362 - * Examines the stack of runs waiting to be merged and merges adjacent runs
1.363 - * until the stack invariants are reestablished:
1.364 - *
1.365 - * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
1.366 - * 2. runLen[i - 2] > runLen[i - 1]
1.367 - *
1.368 - * This method is called each time a new run is pushed onto the stack,
1.369 - * so the invariants are guaranteed to hold for i < stackSize upon
1.370 - * entry to the method.
1.371 - */
1.372 - private void mergeCollapse() {
1.373 - while (stackSize > 1) {
1.374 - int n = stackSize - 2;
1.375 - if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
1.376 - if (runLen[n - 1] < runLen[n + 1])
1.377 - n--;
1.378 - mergeAt(n);
1.379 - } else if (runLen[n] <= runLen[n + 1]) {
1.380 - mergeAt(n);
1.381 - } else {
1.382 - break; // Invariant is established
1.383 - }
1.384 - }
1.385 - }
1.386 -
1.387 - /**
1.388 - * Merges all runs on the stack until only one remains. This method is
1.389 - * called once, to complete the sort.
1.390 - */
1.391 - private void mergeForceCollapse() {
1.392 - while (stackSize > 1) {
1.393 - int n = stackSize - 2;
1.394 - if (n > 0 && runLen[n - 1] < runLen[n + 1])
1.395 - n--;
1.396 - mergeAt(n);
1.397 - }
1.398 - }
1.399 -
1.400 - /**
1.401 - * Merges the two runs at stack indices i and i+1. Run i must be
1.402 - * the penultimate or antepenultimate run on the stack. In other words,
1.403 - * i must be equal to stackSize-2 or stackSize-3.
1.404 - *
1.405 - * @param i stack index of the first of the two runs to merge
1.406 - */
1.407 - @SuppressWarnings("unchecked")
1.408 - private void mergeAt(int i) {
1.409 - assert stackSize >= 2;
1.410 - assert i >= 0;
1.411 - assert i == stackSize - 2 || i == stackSize - 3;
1.412 -
1.413 - int base1 = runBase[i];
1.414 - int len1 = runLen[i];
1.415 - int base2 = runBase[i + 1];
1.416 - int len2 = runLen[i + 1];
1.417 - assert len1 > 0 && len2 > 0;
1.418 - assert base1 + len1 == base2;
1.419 -
1.420 - /*
1.421 - * Record the length of the combined runs; if i is the 3rd-last
1.422 - * run now, also slide over the last run (which isn't involved
1.423 - * in this merge). The current run (i+1) goes away in any case.
1.424 - */
1.425 - runLen[i] = len1 + len2;
1.426 - if (i == stackSize - 3) {
1.427 - runBase[i + 1] = runBase[i + 2];
1.428 - runLen[i + 1] = runLen[i + 2];
1.429 - }
1.430 - stackSize--;
1.431 -
1.432 - /*
1.433 - * Find where the first element of run2 goes in run1. Prior elements
1.434 - * in run1 can be ignored (because they're already in place).
1.435 - */
1.436 - int k = gallopRight((Comparable<Object>) a[base2], a, base1, len1, 0);
1.437 - assert k >= 0;
1.438 - base1 += k;
1.439 - len1 -= k;
1.440 - if (len1 == 0)
1.441 - return;
1.442 -
1.443 - /*
1.444 - * Find where the last element of run1 goes in run2. Subsequent elements
1.445 - * in run2 can be ignored (because they're already in place).
1.446 - */
1.447 - len2 = gallopLeft((Comparable<Object>) a[base1 + len1 - 1], a,
1.448 - base2, len2, len2 - 1);
1.449 - assert len2 >= 0;
1.450 - if (len2 == 0)
1.451 - return;
1.452 -
1.453 - // Merge remaining runs, using tmp array with min(len1, len2) elements
1.454 - if (len1 <= len2)
1.455 - mergeLo(base1, len1, base2, len2);
1.456 - else
1.457 - mergeHi(base1, len1, base2, len2);
1.458 - }
1.459 -
1.460 - /**
1.461 - * Locates the position at which to insert the specified key into the
1.462 - * specified sorted range; if the range contains an element equal to key,
1.463 - * returns the index of the leftmost equal element.
1.464 - *
1.465 - * @param key the key whose insertion point to search for
1.466 - * @param a the array in which to search
1.467 - * @param base the index of the first element in the range
1.468 - * @param len the length of the range; must be > 0
1.469 - * @param hint the index at which to begin the search, 0 <= hint < n.
1.470 - * The closer hint is to the result, the faster this method will run.
1.471 - * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
1.472 - * pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
1.473 - * In other words, key belongs at index b + k; or in other words,
1.474 - * the first k elements of a should precede key, and the last n - k
1.475 - * should follow it.
1.476 - */
1.477 - private static int gallopLeft(Comparable<Object> key, Object[] a,
1.478 - int base, int len, int hint) {
1.479 - assert len > 0 && hint >= 0 && hint < len;
1.480 -
1.481 - int lastOfs = 0;
1.482 - int ofs = 1;
1.483 - if (key.compareTo(a[base + hint]) > 0) {
1.484 - // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
1.485 - int maxOfs = len - hint;
1.486 - while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) > 0) {
1.487 - lastOfs = ofs;
1.488 - ofs = (ofs << 1) + 1;
1.489 - if (ofs <= 0) // int overflow
1.490 - ofs = maxOfs;
1.491 - }
1.492 - if (ofs > maxOfs)
1.493 - ofs = maxOfs;
1.494 -
1.495 - // Make offsets relative to base
1.496 - lastOfs += hint;
1.497 - ofs += hint;
1.498 - } else { // key <= a[base + hint]
1.499 - // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
1.500 - final int maxOfs = hint + 1;
1.501 - while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) <= 0) {
1.502 - lastOfs = ofs;
1.503 - ofs = (ofs << 1) + 1;
1.504 - if (ofs <= 0) // int overflow
1.505 - ofs = maxOfs;
1.506 - }
1.507 - if (ofs > maxOfs)
1.508 - ofs = maxOfs;
1.509 -
1.510 - // Make offsets relative to base
1.511 - int tmp = lastOfs;
1.512 - lastOfs = hint - ofs;
1.513 - ofs = hint - tmp;
1.514 - }
1.515 - assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
1.516 -
1.517 - /*
1.518 - * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
1.519 - * to the right of lastOfs but no farther right than ofs. Do a binary
1.520 - * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
1.521 - */
1.522 - lastOfs++;
1.523 - while (lastOfs < ofs) {
1.524 - int m = lastOfs + ((ofs - lastOfs) >>> 1);
1.525 -
1.526 - if (key.compareTo(a[base + m]) > 0)
1.527 - lastOfs = m + 1; // a[base + m] < key
1.528 - else
1.529 - ofs = m; // key <= a[base + m]
1.530 - }
1.531 - assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
1.532 - return ofs;
1.533 - }
1.534 -
1.535 - /**
1.536 - * Like gallopLeft, except that if the range contains an element equal to
1.537 - * key, gallopRight returns the index after the rightmost equal element.
1.538 - *
1.539 - * @param key the key whose insertion point to search for
1.540 - * @param a the array in which to search
1.541 - * @param base the index of the first element in the range
1.542 - * @param len the length of the range; must be > 0
1.543 - * @param hint the index at which to begin the search, 0 <= hint < n.
1.544 - * The closer hint is to the result, the faster this method will run.
1.545 - * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
1.546 - */
1.547 - private static int gallopRight(Comparable<Object> key, Object[] a,
1.548 - int base, int len, int hint) {
1.549 - assert len > 0 && hint >= 0 && hint < len;
1.550 -
1.551 - int ofs = 1;
1.552 - int lastOfs = 0;
1.553 - if (key.compareTo(a[base + hint]) < 0) {
1.554 - // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
1.555 - int maxOfs = hint + 1;
1.556 - while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) < 0) {
1.557 - lastOfs = ofs;
1.558 - ofs = (ofs << 1) + 1;
1.559 - if (ofs <= 0) // int overflow
1.560 - ofs = maxOfs;
1.561 - }
1.562 - if (ofs > maxOfs)
1.563 - ofs = maxOfs;
1.564 -
1.565 - // Make offsets relative to b
1.566 - int tmp = lastOfs;
1.567 - lastOfs = hint - ofs;
1.568 - ofs = hint - tmp;
1.569 - } else { // a[b + hint] <= key
1.570 - // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
1.571 - int maxOfs = len - hint;
1.572 - while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) >= 0) {
1.573 - lastOfs = ofs;
1.574 - ofs = (ofs << 1) + 1;
1.575 - if (ofs <= 0) // int overflow
1.576 - ofs = maxOfs;
1.577 - }
1.578 - if (ofs > maxOfs)
1.579 - ofs = maxOfs;
1.580 -
1.581 - // Make offsets relative to b
1.582 - lastOfs += hint;
1.583 - ofs += hint;
1.584 - }
1.585 - assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
1.586 -
1.587 - /*
1.588 - * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
1.589 - * the right of lastOfs but no farther right than ofs. Do a binary
1.590 - * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
1.591 - */
1.592 - lastOfs++;
1.593 - while (lastOfs < ofs) {
1.594 - int m = lastOfs + ((ofs - lastOfs) >>> 1);
1.595 -
1.596 - if (key.compareTo(a[base + m]) < 0)
1.597 - ofs = m; // key < a[b + m]
1.598 - else
1.599 - lastOfs = m + 1; // a[b + m] <= key
1.600 - }
1.601 - assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
1.602 - return ofs;
1.603 - }
1.604 -
1.605 - /**
1.606 - * Merges two adjacent runs in place, in a stable fashion. The first
1.607 - * element of the first run must be greater than the first element of the
1.608 - * second run (a[base1] > a[base2]), and the last element of the first run
1.609 - * (a[base1 + len1-1]) must be greater than all elements of the second run.
1.610 - *
1.611 - * For performance, this method should be called only when len1 <= len2;
1.612 - * its twin, mergeHi should be called if len1 >= len2. (Either method
1.613 - * may be called if len1 == len2.)
1.614 - *
1.615 - * @param base1 index of first element in first run to be merged
1.616 - * @param len1 length of first run to be merged (must be > 0)
1.617 - * @param base2 index of first element in second run to be merged
1.618 - * (must be aBase + aLen)
1.619 - * @param len2 length of second run to be merged (must be > 0)
1.620 - */
1.621 - @SuppressWarnings("unchecked")
1.622 - private void mergeLo(int base1, int len1, int base2, int len2) {
1.623 - assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
1.624 -
1.625 - // Copy first run into temp array
1.626 - Object[] a = this.a; // For performance
1.627 - Object[] tmp = ensureCapacity(len1);
1.628 - System.arraycopy(a, base1, tmp, 0, len1);
1.629 -
1.630 - int cursor1 = 0; // Indexes into tmp array
1.631 - int cursor2 = base2; // Indexes int a
1.632 - int dest = base1; // Indexes int a
1.633 -
1.634 - // Move first element of second run and deal with degenerate cases
1.635 - a[dest++] = a[cursor2++];
1.636 - if (--len2 == 0) {
1.637 - System.arraycopy(tmp, cursor1, a, dest, len1);
1.638 - return;
1.639 - }
1.640 - if (len1 == 1) {
1.641 - System.arraycopy(a, cursor2, a, dest, len2);
1.642 - a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
1.643 - return;
1.644 - }
1.645 -
1.646 - int minGallop = this.minGallop; // Use local variable for performance
1.647 - outer:
1.648 - while (true) {
1.649 - int count1 = 0; // Number of times in a row that first run won
1.650 - int count2 = 0; // Number of times in a row that second run won
1.651 -
1.652 - /*
1.653 - * Do the straightforward thing until (if ever) one run starts
1.654 - * winning consistently.
1.655 - */
1.656 - do {
1.657 - assert len1 > 1 && len2 > 0;
1.658 - if (((Comparable) a[cursor2]).compareTo(tmp[cursor1]) < 0) {
1.659 - a[dest++] = a[cursor2++];
1.660 - count2++;
1.661 - count1 = 0;
1.662 - if (--len2 == 0)
1.663 - break outer;
1.664 - } else {
1.665 - a[dest++] = tmp[cursor1++];
1.666 - count1++;
1.667 - count2 = 0;
1.668 - if (--len1 == 1)
1.669 - break outer;
1.670 - }
1.671 - } while ((count1 | count2) < minGallop);
1.672 -
1.673 - /*
1.674 - * One run is winning so consistently that galloping may be a
1.675 - * huge win. So try that, and continue galloping until (if ever)
1.676 - * neither run appears to be winning consistently anymore.
1.677 - */
1.678 - do {
1.679 - assert len1 > 1 && len2 > 0;
1.680 - count1 = gallopRight((Comparable) a[cursor2], tmp, cursor1, len1, 0);
1.681 - if (count1 != 0) {
1.682 - System.arraycopy(tmp, cursor1, a, dest, count1);
1.683 - dest += count1;
1.684 - cursor1 += count1;
1.685 - len1 -= count1;
1.686 - if (len1 <= 1) // len1 == 1 || len1 == 0
1.687 - break outer;
1.688 - }
1.689 - a[dest++] = a[cursor2++];
1.690 - if (--len2 == 0)
1.691 - break outer;
1.692 -
1.693 - count2 = gallopLeft((Comparable) tmp[cursor1], a, cursor2, len2, 0);
1.694 - if (count2 != 0) {
1.695 - System.arraycopy(a, cursor2, a, dest, count2);
1.696 - dest += count2;
1.697 - cursor2 += count2;
1.698 - len2 -= count2;
1.699 - if (len2 == 0)
1.700 - break outer;
1.701 - }
1.702 - a[dest++] = tmp[cursor1++];
1.703 - if (--len1 == 1)
1.704 - break outer;
1.705 - minGallop--;
1.706 - } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
1.707 - if (minGallop < 0)
1.708 - minGallop = 0;
1.709 - minGallop += 2; // Penalize for leaving gallop mode
1.710 - } // End of "outer" loop
1.711 - this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
1.712 -
1.713 - if (len1 == 1) {
1.714 - assert len2 > 0;
1.715 - System.arraycopy(a, cursor2, a, dest, len2);
1.716 - a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
1.717 - } else if (len1 == 0) {
1.718 - throw new IllegalArgumentException(
1.719 - "Comparison method violates its general contract!");
1.720 - } else {
1.721 - assert len2 == 0;
1.722 - assert len1 > 1;
1.723 - System.arraycopy(tmp, cursor1, a, dest, len1);
1.724 - }
1.725 - }
1.726 -
1.727 - /**
1.728 - * Like mergeLo, except that this method should be called only if
1.729 - * len1 >= len2; mergeLo should be called if len1 <= len2. (Either method
1.730 - * may be called if len1 == len2.)
1.731 - *
1.732 - * @param base1 index of first element in first run to be merged
1.733 - * @param len1 length of first run to be merged (must be > 0)
1.734 - * @param base2 index of first element in second run to be merged
1.735 - * (must be aBase + aLen)
1.736 - * @param len2 length of second run to be merged (must be > 0)
1.737 - */
1.738 - @SuppressWarnings("unchecked")
1.739 - private void mergeHi(int base1, int len1, int base2, int len2) {
1.740 - assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
1.741 -
1.742 - // Copy second run into temp array
1.743 - Object[] a = this.a; // For performance
1.744 - Object[] tmp = ensureCapacity(len2);
1.745 - System.arraycopy(a, base2, tmp, 0, len2);
1.746 -
1.747 - int cursor1 = base1 + len1 - 1; // Indexes into a
1.748 - int cursor2 = len2 - 1; // Indexes into tmp array
1.749 - int dest = base2 + len2 - 1; // Indexes into a
1.750 -
1.751 - // Move last element of first run and deal with degenerate cases
1.752 - a[dest--] = a[cursor1--];
1.753 - if (--len1 == 0) {
1.754 - System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
1.755 - return;
1.756 - }
1.757 - if (len2 == 1) {
1.758 - dest -= len1;
1.759 - cursor1 -= len1;
1.760 - System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
1.761 - a[dest] = tmp[cursor2];
1.762 - return;
1.763 - }
1.764 -
1.765 - int minGallop = this.minGallop; // Use local variable for performance
1.766 - outer:
1.767 - while (true) {
1.768 - int count1 = 0; // Number of times in a row that first run won
1.769 - int count2 = 0; // Number of times in a row that second run won
1.770 -
1.771 - /*
1.772 - * Do the straightforward thing until (if ever) one run
1.773 - * appears to win consistently.
1.774 - */
1.775 - do {
1.776 - assert len1 > 0 && len2 > 1;
1.777 - if (((Comparable) tmp[cursor2]).compareTo(a[cursor1]) < 0) {
1.778 - a[dest--] = a[cursor1--];
1.779 - count1++;
1.780 - count2 = 0;
1.781 - if (--len1 == 0)
1.782 - break outer;
1.783 - } else {
1.784 - a[dest--] = tmp[cursor2--];
1.785 - count2++;
1.786 - count1 = 0;
1.787 - if (--len2 == 1)
1.788 - break outer;
1.789 - }
1.790 - } while ((count1 | count2) < minGallop);
1.791 -
1.792 - /*
1.793 - * One run is winning so consistently that galloping may be a
1.794 - * huge win. So try that, and continue galloping until (if ever)
1.795 - * neither run appears to be winning consistently anymore.
1.796 - */
1.797 - do {
1.798 - assert len1 > 0 && len2 > 1;
1.799 - count1 = len1 - gallopRight((Comparable) tmp[cursor2], a, base1, len1, len1 - 1);
1.800 - if (count1 != 0) {
1.801 - dest -= count1;
1.802 - cursor1 -= count1;
1.803 - len1 -= count1;
1.804 - System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
1.805 - if (len1 == 0)
1.806 - break outer;
1.807 - }
1.808 - a[dest--] = tmp[cursor2--];
1.809 - if (--len2 == 1)
1.810 - break outer;
1.811 -
1.812 - count2 = len2 - gallopLeft((Comparable) a[cursor1], tmp, 0, len2, len2 - 1);
1.813 - if (count2 != 0) {
1.814 - dest -= count2;
1.815 - cursor2 -= count2;
1.816 - len2 -= count2;
1.817 - System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
1.818 - if (len2 <= 1)
1.819 - break outer; // len2 == 1 || len2 == 0
1.820 - }
1.821 - a[dest--] = a[cursor1--];
1.822 - if (--len1 == 0)
1.823 - break outer;
1.824 - minGallop--;
1.825 - } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
1.826 - if (minGallop < 0)
1.827 - minGallop = 0;
1.828 - minGallop += 2; // Penalize for leaving gallop mode
1.829 - } // End of "outer" loop
1.830 - this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
1.831 -
1.832 - if (len2 == 1) {
1.833 - assert len1 > 0;
1.834 - dest -= len1;
1.835 - cursor1 -= len1;
1.836 - System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
1.837 - a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
1.838 - } else if (len2 == 0) {
1.839 - throw new IllegalArgumentException(
1.840 - "Comparison method violates its general contract!");
1.841 - } else {
1.842 - assert len1 == 0;
1.843 - assert len2 > 0;
1.844 - System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
1.845 - }
1.846 - }
1.847 -
1.848 - /**
1.849 - * Ensures that the external array tmp has at least the specified
1.850 - * number of elements, increasing its size if necessary. The size
1.851 - * increases exponentially to ensure amortized linear time complexity.
1.852 - *
1.853 - * @param minCapacity the minimum required capacity of the tmp array
1.854 - * @return tmp, whether or not it grew
1.855 - */
1.856 - private Object[] ensureCapacity(int minCapacity) {
1.857 - if (tmp.length < minCapacity) {
1.858 - // Compute smallest power of 2 > minCapacity
1.859 - int newSize = minCapacity;
1.860 - newSize |= newSize >> 1;
1.861 - newSize |= newSize >> 2;
1.862 - newSize |= newSize >> 4;
1.863 - newSize |= newSize >> 8;
1.864 - newSize |= newSize >> 16;
1.865 - newSize++;
1.866 -
1.867 - if (newSize < 0) // Not bloody likely!
1.868 - newSize = minCapacity;
1.869 - else
1.870 - newSize = Math.min(newSize, a.length >>> 1);
1.871 -
1.872 - @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
1.873 - Object[] newArray = new Object[newSize];
1.874 - tmp = newArray;
1.875 - }
1.876 - return tmp;
1.877 - }
1.878 -
1.879 - /**
1.880 - * Checks that fromIndex and toIndex are in range, and throws an
1.881 - * appropriate exception if they aren't.
1.882 - *
1.883 - * @param arrayLen the length of the array
1.884 - * @param fromIndex the index of the first element of the range
1.885 - * @param toIndex the index after the last element of the range
1.886 - * @throws IllegalArgumentException if fromIndex > toIndex
1.887 - * @throws ArrayIndexOutOfBoundsException if fromIndex < 0
1.888 - * or toIndex > arrayLen
1.889 - */
1.890 - private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) {
1.891 - if (fromIndex > toIndex)
1.892 - throw new IllegalArgumentException("fromIndex(" + fromIndex +
1.893 - ") > toIndex(" + toIndex+")");
1.894 - if (fromIndex < 0)
1.895 - throw new ArrayIndexOutOfBoundsException(fromIndex);
1.896 - if (toIndex > arrayLen)
1.897 - throw new ArrayIndexOutOfBoundsException(toIndex);
1.898 - }
1.899 -}