/* | |
* Copyright (C) 2008 The Android Open Source Project | |
* | |
* 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 | |
private int tmpCount; | |
/** 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 performance. 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; | |
tmpCount = 0; | |
/** 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; | |
this.a = null; | |
this.c = null; | |
T[] tmp = this.tmp; | |
for (int i = 0, n = tmpCount; i < n; i++) | |
tmp[i] = null; | |
} | |
/** 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 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[n - 2] > runLen[n - 1] + runLen[n] 2. runLen[n - 1] > runLen[n] | |
* | |
* where n is the index of the last run in runLen. | |
* | |
* This method has been formally verified to be correct after checking the last 4 runs. | |
* Checking for 3 runs results in an exception for large arrays. | |
* (Source: http://envisage-project.eu/proving-android-java-and-python-sorting-algorithm-is-broken-and-how-to-fix-it/) | |
* | |
* 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 >= 1 && runLen[n - 1] <= runLen[n] + runLen[n + 1]) || (n >= 2 && runLen[n - 2] <= runLen[n] + runLen[n - 1])) { | |
if (runLen[n - 1] < runLen[n + 1]) n--; | |
} else if (runLen[n] > runLen[n + 1]) { | |
break; // Invariant is established | |
} | |
mergeAt(n); | |
} | |
} | |
/** 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) { | |
tmpCount = Math.max(tmpCount, 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); | |
} | |
} |