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/* |
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* Copyright (C) 2008 The Android Open Source Project |
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* |
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* Licensed under the Apache License, Version 2.0 (the "License"); |
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* you may not use this file except in compliance with the License. |
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* You may obtain a copy of the License at |
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* |
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* http://www.apache.org/licenses/LICENSE-2.0 |
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* |
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* Unless required by applicable law or agreed to in writing, software |
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* distributed under the License is distributed on an "AS IS" BASIS, |
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* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. |
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* See the License for the specific language governing permissions and |
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* limitations under the License. |
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*/ |
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|
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package java.util; |
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|
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/** |
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* A stable, adaptive, iterative mergesort that requires far fewer than |
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* n lg(n) comparisons when running on partially sorted arrays, while |
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* offering performance comparable to a traditional mergesort when run |
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* on random arrays. Like all proper mergesorts, this sort is stable and |
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* runs O(n log n) time (worst case). In the worst case, this sort requires |
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* temporary storage space for n/2 object references; in the best case, |
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* it requires only a small constant amount of space. |
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* |
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* This implementation was adapted from Tim Peters's list sort for |
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* Python, which is described in detail here: |
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* |
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* http://svn.python.org/projects/python/trunk/Objects/listsort.txt |
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* |
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* Tim's C code may be found here: |
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* |
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* http://svn.python.org/projects/python/trunk/Objects/listobject.c |
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* |
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* The underlying techniques are described in this paper (and may have |
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* even earlier origins): |
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* |
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* "Optimistic Sorting and Information Theoretic Complexity" |
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* Peter McIlroy |
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* SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms), |
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* pp 467-474, Austin, Texas, 25-27 January 1993. |
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* |
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* While the API to this class consists solely of static methods, it is |
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* (privately) instantiable; a TimSort instance holds the state of an ongoing |
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* sort, assuming the input array is large enough to warrant the full-blown |
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* TimSort. Small arrays are sorted in place, using a binary insertion sort. |
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* |
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* @author Josh Bloch |
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*/ |
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class TimSort<T> { |
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/** |
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* This is the minimum sized sequence that will be merged. Shorter |
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* sequences will be lengthened by calling binarySort. If the entire |
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* array is less than this length, no merges will be performed. |
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* |
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* This constant should be a power of two. It was 64 in Tim Peter's C |
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* implementation, but 32 was empirically determined to work better in |
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* this implementation. In the unlikely event that you set this constant |
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* to be a number that's not a power of two, you'll need to change the |
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* {@link #minRunLength} computation. |
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* |
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* If you decrease this constant, you must change the stackLen |
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* computation in the TimSort constructor, or you risk an |
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* ArrayOutOfBounds exception. See listsort.txt for a discussion |
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* of the minimum stack length required as a function of the length |
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* of the array being sorted and the minimum merge sequence length. |
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*/ |
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private static final int MIN_MERGE = 32; |
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|
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/** |
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* The array being sorted. |
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*/ |
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private final T[] a; |
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|
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/** |
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* The comparator for this sort. |
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*/ |
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private final Comparator<? super T> c; |
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|
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/** |
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* When we get into galloping mode, we stay there until both runs win less |
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* often than MIN_GALLOP consecutive times. |
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*/ |
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private static final int MIN_GALLOP = 7; |
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|
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/** |
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* This controls when we get *into* galloping mode. It is initialized |
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* to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for |
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* random data, and lower for highly structured data. |
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*/ |
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private int minGallop = MIN_GALLOP; |
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|
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/** |
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* Maximum initial size of tmp array, which is used for merging. The array |
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* can grow to accommodate demand. |
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* |
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* Unlike Tim's original C version, we do not allocate this much storage |
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* when sorting smaller arrays. This change was required for performance. |
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*/ |
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private static final int INITIAL_TMP_STORAGE_LENGTH = 256; |
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|
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/** |
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* Temp storage for merges. |
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*/ |
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private T[] tmp; // Actual runtime type will be Object[], regardless of T |
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|
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/** |
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* A stack of pending runs yet to be merged. Run i starts at |
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* address base[i] and extends for len[i] elements. It's always |
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* true (so long as the indices are in bounds) that: |
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* |
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* runBase[i] + runLen[i] == runBase[i + 1] |
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* |
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* so we could cut the storage for this, but it's a minor amount, |
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* and keeping all the info explicit simplifies the code. |
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*/ |
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private int stackSize = 0; // Number of pending runs on stack |
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private final int[] runBase; |
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private final int[] runLen; |
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|
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/** |
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* Creates a TimSort instance to maintain the state of an ongoing sort. |
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* |
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* @param a the array to be sorted |
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* @param c the comparator to determine the order of the sort |
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*/ |
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private TimSort(T[] a, Comparator<? super T> c) { |
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this.a = a; |
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this.c = c; |
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|
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// Allocate temp storage (which may be increased later if necessary) |
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int len = a.length; |
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@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) |
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T[] newArray = (T[]) new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ? |
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len >>> 1 : INITIAL_TMP_STORAGE_LENGTH]; |
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tmp = newArray; |
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|
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/* |
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* Allocate runs-to-be-merged stack (which cannot be expanded). The |
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* stack length requirements are described in listsort.txt. The C |
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* version always uses the same stack length (85), but this was |
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* measured to be too expensive when sorting "mid-sized" arrays (e.g., |
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* 100 elements) in Java. Therefore, we use smaller (but sufficiently |
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* large) stack lengths for smaller arrays. The "magic numbers" in the |
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* computation below must be changed if MIN_MERGE is decreased. See |
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* the MIN_MERGE declaration above for more information. |
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*/ |
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int stackLen = (len < 120 ? 5 : |
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len < 1542 ? 10 : |
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len < 119151 ? 19 : 40); |
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runBase = new int[stackLen]; |
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runLen = new int[stackLen]; |
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} |
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|
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/* |
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* The next two methods (which are package private and static) constitute |
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* the entire API of this class. Each of these methods obeys the contract |
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* of the public method with the same signature in java.util.Arrays. |
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*/ |
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|
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static <T> void sort(T[] a, Comparator<? super T> c) { |
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sort(a, 0, a.length, c); |
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} |
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|
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static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c) { |
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if (c == null) { |
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Arrays.sort(a, lo, hi); |
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return; |
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} |
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|
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rangeCheck(a.length, lo, hi); |
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int nRemaining = hi - lo; |
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if (nRemaining < 2) |
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return; // Arrays of size 0 and 1 are always sorted |
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|
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// If array is small, do a "mini-TimSort" with no merges |
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if (nRemaining < MIN_MERGE) { |
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int initRunLen = countRunAndMakeAscending(a, lo, hi, c); |
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binarySort(a, lo, hi, lo + initRunLen, c); |
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return; |
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} |
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|
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/** |
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* March over the array once, left to right, finding natural runs, |
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* extending short natural runs to minRun elements, and merging runs |
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* to maintain stack invariant. |
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*/ |
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TimSort<T> ts = new TimSort<T>(a, c); |
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int minRun = minRunLength(nRemaining); |
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do { |
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// Identify next run |
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int runLen = countRunAndMakeAscending(a, lo, hi, c); |
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|
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// If run is short, extend to min(minRun, nRemaining) |
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if (runLen < minRun) { |
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int force = nRemaining <= minRun ? nRemaining : minRun; |
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binarySort(a, lo, lo + force, lo + runLen, c); |
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runLen = force; |
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} |
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|
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// Push run onto pending-run stack, and maybe merge |
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ts.pushRun(lo, runLen); |
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ts.mergeCollapse(); |
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|
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// Advance to find next run |
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lo += runLen; |
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nRemaining -= runLen; |
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} while (nRemaining != 0); |
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|
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// Merge all remaining runs to complete sort |
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assert lo == hi; |
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ts.mergeForceCollapse(); |
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assert ts.stackSize == 1; |
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} |
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|
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/** |
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* Sorts the specified portion of the specified array using a binary |
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* insertion sort. This is the best method for sorting small numbers |
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* of elements. It requires O(n log n) compares, but O(n^2) data |
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* movement (worst case). |
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* |
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* If the initial part of the specified range is already sorted, |
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* this method can take advantage of it: the method assumes that the |
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* elements from index {@code lo}, inclusive, to {@code start}, |
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* exclusive are already sorted. |
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* |
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* @param a the array in which a range is to be sorted |
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* @param lo the index of the first element in the range to be sorted |
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* @param hi the index after the last element in the range to be sorted |
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* @param start the index of the first element in the range that is |
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* not already known to be sorted ({@code lo <= start <= hi}) |
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* @param c comparator to used for the sort |
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*/ |
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@SuppressWarnings("fallthrough") |
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private static <T> void binarySort(T[] a, int lo, int hi, int start, |
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Comparator<? super T> c) { |
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assert lo <= start && start <= hi; |
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if (start == lo) |
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start++; |
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for ( ; start < hi; start++) { |
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T pivot = a[start]; |
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|
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// Set left (and right) to the index where a[start] (pivot) belongs |
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int left = lo; |
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int right = start; |
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assert left <= right; |
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/* |
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* Invariants: |
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* pivot >= all in [lo, left). |
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* pivot < all in [right, start). |
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*/ |
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while (left < right) { |
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int mid = (left + right) >>> 1; |
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if (c.compare(pivot, a[mid]) < 0) |
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right = mid; |
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else |
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left = mid + 1; |
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} |
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assert left == right; |
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|
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/* |
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* The invariants still hold: pivot >= all in [lo, left) and |
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* pivot < all in [left, start), so pivot belongs at left. Note |
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* that if there are elements equal to pivot, left points to the |
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* first slot after them -- that's why this sort is stable. |
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* Slide elements over to make room to make room for pivot. |
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*/ |
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int n = start - left; // The number of elements to move |
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// Switch is just an optimization for arraycopy in default case |
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switch (n) { |
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case 2: a[left + 2] = a[left + 1]; |
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case 1: a[left + 1] = a[left]; |
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break; |
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default: System.arraycopy(a, left, a, left + 1, n); |
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} |
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a[left] = pivot; |
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} |
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} |
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|
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/** |
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* Returns the length of the run beginning at the specified position in |
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* the specified array and reverses the run if it is descending (ensuring |
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* that the run will always be ascending when the method returns). |
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* |
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* A run is the longest ascending sequence with: |
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* |
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* a[lo] <= a[lo + 1] <= a[lo + 2] <= ... |
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* |
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* or the longest descending sequence with: |
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* |
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* a[lo] > a[lo + 1] > a[lo + 2] > ... |
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* |
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* For its intended use in a stable mergesort, the strictness of the |
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* definition of "descending" is needed so that the call can safely |
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* reverse a descending sequence without violating stability. |
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* |
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* @param a the array in which a run is to be counted and possibly reversed |
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* @param lo index of the first element in the run |
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* @param hi index after the last element that may be contained in the run. |
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It is required that {@code lo < hi}. |
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* @param c the comparator to used for the sort |
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* @return the length of the run beginning at the specified position in |
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* the specified array |
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*/ |
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private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi, |
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Comparator<? super T> c) { |
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assert lo < hi; |
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int runHi = lo + 1; |
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if (runHi == hi) |
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return 1; |
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|
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// Find end of run, and reverse range if descending |
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if (c.compare(a[runHi++], a[lo]) < 0) { // Descending |
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while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0) |
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runHi++; |
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reverseRange(a, lo, runHi); |
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} else { // Ascending |
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while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0) |
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runHi++; |
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} |
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|
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return runHi - lo; |
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} |
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|
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/** |
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* Reverse the specified range of the specified array. |
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* |
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* @param a the array in which a range is to be reversed |
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* @param lo the index of the first element in the range to be reversed |
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* @param hi the index after the last element in the range to be reversed |
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*/ |
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private static void reverseRange(Object[] a, int lo, int hi) { |
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hi--; |
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while (lo < hi) { |
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Object t = a[lo]; |
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a[lo++] = a[hi]; |
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a[hi--] = t; |
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} |
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} |
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|
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/** |
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* Returns the minimum acceptable run length for an array of the specified |
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* length. Natural runs shorter than this will be extended with |
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* {@link #binarySort}. |
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* |
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* Roughly speaking, the computation is: |
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* |
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* If n < MIN_MERGE, return n (it's too small to bother with fancy stuff). |
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* Else if n is an exact power of 2, return MIN_MERGE/2. |
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* Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k |
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* is close to, but strictly less than, an exact power of 2. |
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* |
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* For the rationale, see listsort.txt. |
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* |
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* @param n the length of the array to be sorted |
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* @return the length of the minimum run to be merged |
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*/ |
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private static int minRunLength(int n) { |
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assert n >= 0; |
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int r = 0; // Becomes 1 if any 1 bits are shifted off |
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while (n >= MIN_MERGE) { |
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r |= (n & 1); |
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n >>= 1; |
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} |
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return n + r; |
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} |
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|
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/** |
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* Pushes the specified run onto the pending-run stack. |
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* |
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* @param runBase index of the first element in the run |
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* @param runLen the number of elements in the run |
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*/ |
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private void pushRun(int runBase, int runLen) { |
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this.runBase[stackSize] = runBase; |
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this.runLen[stackSize] = runLen; |
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stackSize++; |
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} |
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|
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/** |
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* Examines the stack of runs waiting to be merged and merges adjacent runs |
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* until the stack invariants are reestablished: |
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* |
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* 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] |
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* 2. runLen[i - 2] > runLen[i - 1] |
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* |
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* This method is called each time a new run is pushed onto the stack, |
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* so the invariants are guaranteed to hold for i < stackSize upon |
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* entry to the method. |
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*/ |
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private void mergeCollapse() { |
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while (stackSize > 1) { |
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int n = stackSize - 2; |
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if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) { |
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if (runLen[n - 1] < runLen[n + 1]) |
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n--; |
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mergeAt(n); |
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} else if (runLen[n] <= runLen[n + 1]) { |
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mergeAt(n); |
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} else { |
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break; // Invariant is established |
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} |
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} |
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} |
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|
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/** |
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* Merges all runs on the stack until only one remains. This method is |
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* called once, to complete the sort. |
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*/ |
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private void mergeForceCollapse() { |
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while (stackSize > 1) { |
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int n = stackSize - 2; |
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if (n > 0 && runLen[n - 1] < runLen[n + 1]) |
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n--; |
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mergeAt(n); |
418 |
} |
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} |
420 |
|
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/** |
422 |
* Merges the two runs at stack indices i and i+1. Run i must be |
423 |
* the penultimate or antepenultimate run on the stack. In other words, |
424 |
* i must be equal to stackSize-2 or stackSize-3. |
425 |
* |
426 |
* @param i stack index of the first of the two runs to merge |
427 |
*/ |
428 |
private void mergeAt(int i) { |
429 |
assert stackSize >= 2; |
430 |
assert i >= 0; |
431 |
assert i == stackSize - 2 || i == stackSize - 3; |
432 |
|
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int base1 = runBase[i]; |
434 |
int len1 = runLen[i]; |
435 |
int base2 = runBase[i + 1]; |
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int len2 = runLen[i + 1]; |
437 |
assert len1 > 0 && len2 > 0; |
438 |
assert base1 + len1 == base2; |
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|
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/* |
441 |
* Record the length of the combined runs; if i is the 3rd-last |
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* run now, also slide over the last run (which isn't involved |
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* in this merge). The current run (i+1) goes away in any case. |
444 |
*/ |
445 |
runLen[i] = len1 + len2; |
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if (i == stackSize - 3) { |
447 |
runBase[i + 1] = runBase[i + 2]; |
448 |
runLen[i + 1] = runLen[i + 2]; |
449 |
} |
450 |
stackSize--; |
451 |
|
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/* |
453 |
* Find where the first element of run2 goes in run1. Prior elements |
454 |
* in run1 can be ignored (because they're already in place). |
455 |
*/ |
456 |
int k = gallopRight(a[base2], a, base1, len1, 0, c); |
457 |
assert k >= 0; |
458 |
base1 += k; |
459 |
len1 -= k; |
460 |
if (len1 == 0) |
461 |
return; |
462 |
|
463 |
/* |
464 |
* Find where the last element of run1 goes in run2. Subsequent elements |
465 |
* in run2 can be ignored (because they're already in place). |
466 |
*/ |
467 |
len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c); |
468 |
assert len2 >= 0; |
469 |
if (len2 == 0) |
470 |
return; |
471 |
|
472 |
// Merge remaining runs, using tmp array with min(len1, len2) elements |
473 |
if (len1 <= len2) |
474 |
mergeLo(base1, len1, base2, len2); |
475 |
else |
476 |
mergeHi(base1, len1, base2, len2); |
477 |
} |
478 |
|
479 |
/** |
480 |
* Locates the position at which to insert the specified key into the |
481 |
* specified sorted range; if the range contains an element equal to key, |
482 |
* returns the index of the leftmost equal element. |
483 |
* |
484 |
* @param key the key whose insertion point to search for |
485 |
* @param a the array in which to search |
486 |
* @param base the index of the first element in the range |
487 |
* @param len the length of the range; must be > 0 |
488 |
* @param hint the index at which to begin the search, 0 <= hint < n. |
489 |
* The closer hint is to the result, the faster this method will run. |
490 |
* @param c the comparator used to order the range, and to search |
491 |
* @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k], |
492 |
* pretending that a[b - 1] is minus infinity and a[b + n] is infinity. |
493 |
* In other words, key belongs at index b + k; or in other words, |
494 |
* the first k elements of a should precede key, and the last n - k |
495 |
* should follow it. |
496 |
*/ |
497 |
private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint, |
498 |
Comparator<? super T> c) { |
499 |
assert len > 0 && hint >= 0 && hint < len; |
500 |
int lastOfs = 0; |
501 |
int ofs = 1; |
502 |
if (c.compare(key, a[base + hint]) > 0) { |
503 |
// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs] |
504 |
int maxOfs = len - hint; |
505 |
while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) { |
506 |
lastOfs = ofs; |
507 |
ofs = (ofs << 1) + 1; |
508 |
if (ofs <= 0) // int overflow |
509 |
ofs = maxOfs; |
510 |
} |
511 |
if (ofs > maxOfs) |
512 |
ofs = maxOfs; |
513 |
|
514 |
// Make offsets relative to base |
515 |
lastOfs += hint; |
516 |
ofs += hint; |
517 |
} else { // key <= a[base + hint] |
518 |
// Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs] |
519 |
final int maxOfs = hint + 1; |
520 |
while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) { |
521 |
lastOfs = ofs; |
522 |
ofs = (ofs << 1) + 1; |
523 |
if (ofs <= 0) // int overflow |
524 |
ofs = maxOfs; |
525 |
} |
526 |
if (ofs > maxOfs) |
527 |
ofs = maxOfs; |
528 |
|
529 |
// Make offsets relative to base |
530 |
int tmp = lastOfs; |
531 |
lastOfs = hint - ofs; |
532 |
ofs = hint - tmp; |
533 |
} |
534 |
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; |
535 |
|
536 |
/* |
537 |
* Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere |
538 |
* to the right of lastOfs but no farther right than ofs. Do a binary |
539 |
* search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs]. |
540 |
*/ |
541 |
lastOfs++; |
542 |
while (lastOfs < ofs) { |
543 |
int m = lastOfs + ((ofs - lastOfs) >>> 1); |
544 |
|
545 |
if (c.compare(key, a[base + m]) > 0) |
546 |
lastOfs = m + 1; // a[base + m] < key |
547 |
else |
548 |
ofs = m; // key <= a[base + m] |
549 |
} |
550 |
assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs] |
551 |
return ofs; |
552 |
} |
553 |
|
554 |
/** |
555 |
* Like gallopLeft, except that if the range contains an element equal to |
556 |
* key, gallopRight returns the index after the rightmost equal element. |
557 |
* |
558 |
* @param key the key whose insertion point to search for |
559 |
* @param a the array in which to search |
560 |
* @param base the index of the first element in the range |
561 |
* @param len the length of the range; must be > 0 |
562 |
* @param hint the index at which to begin the search, 0 <= hint < n. |
563 |
* The closer hint is to the result, the faster this method will run. |
564 |
* @param c the comparator used to order the range, and to search |
565 |
* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k] |
566 |
*/ |
567 |
private static <T> int gallopRight(T key, T[] a, int base, int len, |
568 |
int hint, Comparator<? super T> c) { |
569 |
assert len > 0 && hint >= 0 && hint < len; |
570 |
|
571 |
int ofs = 1; |
572 |
int lastOfs = 0; |
573 |
if (c.compare(key, a[base + hint]) < 0) { |
574 |
// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs] |
575 |
int maxOfs = hint + 1; |
576 |
while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) { |
577 |
lastOfs = ofs; |
578 |
ofs = (ofs << 1) + 1; |
579 |
if (ofs <= 0) // int overflow |
580 |
ofs = maxOfs; |
581 |
} |
582 |
if (ofs > maxOfs) |
583 |
ofs = maxOfs; |
584 |
|
585 |
// Make offsets relative to b |
586 |
int tmp = lastOfs; |
587 |
lastOfs = hint - ofs; |
588 |
ofs = hint - tmp; |
589 |
} else { // a[b + hint] <= key |
590 |
// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs] |
591 |
int maxOfs = len - hint; |
592 |
while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) { |
593 |
lastOfs = ofs; |
594 |
ofs = (ofs << 1) + 1; |
595 |
if (ofs <= 0) // int overflow |
596 |
ofs = maxOfs; |
597 |
} |
598 |
if (ofs > maxOfs) |
599 |
ofs = maxOfs; |
600 |
|
601 |
// Make offsets relative to b |
602 |
lastOfs += hint; |
603 |
ofs += hint; |
604 |
} |
605 |
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; |
606 |
|
607 |
/* |
608 |
* Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to |
609 |
* the right of lastOfs but no farther right than ofs. Do a binary |
610 |
* search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs]. |
611 |
*/ |
612 |
lastOfs++; |
613 |
while (lastOfs < ofs) { |
614 |
int m = lastOfs + ((ofs - lastOfs) >>> 1); |
615 |
|
616 |
if (c.compare(key, a[base + m]) < 0) |
617 |
ofs = m; // key < a[b + m] |
618 |
else |
619 |
lastOfs = m + 1; // a[b + m] <= key |
620 |
} |
621 |
assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs] |
622 |
return ofs; |
623 |
} |
624 |
|
625 |
/** |
626 |
* Merges two adjacent runs in place, in a stable fashion. The first |
627 |
* element of the first run must be greater than the first element of the |
628 |
* second run (a[base1] > a[base2]), and the last element of the first run |
629 |
* (a[base1 + len1-1]) must be greater than all elements of the second run. |
630 |
* |
631 |
* For performance, this method should be called only when len1 <= len2; |
632 |
* its twin, mergeHi should be called if len1 >= len2. (Either method |
633 |
* may be called if len1 == len2.) |
634 |
* |
635 |
* @param base1 index of first element in first run to be merged |
636 |
* @param len1 length of first run to be merged (must be > 0) |
637 |
* @param base2 index of first element in second run to be merged |
638 |
* (must be aBase + aLen) |
639 |
* @param len2 length of second run to be merged (must be > 0) |
640 |
*/ |
641 |
private void mergeLo(int base1, int len1, int base2, int len2) { |
642 |
assert len1 > 0 && len2 > 0 && base1 + len1 == base2; |
643 |
|
644 |
// Copy first run into temp array |
645 |
T[] a = this.a; // For performance |
646 |
T[] tmp = ensureCapacity(len1); |
647 |
System.arraycopy(a, base1, tmp, 0, len1); |
648 |
|
649 |
int cursor1 = 0; // Indexes into tmp array |
650 |
int cursor2 = base2; // Indexes int a |
651 |
int dest = base1; // Indexes int a |
652 |
|
653 |
// Move first element of second run and deal with degenerate cases |
654 |
a[dest++] = a[cursor2++]; |
655 |
if (--len2 == 0) { |
656 |
System.arraycopy(tmp, cursor1, a, dest, len1); |
657 |
return; |
658 |
} |
659 |
if (len1 == 1) { |
660 |
System.arraycopy(a, cursor2, a, dest, len2); |
661 |
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge |
662 |
return; |
663 |
} |
664 |
|
665 |
Comparator<? super T> c = this.c; // Use local variable for performance |
666 |
int minGallop = this.minGallop; // " " " " " |
667 |
outer: |
668 |
while (true) { |
669 |
int count1 = 0; // Number of times in a row that first run won |
670 |
int count2 = 0; // Number of times in a row that second run won |
671 |
|
672 |
/* |
673 |
* Do the straightforward thing until (if ever) one run starts |
674 |
* winning consistently. |
675 |
*/ |
676 |
do { |
677 |
assert len1 > 1 && len2 > 0; |
678 |
if (c.compare(a[cursor2], tmp[cursor1]) < 0) { |
679 |
a[dest++] = a[cursor2++]; |
680 |
count2++; |
681 |
count1 = 0; |
682 |
if (--len2 == 0) |
683 |
break outer; |
684 |
} else { |
685 |
a[dest++] = tmp[cursor1++]; |
686 |
count1++; |
687 |
count2 = 0; |
688 |
if (--len1 == 1) |
689 |
break outer; |
690 |
} |
691 |
} while ((count1 | count2) < minGallop); |
692 |
|
693 |
/* |
694 |
* One run is winning so consistently that galloping may be a |
695 |
* huge win. So try that, and continue galloping until (if ever) |
696 |
* neither run appears to be winning consistently anymore. |
697 |
*/ |
698 |
do { |
699 |
assert len1 > 1 && len2 > 0; |
700 |
count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c); |
701 |
if (count1 != 0) { |
702 |
System.arraycopy(tmp, cursor1, a, dest, count1); |
703 |
dest += count1; |
704 |
cursor1 += count1; |
705 |
len1 -= count1; |
706 |
if (len1 <= 1) // len1 == 1 || len1 == 0 |
707 |
break outer; |
708 |
} |
709 |
a[dest++] = a[cursor2++]; |
710 |
if (--len2 == 0) |
711 |
break outer; |
712 |
|
713 |
count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c); |
714 |
if (count2 != 0) { |
715 |
System.arraycopy(a, cursor2, a, dest, count2); |
716 |
dest += count2; |
717 |
cursor2 += count2; |
718 |
len2 -= count2; |
719 |
if (len2 == 0) |
720 |
break outer; |
721 |
} |
722 |
a[dest++] = tmp[cursor1++]; |
723 |
if (--len1 == 1) |
724 |
break outer; |
725 |
minGallop--; |
726 |
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); |
727 |
if (minGallop < 0) |
728 |
minGallop = 0; |
729 |
minGallop += 2; // Penalize for leaving gallop mode |
730 |
} // End of "outer" loop |
731 |
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field |
732 |
|
733 |
if (len1 == 1) { |
734 |
assert len2 > 0; |
735 |
System.arraycopy(a, cursor2, a, dest, len2); |
736 |
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge |
737 |
} else if (len1 == 0) { |
738 |
throw new IllegalArgumentException( |
739 |
"Comparison method violates its general contract!"); |
740 |
} else { |
741 |
assert len2 == 0; |
742 |
assert len1 > 1; |
743 |
System.arraycopy(tmp, cursor1, a, dest, len1); |
744 |
} |
745 |
} |
746 |
|
747 |
/** |
748 |
* Like mergeLo, except that this method should be called only if |
749 |
* len1 >= len2; mergeLo should be called if len1 <= len2. (Either method |
750 |
* may be called if len1 == len2.) |
751 |
* |
752 |
* @param base1 index of first element in first run to be merged |
753 |
* @param len1 length of first run to be merged (must be > 0) |
754 |
* @param base2 index of first element in second run to be merged |
755 |
* (must be aBase + aLen) |
756 |
* @param len2 length of second run to be merged (must be > 0) |
757 |
*/ |
758 |
private void mergeHi(int base1, int len1, int base2, int len2) { |
759 |
assert len1 > 0 && len2 > 0 && base1 + len1 == base2; |
760 |
|
761 |
// Copy second run into temp array |
762 |
T[] a = this.a; // For performance |
763 |
T[] tmp = ensureCapacity(len2); |
764 |
System.arraycopy(a, base2, tmp, 0, len2); |
765 |
|
766 |
int cursor1 = base1 + len1 - 1; // Indexes into a |
767 |
int cursor2 = len2 - 1; // Indexes into tmp array |
768 |
int dest = base2 + len2 - 1; // Indexes into a |
769 |
|
770 |
// Move last element of first run and deal with degenerate cases |
771 |
a[dest--] = a[cursor1--]; |
772 |
if (--len1 == 0) { |
773 |
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2); |
774 |
return; |
775 |
} |
776 |
if (len2 == 1) { |
777 |
dest -= len1; |
778 |
cursor1 -= len1; |
779 |
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); |
780 |
a[dest] = tmp[cursor2]; |
781 |
return; |
782 |
} |
783 |
|
784 |
Comparator<? super T> c = this.c; // Use local variable for performance |
785 |
int minGallop = this.minGallop; // " " " " " |
786 |
outer: |
787 |
while (true) { |
788 |
int count1 = 0; // Number of times in a row that first run won |
789 |
int count2 = 0; // Number of times in a row that second run won |
790 |
|
791 |
/* |
792 |
* Do the straightforward thing until (if ever) one run |
793 |
* appears to win consistently. |
794 |
*/ |
795 |
do { |
796 |
assert len1 > 0 && len2 > 1; |
797 |
if (c.compare(tmp[cursor2], a[cursor1]) < 0) { |
798 |
a[dest--] = a[cursor1--]; |
799 |
count1++; |
800 |
count2 = 0; |
801 |
if (--len1 == 0) |
802 |
break outer; |
803 |
} else { |
804 |
a[dest--] = tmp[cursor2--]; |
805 |
count2++; |
806 |
count1 = 0; |
807 |
if (--len2 == 1) |
808 |
break outer; |
809 |
} |
810 |
} while ((count1 | count2) < minGallop); |
811 |
|
812 |
/* |
813 |
* One run is winning so consistently that galloping may be a |
814 |
* huge win. So try that, and continue galloping until (if ever) |
815 |
* neither run appears to be winning consistently anymore. |
816 |
*/ |
817 |
do { |
818 |
assert len1 > 0 && len2 > 1; |
819 |
count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c); |
820 |
if (count1 != 0) { |
821 |
dest -= count1; |
822 |
cursor1 -= count1; |
823 |
len1 -= count1; |
824 |
System.arraycopy(a, cursor1 + 1, a, dest + 1, count1); |
825 |
if (len1 == 0) |
826 |
break outer; |
827 |
} |
828 |
a[dest--] = tmp[cursor2--]; |
829 |
if (--len2 == 1) |
830 |
break outer; |
831 |
|
832 |
count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c); |
833 |
if (count2 != 0) { |
834 |
dest -= count2; |
835 |
cursor2 -= count2; |
836 |
len2 -= count2; |
837 |
System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2); |
838 |
if (len2 <= 1) // len2 == 1 || len2 == 0 |
839 |
break outer; |
840 |
} |
841 |
a[dest--] = a[cursor1--]; |
842 |
if (--len1 == 0) |
843 |
break outer; |
844 |
minGallop--; |
845 |
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); |
846 |
if (minGallop < 0) |
847 |
minGallop = 0; |
848 |
minGallop += 2; // Penalize for leaving gallop mode |
849 |
} // End of "outer" loop |
850 |
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field |
851 |
|
852 |
if (len2 == 1) { |
853 |
assert len1 > 0; |
854 |
dest -= len1; |
855 |
cursor1 -= len1; |
856 |
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); |
857 |
a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge |
858 |
} else if (len2 == 0) { |
859 |
throw new IllegalArgumentException( |
860 |
"Comparison method violates its general contract!"); |
861 |
} else { |
862 |
assert len1 == 0; |
863 |
assert len2 > 0; |
864 |
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2); |
865 |
} |
866 |
} |
867 |
|
868 |
/** |
869 |
* Ensures that the external array tmp has at least the specified |
870 |
* number of elements, increasing its size if necessary. The size |
871 |
* increases exponentially to ensure amortized linear time complexity. |
872 |
* |
873 |
* @param minCapacity the minimum required capacity of the tmp array |
874 |
* @return tmp, whether or not it grew |
875 |
*/ |
876 |
private T[] ensureCapacity(int minCapacity) { |
877 |
if (tmp.length < minCapacity) { |
878 |
// Compute smallest power of 2 > minCapacity |
879 |
int newSize = minCapacity; |
880 |
newSize |= newSize >> 1; |
881 |
newSize |= newSize >> 2; |
882 |
newSize |= newSize >> 4; |
883 |
newSize |= newSize >> 8; |
884 |
newSize |= newSize >> 16; |
885 |
newSize++; |
886 |
|
887 |
if (newSize < 0) // Not bloody likely! |
888 |
newSize = minCapacity; |
889 |
else |
890 |
newSize = Math.min(newSize, a.length >>> 1); |
891 |
|
892 |
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) |
893 |
T[] newArray = (T[]) new Object[newSize]; |
894 |
tmp = newArray; |
895 |
} |
896 |
return tmp; |
897 |
} |
898 |
|
899 |
/** |
900 |
* Checks that fromIndex and toIndex are in range, and throws an |
901 |
* appropriate exception if they aren't. |
902 |
* |
903 |
* @param arrayLen the length of the array |
904 |
* @param fromIndex the index of the first element of the range |
905 |
* @param toIndex the index after the last element of the range |
906 |
* @throws IllegalArgumentException if fromIndex > toIndex |
907 |
* @throws ArrayIndexOutOfBoundsException if fromIndex < 0 |
908 |
* or toIndex > arrayLen |
909 |
*/ |
910 |
private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) { |
911 |
if (fromIndex > toIndex) |
912 |
throw new IllegalArgumentException("fromIndex(" + fromIndex + |
913 |
") > toIndex(" + toIndex+")"); |
914 |
if (fromIndex < 0) |
915 |
throw new ArrayIndexOutOfBoundsException(fromIndex); |
916 |
if (toIndex > arrayLen) |
917 |
throw new ArrayIndexOutOfBoundsException(toIndex); |
918 |
} |
919 |
} |