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/* * Copyright (C) 2008 The Android Open Source Project * // w w w . j a v a 2s . com * Licensed under the Apache License, Version 2.0 (the "License"); you may not use this file except in compliance with the * License. You may obtain a copy of the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an "AS IS" * BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language * governing permissions and limitations under the License. */ package com.badlogic.gdx.utils; import java.util.Arrays; import java.util.Comparator; /** * 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. */ 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 T[] a; /** The comparator for this sort. */ private 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; /** * Asserts have been placed in if-statements for performace. To enable them, * set this field to true and enable them in VM with a command line flag. If * you modify this class, please do test the asserts! */ private static final boolean DEBUG = false; TimSort() { tmp = (T[]) new Object[INITIAL_TMP_STORAGE_LENGTH]; runBase = new int[40]; runLen = new int[40]; } public void doSort(T[] a, Comparator<T> c, int lo, int hi) { stackSize = 0; 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; } this.a = a; this.c = c; /** * 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. */ 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 pushRun(lo, runLen); mergeCollapse(); // Advance to find next run lo += runLen; nRemaining -= runLen; } while (nRemaining != 0); // Merge all remaining runs to complete sort if (DEBUG) assert lo == hi; mergeForceCollapse(); if (DEBUG) assert stackSize == 1; } /** * 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; 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. */ static <T> void sort(T[] a, Comparator<? super T> c) { sort(a, 0, a.length, c); } 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 if (DEBUG) assert lo == hi; ts.mergeForceCollapse(); if (DEBUG) 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) { if (DEBUG) 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; if (DEBUG) 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; } if (DEBUG) 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) { if (DEBUG) 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) { if (DEBUG) 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) { if (DEBUG) assert stackSize >= 2; if (DEBUG) assert i >= 0; if (DEBUG) 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]; if (DEBUG) assert len1 > 0 && len2 > 0; if (DEBUG) 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); if (DEBUG) 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); if (DEBUG) 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) { if (DEBUG) 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; } if (DEBUG) 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] } if (DEBUG) 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) { if (DEBUG) 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; } if (DEBUG) 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 } if (DEBUG) 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) { if (DEBUG) 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 { if (DEBUG) 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 { if (DEBUG) 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) { if (DEBUG) 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 { if (DEBUG) assert len2 == 0; if (DEBUG) 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) { if (DEBUG) 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 { if (DEBUG) 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 { if (DEBUG) 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) { if (DEBUG) 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 { if (DEBUG) assert len1 == 0; if (DEBUG) 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); 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); } }