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ComparableTimSort.java
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805 lines (723 loc) · 29.3 KB
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/*
* 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;
/** This is a near duplicate of {@link TimSort}, modified for use with arrays of objects that implement {@link Comparable},
* instead of using explicit comparators.
*
* <p>
* If you are using an optimizing VM, you may find that ComparableTimSort offers no performance benefit over TimSort in
* conjunction with a comparator that simply returns {@code ((Comparable)first).compareTo(Second)}. If this is the case, you are
* better off deleting ComparableTimSort to eliminate the code duplication. (See Arrays.java for details.) */
class ComparableTimSort {
/** 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 Object[] a;
/** 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 Object[] tmp;
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 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;
ComparableTimSort () {
tmp = new Object[INITIAL_TMP_STORAGE_LENGTH];
runBase = new int[40];
runLen = new int[40];
}
public void doSort (Object[] a, 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);
binarySort(a, lo, hi, lo + initRunLen);
return;
}
this.a = a;
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);
// 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);
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;
Object[] 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 */
private ComparableTimSort (Object[] a) {
this.a = a;
// Allocate temp storage (which may be increased later if necessary)
int len = a.length;
Object[] newArray = 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 void sort (Object[] a) {
sort(a, 0, a.length);
}
static void sort (Object[] a, int lo, int hi) {
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);
binarySort(a, lo, hi, lo + initRunLen);
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. */
ComparableTimSort ts = new ComparableTimSort(a);
int minRun = minRunLength(nRemaining);
do {
// Identify next run
int runLen = countRunAndMakeAscending(a, lo, hi);
// 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);
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} */
@SuppressWarnings("fallthrough")
private static void binarySort (Object[] a, int lo, int hi, int start) {
if (DEBUG) assert lo <= start && start <= hi;
if (start == lo) start++;
for (; start < hi; start++) {
@SuppressWarnings("unchecked")
Comparable<Object> pivot = (Comparable)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 (pivot.compareTo(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}.
* @return the length of the run beginning at the specified position in the specified array */
@SuppressWarnings("unchecked")
private static int countRunAndMakeAscending (Object[] a, int lo, int hi) {
if (DEBUG) assert lo < hi;
int runHi = lo + 1;
if (runHi == hi) return 1;
// Find end of run, and reverse range if descending
if (((Comparable)a[runHi++]).compareTo(a[lo]) < 0) { // Descending
while (runHi < hi && ((Comparable)a[runHi]).compareTo(a[runHi - 1]) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else { // Ascending
while (runHi < hi && ((Comparable)a[runHi]).compareTo(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 */
@SuppressWarnings("unchecked")
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((Comparable<Object>)a[base2], a, base1, len1, 0);
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((Comparable<Object>)a[base1 + len1 - 1], a, base2, len2, len2 - 1);
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.
* @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 int gallopLeft (Comparable<Object> key, Object[] a, int base, int len, int hint) {
if (DEBUG) assert len > 0 && hint >= 0 && hint < len;
int lastOfs = 0;
int ofs = 1;
if (key.compareTo(a[base + hint]) > 0) {
// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && key.compareTo(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 && key.compareTo(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 (key.compareTo(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.
* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k] */
private static int gallopRight (Comparable<Object> key, Object[] a, int base, int len, int hint) {
if (DEBUG) assert len > 0 && hint >= 0 && hint < len;
int ofs = 1;
int lastOfs = 0;
if (key.compareTo(a[base + hint]) < 0) {
// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
int maxOfs = hint + 1;
while (ofs < maxOfs && key.compareTo(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 && key.compareTo(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 (key.compareTo(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) */
@SuppressWarnings("unchecked")
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
Object[] a = this.a; // For performance
Object[] 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;
}
int minGallop = this.minGallop; // Use local variable for performance
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 (((Comparable)a[cursor2]).compareTo(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((Comparable)a[cursor2], tmp, cursor1, len1, 0);
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((Comparable)tmp[cursor1], a, cursor2, len2, 0);
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) */
@SuppressWarnings("unchecked")
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
Object[] a = this.a; // For performance
Object[] 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;
}
int minGallop = this.minGallop; // Use local variable for performance
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 (((Comparable)tmp[cursor2]).compareTo(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((Comparable)tmp[cursor2], a, base1, len1, len1 - 1);
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((Comparable)a[cursor1], tmp, 0, len2, len2 - 1);
if (count2 != 0) {
dest -= count2;
cursor2 -= count2;
len2 -= count2;
System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
if (len2 <= 1) break outer; // len2 == 1 || len2 == 0
}
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 Object[] 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);
Object[] newArray = 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);
}
}