/*
* Copyright 2009 Google Inc. All Rights Reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation. Oracle designates this
* particular file as subject to the "Classpath" exception as provided
* by Oracle in the LICENSE file that accompanied this code.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
* or visit www.oracle.com if you need additional information or have any
* questions.
*/
package com.WazaBe.HoloEverywhere.util;
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.
*
* @author Josh Bloch
*/
@SuppressWarnings({ "unchecked" })
class TimSort<T> {
/**
* 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;
/**
* 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 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;
/**
* Sorts the specified portion of the specified array using a binary
* insertion sort. This is the best method for sorting small numbers of
* elements. It requires O(n log n) compares, but O(n^2) data movement
* (worst case).
*
* If the initial part of the specified range is already sorted, this method
* can take advantage of it: the method assumes that the elements from index
* {@code lo}, inclusive, to {@code start}, exclusive are already sorted.
*
* @param a
* the array in which a range is to be sorted
* @param lo
* the index of the first element in the range to be sorted
* @param hi
* the index after the last element in the range to be sorted
* @param start
* the index of the first element in the range that is not
* already known to be sorted ({@code lo <= start <= hi})
* @param c
* comparator to used for the sort
*/
@SuppressWarnings("fallthrough")
private static <T> void binarySort(T[] a, int lo, int hi, int start,
Comparator<? super T> c) {
assert lo <= start && start <= hi;
if (start == lo) {
start++;
}
for (; start < hi; start++) {
T pivot = a[start];
// Set left (and right) to the index where a[start] (pivot) belongs
int left = lo;
int right = start;
assert left <= right;
/*
* Invariants: pivot >= all in [lo, left). pivot < all in [right,
* start).
*/
while (left < right) {
int mid = left + right >>> 1;
if (c.compare(pivot, a[mid]) < 0) {
right = mid;
} else {
left = mid + 1;
}
}
assert left == right;
/*
* The invariants still hold: pivot >= all in [lo, left) and pivot <
* all in [left, start), so pivot belongs at left. Note that if
* there are elements equal to pivot, left points to the first slot
* after them -- that's why this sort is stable. Slide elements over
* to make room for pivot.
*/
int n = start - left; // The number of elements to move
// Switch is just an optimization for arraycopy in default case
switch (n) {
case 2:
a[left + 2] = a[left + 1];
case 1:
a[left + 1] = a[left];
break;
default:
System.arraycopy(a, left, a, left + 1, n);
}
a[left] = pivot;
}
}
/**
* Returns the length of the run beginning at the specified position in the
* specified array and reverses the run if it is descending (ensuring that
* the run will always be ascending when the method returns).
*
* A run is the longest ascending sequence with:
*
* a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
*
* or the longest descending sequence with:
*
* a[lo] > a[lo + 1] > a[lo + 2] > ...
*
* For its intended use in a stable mergesort, the strictness of the
* definition of "descending" is needed so that the call can safely reverse
* a descending sequence without violating stability.
*
* @param a
* the array in which a run is to be counted and possibly
* reversed
* @param lo
* index of the first element in the run
* @param hi
* index after the last element that may be contained in the run.
* It is required that {@code lo < hi}.
* @param c
* the comparator to used for the sort
* @return the length of the run beginning at the specified position in the
* specified array
*/
private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,
Comparator<? super T> c) {
assert lo < hi;
int runHi = lo + 1;
if (runHi == hi) {
return 1;
}
// Find end of run, and reverse range if descending
if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0) {
runHi++;
}
reverseRange(a, lo, runHi);
} else { // Ascending
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0) {
runHi++;
}
}
return runHi - lo;
}
/**
* Locates the position at which to insert the specified key into the
* specified sorted range; if the range contains an element equal to key,
* returns the index of the leftmost equal element.
*
* @param key
* the key whose insertion point to search for
* @param a
* the array in which to search
* @param base
* the index of the first element in the range
* @param len
* the length of the range; must be > 0
* @param hint
* the index at which to begin the search, 0 <= hint < n. The
* closer hint is to the result, the faster this method will run.
* @param c
* the comparator used to order the range, and to search
* @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
* pretending that a[b - 1] is minus infinity and a[b + n] is
* infinity. In other words, key belongs at index b + k; or in other
* words, the first k elements of a should precede key, and the last
* n - k should follow it.
*/
private static <T> int gallopLeft(T key, T[] a, int base, int len,
int hint, Comparator<? super T> c) {
assert len > 0 && hint >= 0 && hint < len;
int lastOfs = 0;
int ofs = 1;
if (c.compare(key, a[base + hint]) > 0) {
// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) {
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) {
ofs = maxOfs;
}
}
if (ofs > maxOfs) {
ofs = maxOfs;
}
// Make offsets relative to base
int tmp = lastOfs;
lastOfs = hint - ofs;
ofs = hint - tmp;
}
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
/*
* Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere to
* the right of lastOfs but no farther right than ofs. Do a binary
* search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
*/
lastOfs++;
while (lastOfs < ofs) {
int m = lastOfs + (ofs - lastOfs >>> 1);
if (c.compare(key, a[base + m]) > 0) {
lastOfs = m + 1; // a[base + m] < key
} else {
ofs = m; // key <= a[base + m]
}
}
assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
return ofs;
}
/**
* Like gallopLeft, except that if the range contains an element equal to
* key, gallopRight returns the index after the rightmost equal element.
*
* @param key
* the key whose insertion point to search for
* @param a
* the array in which to search
* @param base
* the index of the first element in the range
* @param len
* the length of the range; must be > 0
* @param hint
* the index at which to begin the search, 0 <= hint < n. The
* closer hint is to the result, the faster this method will run.
* @param c
* the comparator used to order the range, and to search
* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
*/
private static <T> int gallopRight(T key, T[] a, int base, int len,
int hint, Comparator<? super T> c) {
assert len > 0 && hint >= 0 && hint < len;
int ofs = 1;
int lastOfs = 0;
if (c.compare(key, a[base + hint]) < 0) {
// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
int maxOfs = hint + 1;
while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) {
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) {
ofs = maxOfs;
}
}
if (ofs > maxOfs) {
ofs = maxOfs;
}
// Make offsets relative to b
lastOfs += hint;
ofs += hint;
}
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
/*
* Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
* the right of lastOfs but no farther right than ofs. Do a binary
* search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
*/
lastOfs++;
while (lastOfs < ofs) {
int m = lastOfs + (ofs - lastOfs >>> 1);
if (c.compare(key, a[base + m]) < 0) {
ofs = m; // key < a[b + m]
} else {
lastOfs = m + 1; // a[b + m] <= key
}
}
assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
return ofs;
}
/**
* Returns the minimum acceptable run length for an array of the specified
* length. Natural runs shorter than this will be extended with
* {@link #binarySort}.
*
* Roughly speaking, the computation is:
*
* If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
* Else if n is an exact power of 2, return MIN_MERGE/2. Else return an int
* k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k is close to, but strictly
* less than, an exact power of 2.
*
* For the rationale, see listsort.txt.
*
* @param n
* the length of the array to be sorted
* @return the length of the minimum run to be merged
*/
private static int minRunLength(int n) {
assert n >= 0;
int r = 0; // Becomes 1 if any 1 bits are shifted off
while (n >= MIN_MERGE) {
r |= n & 1;
n >>= 1;
}
return n + r;
}
/**
* 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);
}
}
/**
* 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;
}
}
static <T> void sort(T[] a, Comparator<? super T> c) {
sort(a, 0, a.length, c);
}
/*
* 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, int lo, int hi, Comparator<? super T> c) {
if (c == null) {
Arrays.sort(a, lo, hi);
return;
}
rangeCheck(a.length, lo, hi);
int nRemaining = hi - lo;
if (nRemaining < 2) {
return; // Arrays of size 0 and 1 are always sorted
}
// If array is small, do a "mini-TimSort" with no merges
if (nRemaining < MIN_MERGE) {
int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
binarySort(a, lo, hi, lo + initRunLen, c);
return;
}
/**
* March over the array once, left to right, finding natural runs,
* extending short natural runs to minRun elements, and merging runs to
* maintain stack invariant.
*/
TimSort<T> ts = new TimSort<T>(a, c);
int minRun = minRunLength(nRemaining);
do {
// Identify next run
int runLen = countRunAndMakeAscending(a, lo, hi, c);
// If run is short, extend to min(minRun, nRemaining)
if (runLen < minRun) {
int force = nRemaining <= minRun ? nRemaining : minRun;
binarySort(a, lo, lo + force, lo + runLen, c);
runLen = force;
}
// Push run onto pending-run stack, and maybe merge
ts.pushRun(lo, runLen);
ts.mergeCollapse();
// Advance to find next run
lo += runLen;
nRemaining -= runLen;
} while (nRemaining != 0);
// Merge all remaining runs to complete sort
assert lo == hi;
ts.mergeForceCollapse();
assert ts.stackSize == 1;
}
/**
* The array being sorted.
*/
private final T[] a;
/**
* The comparator for this sort.
*/
private final Comparator<? super T> c;
/**
* 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;
private final int[] runBase;
private final int[] runLen;
/**
* 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
/**
* Temp storage for merges.
*/
private T[] tmp; // Actual runtime type will be Object[], regardless of T
/**
* 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];
}
/**
* 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) {
newSize = minCapacity;
} else {
newSize = Math.min(newSize, a.length >>> 1);
}
T[] newArray = (T[]) new Object[newSize];
tmp = newArray;
}
return tmp;
}
/**
* Merges the two runs at stack indices i and i+1. Run i must be the
* penultimate or antepenultimate run on the stack. In other words, i must
* be equal to stackSize-2 or stackSize-3.
*
* @param i
* stack index of the first of the two runs to merge
*/
private void mergeAt(int i) {
assert stackSize >= 2;
assert i >= 0;
assert i == stackSize - 2 || i == stackSize - 3;
int base1 = runBase[i];
int len1 = runLen[i];
int base2 = runBase[i + 1];
int len2 = runLen[i + 1];
assert len1 > 0 && len2 > 0;
assert base1 + len1 == base2;
/*
* Record the length of the combined runs; if i is the 3rd-last run now,
* also slide over the last run (which isn't involved in this merge).
* The current run (i+1) goes away in any case.
*/
runLen[i] = len1 + len2;
if (i == stackSize - 3) {
runBase[i + 1] = runBase[i + 2];
runLen[i + 1] = runLen[i + 2];
}
stackSize--;
/*
* Find where the first element of run2 goes in run1. Prior elements in
* run1 can be ignored (because they're already in place).
*/
int k = gallopRight(a[base2], a, base1, len1, 0, c);
assert k >= 0;
base1 += k;
len1 -= k;
if (len1 == 0) {
return;
}
/*
* Find where the last element of run1 goes in run2. Subsequent elements
* in run2 can be ignored (because they're already in place).
*/
len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
assert len2 >= 0;
if (len2 == 0) {
return;
}
// Merge remaining runs, using tmp array with min(len1, len2) elements
if (len1 <= len2) {
mergeLo(base1, len1, base2, len2);
} else {
mergeHi(base1, len1, base2, len2);
}
}
/**
* 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);
}
}
/**
* Like mergeLo, except that this method should be called only if len1 >=
* len2; mergeLo should be called if len1 <= len2. (Either method may be
* called if len1 == len2.)
*
* @param base1
* index of first element in first run to be merged
* @param len1
* length of first run to be merged (must be > 0)
* @param base2
* index of first element in second run to be merged (must be
* aBase + aLen)
* @param len2
* length of second run to be merged (must be > 0)
*/
private void mergeHi(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy second run into temp array
T[] a = this.a; // For performance
T[] tmp = ensureCapacity(len2);
System.arraycopy(a, base2, tmp, 0, len2);
int cursor1 = base1 + len1 - 1; // Indexes into a
int cursor2 = len2 - 1; // Indexes into tmp array
int dest = base2 + len2 - 1; // Indexes into a
// Move last element of first run and deal with degenerate cases
a[dest--] = a[cursor1--];
if (--len1 == 0) {
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
return;
}
if (len2 == 1) {
dest -= len1;
cursor1 -= len1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
a[dest] = tmp[cursor2];
return;
}
Comparator<? super T> c = this.c; // Use local variable for performance
int minGallop = this.minGallop; // " " " " "
outer: while (true) {
int count1 = 0; // Number of times in a row that first run won
int count2 = 0; // Number of times in a row that second run won
/*
* Do the straightforward thing until (if ever) one run appears to
* win consistently.
*/
do {
assert len1 > 0 && len2 > 1;
if (c.compare(tmp[cursor2], a[cursor1]) < 0) {
a[dest--] = a[cursor1--];
count1++;
count2 = 0;
if (--len1 == 0) {
break outer;
}
} else {
a[dest--] = tmp[cursor2--];
count2++;
count1 = 0;
if (--len2 == 1) {
break outer;
}
}
} while ((count1 | count2) < minGallop);
/*
* One run is winning so consistently that galloping may be a huge
* win. So try that, and continue galloping until (if ever) neither
* run appears to be winning consistently anymore.
*/
do {
assert len1 > 0 && len2 > 1;
count1 = len1
- gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
if (count1 != 0) {
dest -= count1;
cursor1 -= count1;
len1 -= count1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
if (len1 == 0) {
break outer;
}
}
a[dest--] = tmp[cursor2--];
if (--len2 == 1) {
break outer;
}
count2 = len2
- gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);
if (count2 != 0) {
dest -= count2;
cursor2 -= count2;
len2 -= count2;
System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
if (len2 <= 1) {
break outer;
}
}
a[dest--] = a[cursor1--];
if (--len1 == 0) {
break outer;
}
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0) {
minGallop = 0;
}
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
if (len2 == 1) {
assert len1 > 0;
dest -= len1;
cursor1 -= len1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
} else if (len2 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
assert len1 == 0;
assert len2 > 0;
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
}
}
/**
* Merges two adjacent runs in place, in a stable fashion. The first element
* of the first run must be greater than the first element of the second run
* (a[base1] > a[base2]), and the last element of the first run (a[base1 +
* len1-1]) must be greater than all elements of the second run.
*
* For performance, this method should be called only when len1 <= len2; its
* twin, mergeHi should be called if len1 >= len2. (Either method may be
* called if len1 == len2.)
*
* @param base1
* index of first element in first run to be merged
* @param len1
* length of first run to be merged (must be > 0)
* @param base2
* index of first element in second run to be merged (must be
* aBase + aLen)
* @param len2
* length of second run to be merged (must be > 0)
*/
private void mergeLo(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy first run into temp array
T[] a = this.a; // For performance
T[] tmp = ensureCapacity(len1);
System.arraycopy(a, base1, tmp, 0, len1);
int cursor1 = 0; // Indexes into tmp array
int cursor2 = base2; // Indexes int a
int dest = base1; // Indexes int a
// Move first element of second run and deal with degenerate cases
a[dest++] = a[cursor2++];
if (--len2 == 0) {
System.arraycopy(tmp, cursor1, a, dest, len1);
return;
}
if (len1 == 1) {
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
return;
}
Comparator<? super T> c = this.c; // Use local variable for performance
int minGallop = this.minGallop; // " " " " "
outer: while (true) {
int count1 = 0; // Number of times in a row that first run won
int count2 = 0; // Number of times in a row that second run won
/*
* Do the straightforward thing until (if ever) one run starts
* winning consistently.
*/
do {
assert len1 > 1 && len2 > 0;
if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
a[dest++] = a[cursor2++];
count2++;
count1 = 0;
if (--len2 == 0) {
break outer;
}
} else {
a[dest++] = tmp[cursor1++];
count1++;
count2 = 0;
if (--len1 == 1) {
break outer;
}
}
} while ((count1 | count2) < minGallop);
/*
* One run is winning so consistently that galloping may be a huge
* win. So try that, and continue galloping until (if ever) neither
* run appears to be winning consistently anymore.
*/
do {
assert len1 > 1 && len2 > 0;
count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
if (count1 != 0) {
System.arraycopy(tmp, cursor1, a, dest, count1);
dest += count1;
cursor1 += count1;
len1 -= count1;
if (len1 <= 1) {
break outer;
}
}
a[dest++] = a[cursor2++];
if (--len2 == 0) {
break outer;
}
count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
if (count2 != 0) {
System.arraycopy(a, cursor2, a, dest, count2);
dest += count2;
cursor2 += count2;
len2 -= count2;
if (len2 == 0) {
break outer;
}
}
a[dest++] = tmp[cursor1++];
if (--len1 == 1) {
break outer;
}
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0) {
minGallop = 0;
}
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
if (len1 == 1) {
assert len2 > 0;
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
} else if (len1 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
assert len2 == 0;
assert len1 > 1;
System.arraycopy(tmp, cursor1, a, dest, len1);
}
}
/**
* 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++;
}
}