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- /*
- * Copyright 2009 Google Inc. All Rights Reserved.
- * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
- *
- * This code is free software; you can redistribute it and/or modify it
- * under the terms of the GNU General Public License version 2 only, as
- * published by the Free Software Foundation. Sun designates this
- * particular file as subject to the "Classpath" exception as provided
- * by Sun in the LICENSE file that accompanied this code.
- *
- * This code is distributed in the hope that it will be useful, but WITHOUT
- * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
- * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
- * version 2 for more details (a copy is included in the LICENSE file that
- * accompanied this code).
- *
- * You should have received a copy of the GNU General Public License version
- * 2 along with this work; if not, write to the Free Software Foundation,
- * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
- *
- * Please contact Sun Microsystems, Inc., 4150 Network Circle, Santa Clara,
- * CA 95054 USA or visit www.sun.com if you need additional information or
- * have any questions.
- */
- // glowcoder commented
- // package java.util;
- import java.util.*;
- /**
- * A stable, adaptive, iterative mergesort that requires far fewer than
- * n lg(n) comparisons when running on partially sorted arrays, while
- * offering performance comparable to a traditional mergesort when run
- * on random arrays. Like all proper mergesorts, this sort is stable and
- * runs O(n log n) time (worst case). In the worst case, this sort requires
- * temporary storage space for n/2 object references; in the best case,
- * it requires only a small constant amount of space.
- *
- * This implementation was adapted from Tim Peters's list sort for
- * Python, which is described in detail here:
- *
- * http://svn.python.org/projects/python/trunk/Objects/listsort.txt
- *
- * Tim's C code may be found here:
- *
- * http://svn.python.org/projects/python/trunk/Objects/listobject.c
- *
- * The underlying techniques are described in this paper (and may have
- * even earlier origins):
- *
- * "Optimistic Sorting and Information Theoretic Complexity"
- * Peter McIlroy
- * SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),
- * pp 467-474, Austin, Texas, 25-27 January 1993.
- *
- * While the API to this class consists solely of static methods, it is
- * (privately) instantiable; a TimSort instance holds the state of an ongoing
- * sort, assuming the input array is large enough to warrant the full-blown
- * TimSort. Small arrays are sorted in place, using a binary insertion sort.
- *
- * @author Josh Bloch
- */
- class TimSort<T> {
- /**
- * This is the minimum sized sequence that will be merged. Shorter
- * sequences will be lengthened by calling binarySort. If the entire
- * array is less than this length, no merges will be performed.
- *
- * This constant should be a power of two. It was 64 in Tim Peter's C
- * implementation, but 32 was empirically determined to work better in
- * this implementation. In the unlikely event that you set this constant
- * to be a number that's not a power of two, you'll need to change the
- * {@link #minRunLength} computation.
- *
- * If you decrease this constant, you must change the stackLen
- * computation in the TimSort constructor, or you risk an
- * ArrayOutOfBounds exception. See listsort.txt for a discussion
- * of the minimum stack length required as a function of the length
- * of the array being sorted and the minimum merge sequence length.
- */
- private static final int MIN_MERGE = 32;
- /**
- * The array being sorted.
- */
- private final T[] a;
- /**
- * The comparator for this sort.
- */
- private final Comparator<? super T> c;
- /**
- * When we get into galloping mode, we stay there until both runs win less
- * often than MIN_GALLOP consecutive times.
- */
- private static final int MIN_GALLOP = 7;
- /**
- * This controls when we get *into* galloping mode. It is initialized
- * to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
- * random data, and lower for highly structured data.
- */
- private int minGallop = MIN_GALLOP;
- /**
- * Maximum initial size of tmp array, which is used for merging. The array
- * can grow to accommodate demand.
- *
- * Unlike Tim's original C version, we do not allocate this much storage
- * when sorting smaller arrays. This change was required for performance.
- */
- private static final int INITIAL_TMP_STORAGE_LENGTH = 256;
- /**
- * Temp storage for merges.
- */
- private T[] tmp; // Actual runtime type will be Object[], regardless of T
- /**
- * A stack of pending runs yet to be merged. Run i starts at
- * address base[i] and extends for len[i] elements. It's always
- * true (so long as the indices are in bounds) that:
- *
- * runBase[i] + runLen[i] == runBase[i + 1]
- *
- * so we could cut the storage for this, but it's a minor amount,
- * and keeping all the info explicit simplifies the code.
- */
- private int stackSize = 0; // Number of pending runs on stack
- private final int[] runBase;
- private final int[] runLen;
- /**
- * Creates a TimSort instance to maintain the state of an ongoing sort.
- *
- * @param a the array to be sorted
- * @param c the comparator to determine the order of the sort
- */
- private TimSort(T[] a, Comparator<? super T> c) {
- this.a = a;
- this.c = c;
- // Allocate temp storage (which may be increased later if necessary)
- int len = a.length;
- @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
- T[] newArray = (T[]) new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ?
- len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];
- tmp = newArray;
- /*
- * Allocate runs-to-be-merged stack (which cannot be expanded). The
- * stack length requirements are described in listsort.txt. The C
- * version always uses the same stack length (85), but this was
- * measured to be too expensive when sorting "mid-sized" arrays (e.g.,
- * 100 elements) in Java. Therefore, we use smaller (but sufficiently
- * large) stack lengths for smaller arrays. The "magic numbers" in the
- * computation below must be changed if MIN_MERGE is decreased. See
- * the MIN_MERGE declaration above for more information.
- */
- int stackLen = (len < 120 ? 5 :
- len < 1542 ? 10 :
- len < 119151 ? 19 : 40);
- runBase = new int[stackLen];
- runLen = new int[stackLen];
- }
- /*
- * The next two methods (which are package private and static) constitute
- * the entire API of this class. Each of these methods obeys the contract
- * of the public method with the same signature in java.util.Arrays.
- */
- // glowcoder added "public"
- public static <T> void sort(T[] a, Comparator<? super T> c) {
- sort(a, 0, a.length, c);
- }
- // glowcoder added "public"
- static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c) {
- if (c == null) {
- Arrays.sort(a, lo, hi);
- return;
- }
- rangeCheck(a.length, lo, hi);
- int nRemaining = hi - lo;
- if (nRemaining < 2)
- return; // Arrays of size 0 and 1 are always sorted
- // If array is small, do a "mini-TimSort" with no merges
- if (nRemaining < MIN_MERGE) {
- int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
- binarySort(a, lo, hi, lo + initRunLen, c);
- return;
- }
- /**
- * March over the array once, left to right, finding natural runs,
- * extending short natural runs to minRun elements, and merging runs
- * to maintain stack invariant.
- */
- TimSort<T> ts = new TimSort<T>(a, c);
- int minRun = minRunLength(nRemaining);
- do {
- // Identify next run
- int runLen = countRunAndMakeAscending(a, lo, hi, c);
- // If run is short, extend to min(minRun, nRemaining)
- if (runLen < minRun) {
- int force = nRemaining <= minRun ? nRemaining : minRun;
- binarySort(a, lo, lo + force, lo + runLen, c);
- runLen = force;
- }
- // Push run onto pending-run stack, and maybe merge
- ts.pushRun(lo, runLen);
- ts.mergeCollapse();
- // Advance to find next run
- lo += runLen;
- nRemaining -= runLen;
- } while (nRemaining != 0);
- // Merge all remaining runs to complete sort
- assert lo == hi;
- ts.mergeForceCollapse();
- assert ts.stackSize == 1;
- }
- /**
- * Sorts the specified portion of the specified array using a binary
- * insertion sort. This is the best method for sorting small numbers
- * of elements. It requires O(n log n) compares, but O(n^2) data
- * movement (worst case).
- *
- * If the initial part of the specified range is already sorted,
- * this method can take advantage of it: the method assumes that the
- * elements from index {@code lo}, inclusive, to {@code start},
- * exclusive are already sorted.
- *
- * @param a the array in which a range is to be sorted
- * @param lo the index of the first element in the range to be sorted
- * @param hi the index after the last element in the range to be sorted
- * @param start the index of the first element in the range that is
- * not already known to be sorted (@code lo <= start <= hi}
- * @param c comparator to used for the sort
- */
- @SuppressWarnings("fallthrough")
- private static <T> void binarySort(T[] a, int lo, int hi, int start,
- Comparator<? super T> c) {
- assert lo <= start && start <= hi;
- if (start == lo)
- start++;
- for ( ; start < hi; start++) {
- T pivot = a[start];
- // Set left (and right) to the index where a[start] (pivot) belongs
- int left = lo;
- int right = start;
- assert left <= right;
- /*
- * Invariants:
- * pivot >= all in [lo, left).
- * pivot < all in [right, start).
- */
- while (left < right) {
- int mid = (left + right) >>> 1;
- if (c.compare(pivot, a[mid]) < 0)
- right = mid;
- else
- left = mid + 1;
- }
- assert left == right;
- /*
- * The invariants still hold: pivot >= all in [lo, left) and
- * pivot < all in [left, start), so pivot belongs at left. Note
- * that if there are elements equal to pivot, left points to the
- * first slot after them -- that's why this sort is stable.
- * Slide elements over to make room to make room for pivot.
- */
- int n = start - left; // The number of elements to move
- // Switch is just an optimization for arraycopy in default case
- switch(n) {
- case 2: a[left + 2] = a[left + 1];
- case 1: a[left + 1] = a[left];
- break;
- default: System.arraycopy(a, left, a, left + 1, n);
- }
- a[left] = pivot;
- }
- }
- /**
- * Returns the length of the run beginning at the specified position in
- * the specified array and reverses the run if it is descending (ensuring
- * that the run will always be ascending when the method returns).
- *
- * A run is the longest ascending sequence with:
- *
- * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
- *
- * or the longest descending sequence with:
- *
- * a[lo] > a[lo + 1] > a[lo + 2] > ...
- *
- * For its intended use in a stable mergesort, the strictness of the
- * definition of "descending" is needed so that the call can safely
- * reverse a descending sequence without violating stability.
- *
- * @param a the array in which a run is to be counted and possibly reversed
- * @param lo index of the first element in the run
- * @param hi index after the last element that may be contained in the run.
- It is required that @code{lo < hi}.
- * @param c the comparator to used for the sort
- * @return the length of the run beginning at the specified position in
- * the specified array
- */
- private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,
- Comparator<? super T> c) {
- assert lo < hi;
- int runHi = lo + 1;
- if (runHi == hi)
- return 1;
- // Find end of run, and reverse range if descending
- if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
- while(runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
- runHi++;
- reverseRange(a, lo, runHi);
- } else { // Ascending
- while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
- runHi++;
- }
- return runHi - lo;
- }
- /**
- * Reverse the specified range of the specified array.
- *
- * @param a the array in which a range is to be reversed
- * @param lo the index of the first element in the range to be reversed
- * @param hi the index after the last element in the range to be reversed
- */
- private static void reverseRange(Object[] a, int lo, int hi) {
- hi--;
- while (lo < hi) {
- Object t = a[lo];
- a[lo++] = a[hi];
- a[hi--] = t;
- }
- }
- /**
- * Returns the minimum acceptable run length for an array of the specified
- * length. Natural runs shorter than this will be extended with
- * {@link #binarySort}.
- *
- * Roughly speaking, the computation is:
- *
- * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
- * Else if n is an exact power of 2, return MIN_MERGE/2.
- * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
- * is close to, but strictly less than, an exact power of 2.
- *
- * For the rationale, see listsort.txt.
- *
- * @param n the length of the array to be sorted
- * @return the length of the minimum run to be merged
- */
- private static int minRunLength(int n) {
- assert n >= 0;
- int r = 0; // Becomes 1 if any 1 bits are shifted off
- while (n >= MIN_MERGE) {
- r |= (n & 1);
- n >>= 1;
- }
- return n + r;
- }
- /**
- * Pushes the specified run onto the pending-run stack.
- *
- * @param runBase index of the first element in the run
- * @param runLen the number of elements in the run
- */
- private void pushRun(int runBase, int runLen) {
- this.runBase[stackSize] = runBase;
- this.runLen[stackSize] = runLen;
- stackSize++;
- }
- /**
- * Examines the stack of runs waiting to be merged and merges adjacent runs
- * until the stack invariants are reestablished:
- *
- * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
- * 2. runLen[i - 2] > runLen[i - 1]
- *
- * This method is called each time a new run is pushed onto the stack,
- * so the invariants are guaranteed to hold for i < stackSize upon
- * entry to the method.
- */
- private void mergeCollapse() {
- while (stackSize > 1) {
- int n = stackSize - 2;
- if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
- if (runLen[n - 1] < runLen[n + 1])
- n--;
- mergeAt(n);
- } else if (runLen[n] <= runLen[n + 1]) {
- mergeAt(n);
- } else {
- break; // Invariant is established
- }
- }
- }
- /**
- * Merges all runs on the stack until only one remains. This method is
- * called once, to complete the sort.
- */
- private void mergeForceCollapse() {
- while (stackSize > 1) {
- int n = stackSize - 2;
- if (n > 0 && runLen[n - 1] < runLen[n + 1])
- n--;
- mergeAt(n);
- }
- }
- /**
- * Merges the two runs at stack indices i and i+1. Run i must be
- * the penultimate or antepenultimate run on the stack. In other words,
- * i must be equal to stackSize-2 or stackSize-3.
- *
- * @param i stack index of the first of the two runs to merge
- */
- private void mergeAt(int i) {
- assert stackSize >= 2;
- assert i >= 0;
- assert i == stackSize - 2 || i == stackSize - 3;
- int base1 = runBase[i];
- int len1 = runLen[i];
- int base2 = runBase[i + 1];
- int len2 = runLen[i + 1];
- assert len1 > 0 && len2 > 0;
- assert base1 + len1 == base2;
- /*
- * Record the length of the combined runs; if i is the 3rd-last
- * run now, also slide over the last run (which isn't involved
- * in this merge). The current run (i+1) goes away in any case.
- */
- runLen[i] = len1 + len2;
- if (i == stackSize - 3) {
- runBase[i + 1] = runBase[i + 2];
- runLen[i + 1] = runLen[i + 2];
- }
- stackSize--;
- /*
- * Find where the first element of run2 goes in run1. Prior elements
- * in run1 can be ignored (because they're already in place).
- */
- int k = gallopRight(a[base2], a, base1, len1, 0, c);
- assert k >= 0;
- base1 += k;
- len1 -= k;
- if (len1 == 0)
- return;
- /*
- * Find where the last element of run1 goes in run2. Subsequent elements
- * in run2 can be ignored (because they're already in place).
- */
- len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
- assert len2 >= 0;
- if (len2 == 0)
- return;
- // Merge remaining runs, using tmp array with min(len1, len2) elements
- if (len1 <= len2)
- mergeLo(base1, len1, base2, len2);
- else
- mergeHi(base1, len1, base2, len2);
- }
- /**
- * Locates the position at which to insert the specified key into the
- * specified sorted range; if the range contains an element equal to key,
- * returns the index of the leftmost equal element.
- *
- * @param key the key whose insertion point to search for
- * @param a the array in which to search
- * @param base the index of the first element in the range
- * @param len the length of the range; must be > 0
- * @param hint the index at which to begin the search, 0 <= hint < n.
- * The closer hint is to the result, the faster this method will run.
- * @param c the comparator used to order the range, and to search
- * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
- * pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
- * In other words, key belongs at index b + k; or in other words,
- * the first k elements of a should precede key, and the last n - k
- * should follow it.
- */
- private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint,
- Comparator<? super T> c) {
- assert len > 0 && hint >= 0 && hint < len;
- int lastOfs = 0;
- int ofs = 1;
- if (c.compare(key, a[base + hint]) > 0) {
- // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
- int maxOfs = len - hint;
- while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) {
- lastOfs = ofs;
- ofs = (ofs << 1) + 1;
- if (ofs <= 0) // int overflow
- ofs = maxOfs;
- }
- if (ofs > maxOfs)
- ofs = maxOfs;
- // Make offsets relative to base
- lastOfs += hint;
- ofs += hint;
- } else { // key <= a[base + hint]
- // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
- final int maxOfs = hint + 1;
- while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) {
- lastOfs = ofs;
- ofs = (ofs << 1) + 1;
- if (ofs <= 0) // int overflow
- ofs = maxOfs;
- }
- if (ofs > maxOfs)
- ofs = maxOfs;
- // Make offsets relative to base
- int tmp = lastOfs;
- lastOfs = hint - ofs;
- ofs = hint - tmp;
- }
- assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
- /*
- * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
- * to the right of lastOfs but no farther right than ofs. Do a binary
- * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
- */
- lastOfs++;
- while (lastOfs < ofs) {
- int m = lastOfs + ((ofs - lastOfs) >>> 1);
- if (c.compare(key, a[base + m]) > 0)
- lastOfs = m + 1; // a[base + m] < key
- else
- ofs = m; // key <= a[base + m]
- }
- assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
- return ofs;
- }
- /**
- * Like gallopLeft, except that if the range contains an element equal to
- * key, gallopRight returns the index after the rightmost equal element.
- *
- * @param key the key whose insertion point to search for
- * @param a the array in which to search
- * @param base the index of the first element in the range
- * @param len the length of the range; must be > 0
- * @param hint the index at which to begin the search, 0 <= hint < n.
- * The closer hint is to the result, the faster this method will run.
- * @param c the comparator used to order the range, and to search
- * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
- */
- private static <T> int gallopRight(T key, T[] a, int base, int len,
- int hint, Comparator<? super T> c) {
- assert len > 0 && hint >= 0 && hint < len;
- int ofs = 1;
- int lastOfs = 0;
- if (c.compare(key, a[base + hint]) < 0) {
- // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
- int maxOfs = hint + 1;
- while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) {
- lastOfs = ofs;
- ofs = (ofs << 1) + 1;
- if (ofs <= 0) // int overflow
- ofs = maxOfs;
- }
- if (ofs > maxOfs)
- ofs = maxOfs;
- // Make offsets relative to b
- int tmp = lastOfs;
- lastOfs = hint - ofs;
- ofs = hint - tmp;
- } else { // a[b + hint] <= key
- // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
- int maxOfs = len - hint;
- while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) {
- lastOfs = ofs;
- ofs = (ofs << 1) + 1;
- if (ofs <= 0) // int overflow
- ofs = maxOfs;
- }
- if (ofs > maxOfs)
- ofs = maxOfs;
- // Make offsets relative to b
- lastOfs += hint;
- ofs += hint;
- }
- assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
- /*
- * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
- * the right of lastOfs but no farther right than ofs. Do a binary
- * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
- */
- lastOfs++;
- while (lastOfs < ofs) {
- int m = lastOfs + ((ofs - lastOfs) >>> 1);
- if (c.compare(key, a[base + m]) < 0)
- ofs = m; // key < a[b + m]
- else
- lastOfs = m + 1; // a[b + m] <= key
- }
- assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
- return ofs;
- }
- /**
- * Merges two adjacent runs in place, in a stable fashion. The first
- * element of the first run must be greater than the first element of the
- * second run (a[base1] > a[base2]), and the last element of the first run
- * (a[base1 + len1-1]) must be greater than all elements of the second run.
- *
- * For performance, this method should be called only when len1 <= len2;
- * its twin, mergeHi should be called if len1 >= len2. (Either method
- * may be called if len1 == len2.)
- *
- * @param base1 index of first element in first run to be merged
- * @param len1 length of first run to be merged (must be > 0)
- * @param base2 index of first element in second run to be merged
- * (must be aBase + aLen)
- * @param len2 length of second run to be merged (must be > 0)
- */
- private void mergeLo(int base1, int len1, int base2, int len2) {
- assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
- // Copy first run into temp array
- T[] a = this.a; // For performance
- T[] tmp = ensureCapacity(len1);
- System.arraycopy(a, base1, tmp, 0, len1);
- int cursor1 = 0; // Indexes into tmp array
- int cursor2 = base2; // Indexes int a
- int dest = base1; // Indexes int a
- // Move first element of second run and deal with degenerate cases
- a[dest++] = a[cursor2++];
- if (--len2 == 0) {
- System.arraycopy(tmp, cursor1, a, dest, len1);
- return;
- }
- if (len1 == 1) {
- System.arraycopy(a, cursor2, a, dest, len2);
- a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
- return;
- }
- Comparator<? super T> c = this.c; // Use local variable for performance
- int minGallop = this.minGallop; // " " " " "
- outer:
- while (true) {
- int count1 = 0; // Number of times in a row that first run won
- int count2 = 0; // Number of times in a row that second run won
- /*
- * Do the straightforward thing until (if ever) one run starts
- * winning consistently.
- */
- do {
- assert len1 > 1 && len2 > 0;
- if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
- a[dest++] = a[cursor2++];
- count2++;
- count1 = 0;
- if (--len2 == 0)
- break outer;
- } else {
- a[dest++] = tmp[cursor1++];
- count1++;
- count2 = 0;
- if (--len1 == 1)
- break outer;
- }
- } while ((count1 | count2) < minGallop);
- /*
- * One run is winning so consistently that galloping may be a
- * huge win. So try that, and continue galloping until (if ever)
- * neither run appears to be winning consistently anymore.
- */
- do {
- assert len1 > 1 && len2 > 0;
- count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
- if (count1 != 0) {
- System.arraycopy(tmp, cursor1, a, dest, count1);
- dest += count1;
- cursor1 += count1;
- len1 -= count1;
- if (len1 <= 1) // len1 == 1 || len1 == 0
- break outer;
- }
- a[dest++] = a[cursor2++];
- if (--len2 == 0)
- break outer;
- count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
- if (count2 != 0) {
- System.arraycopy(a, cursor2, a, dest, count2);
- dest += count2;
- cursor2 += count2;
- len2 -= count2;
- if (len2 == 0)
- break outer;
- }
- a[dest++] = tmp[cursor1++];
- if (--len1 == 1)
- break outer;
- minGallop--;
- } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
- if (minGallop < 0)
- minGallop = 0;
- minGallop += 2; // Penalize for leaving gallop mode
- } // End of "outer" loop
- this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
- if (len1 == 1) {
- assert len2 > 0;
- System.arraycopy(a, cursor2, a, dest, len2);
- a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
- } else if (len1 == 0) {
- throw new IllegalArgumentException(
- "Comparison method violates its general contract!");
- } else {
- assert len2 == 0;
- assert len1 > 1;
- System.arraycopy(tmp, cursor1, a, dest, len1);
- }
- }
- /**
- * Like mergeLo, except that this method should be called only if
- * len1 >= len2; mergeLo should be called if len1 <= len2. (Either method
- * may be called if len1 == len2.)
- *
- * @param base1 index of first element in first run to be merged
- * @param len1 length of first run to be merged (must be > 0)
- * @param base2 index of first element in second run to be merged
- * (must be aBase + aLen)
- * @param len2 length of second run to be merged (must be > 0)
- */
- private void mergeHi(int base1, int len1, int base2, int len2) {
- assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
- // Copy second run into temp array
- T[] a = this.a; // For performance
- T[] tmp = ensureCapacity(len2);
- System.arraycopy(a, base2, tmp, 0, len2);
- int cursor1 = base1 + len1 - 1; // Indexes into a
- int cursor2 = len2 - 1; // Indexes into tmp array
- int dest = base2 + len2 - 1; // Indexes into a
- // Move last element of first run and deal with degenerate cases
- a[dest--] = a[cursor1--];
- if (--len1 == 0) {
- System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
- return;
- }
- if (len2 == 1) {
- dest -= len1;
- cursor1 -= len1;
- System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
- a[dest] = tmp[cursor2];
- return;
- }
- Comparator<? super T> c = this.c; // Use local variable for performance
- int minGallop = this.minGallop; // " " " " "
- outer:
- while (true) {
- int count1 = 0; // Number of times in a row that first run won
- int count2 = 0; // Number of times in a row that second run won
- /*
- * Do the straightforward thing until (if ever) one run
- * appears to win consistently.
- */
- do {
- assert len1 > 0 && len2 > 1;
- if (c.compare(tmp[cursor2], a[cursor1]) < 0) {
- a[dest--] = a[cursor1--];
- count1++;
- count2 = 0;
- if (--len1 == 0)
- break outer;
- } else {
- a[dest--] = tmp[cursor2--];
- count2++;
- count1 = 0;
- if (--len2 == 1)
- break outer;
- }
- } while ((count1 | count2) < minGallop);
- /*
- * One run is winning so consistently that galloping may be a
- * huge win. So try that, and continue galloping until (if ever)
- * neither run appears to be winning consistently anymore.
- */
- do {
- assert len1 > 0 && len2 > 1;
- count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
- if (count1 != 0) {
- dest -= count1;
- cursor1 -= count1;
- len1 -= count1;
- System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
- if (len1 == 0)
- break outer;
- }
- a[dest--] = tmp[cursor2--];
- if (--len2 == 1)
- break outer;
- count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);
- if (count2 != 0) {
- dest -= count2;
- cursor2 -= count2;
- len2 -= count2;
- System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
- if (len2 <= 1) // len2 == 1 || len2 == 0
- break outer;
- }
- a[dest--] = a[cursor1--];
- if (--len1 == 0)
- break outer;
- minGallop--;
- } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
- if (minGallop < 0)
- minGallop = 0;
- minGallop += 2; // Penalize for leaving gallop mode
- } // End of "outer" loop
- this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
- if (len2 == 1) {
- assert len1 > 0;
- dest -= len1;
- cursor1 -= len1;
- System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
- a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
- } else if (len2 == 0) {
- throw new IllegalArgumentException(
- "Comparison method violates its general contract!");
- } else {
- assert len1 == 0;
- assert len2 > 0;
- System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
- }
- }
- /**
- * Ensures that the external array tmp has at least the specified
- * number of elements, increasing its size if necessary. The size
- * increases exponentially to ensure amortized linear time complexity.
- *
- * @param minCapacity the minimum required capacity of the tmp array
- * @return tmp, whether or not it grew
- */
- private T[] ensureCapacity(int minCapacity) {
- if (tmp.length < minCapacity) {
- // Compute smallest power of 2 > minCapacity
- int newSize = minCapacity;
- newSize |= newSize >> 1;
- newSize |= newSize >> 2;
- newSize |= newSize >> 4;
- newSize |= newSize >> 8;
- newSize |= newSize >> 16;
- newSize++;
- if (newSize < 0) // Not bloody likely!
- newSize = minCapacity;
- else
- newSize = Math.min(newSize, a.length >>> 1);
- @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
- T[] newArray = (T[]) new Object[newSize];
- tmp = newArray;
- }
- return tmp;
- }
- /**
- * Checks that fromIndex and toIndex are in range, and throws an
- * appropriate exception if they aren't.
- *
- * @param arrayLen the length of the array
- * @param fromIndex the index of the first element of the range
- * @param toIndex the index after the last element of the range
- * @throws IllegalArgumentException if fromIndex > toIndex
- * @throws ArrayIndexOutOfBoundsException if fromIndex < 0
- * or toIndex > arrayLen
- */
- private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) {
- if (fromIndex > toIndex)
- throw new IllegalArgumentException("fromIndex(" + fromIndex +
- ") > toIndex(" + toIndex+")");
- if (fromIndex < 0)
- throw new ArrayIndexOutOfBoundsException(fromIndex);
- if (toIndex > arrayLen)
- throw new ArrayIndexOutOfBoundsException(toIndex);
- }
- }
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