spark TimSort 源码
spark TimSort 代码
文件路径：/core/src/main/java/org/apache/spark/util/collection/TimSort.java
/*
* Licensed to the Apache Software Foundation (ASF) under one or more
* contributor license agreements. See the NOTICE file distributed with
* this work for additional information regarding copyright ownership.
* The ASF licenses this file to You 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.
*/
/*
* Based on TimSort.java from the Android Open Source Project
*
* 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 org.apache.spark.util.collection;
import java.util.Comparator;
/**
* A port of the Android TimSort class, which utilizes a "stable, adaptive, iterative mergesort."
* See the method comment on sort() for more details.
*
* This has been kept in Java with the original style in order to match very closely with the
* Android source code, and thus be easy to verify correctness. The class is package private. We put
* a simple Scala wrapper {@link org.apache.spark.util.collection.Sorter}, which is available to
* package org.apache.spark.
*
* The purpose of the port is to generalize the interface to the sort to accept input data formats
* besides simple arrays where every element is sorted individually. For instance, the AppendOnlyMap
* uses this to sort an Array with alternating elements of the form [key, value, key, value].
* This generalization comes with minimal overhead -- see SortDataFormat for more information.
*
* We allow key reuse to prevent creating many key objects -- see SortDataFormat.
*
* @see org.apache.spark.util.collection.SortDataFormat
* @see org.apache.spark.util.collection.Sorter
*/
class TimSort<K, Buffer> {
/**
* 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
* 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;
private final SortDataFormat<K, Buffer> s;
public TimSort(SortDataFormat<K, Buffer> sortDataFormat) {
this.s = sortDataFormat;
}
/**
* 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
*/
public void sort(Buffer a, int lo, int hi, Comparator<? super K> c) {
assert c != null;
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.
*/
SortState sortState = new SortState(a, c, hi - lo);
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
sortState.pushRun(lo, runLen);
sortState.mergeCollapse();
// Advance to find next run
lo += runLen;
nRemaining -= runLen;
} while (nRemaining != 0);
// Merge all remaining runs to complete sort
assert lo == hi;
sortState.mergeForceCollapse();
assert sortState.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 void binarySort(Buffer a, int lo, int hi, int start, Comparator<? super K> c) {
assert lo <= start && start <= hi;
if (start == lo)
start++;
K key0 = s.newKey();
K key1 = s.newKey();
Buffer pivotStore = s.allocate(1);
for ( ; start < hi; start++) {
s.copyElement(a, start, pivotStore, 0);
K pivot = s.getKey(pivotStore, 0, key0);
// 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, s.getKey(a, mid, key1)) < 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: s.copyElement(a, left + 1, a, left + 2);
case 1: s.copyElement(a, left, a, left + 1);
break;
default: s.copyRange(a, left, a, left + 1, n);
}
s.copyElement(pivotStore, 0, a, left);
}
}
/**
* 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 int countRunAndMakeAscending(Buffer a, int lo, int hi, Comparator<? super K> c) {
assert lo < hi;
int runHi = lo + 1;
if (runHi == hi)
return 1;
K key0 = s.newKey();
K key1 = s.newKey();
// Find end of run, and reverse range if descending
if (c.compare(s.getKey(a, runHi++, key0), s.getKey(a, lo, key1)) < 0) { // Descending
while (runHi < hi && c.compare(s.getKey(a, runHi, key0), s.getKey(a, runHi - 1, key1)) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else { // Ascending
while (runHi < hi && c.compare(s.getKey(a, runHi, key0), s.getKey(a, runHi - 1, key1)) >= 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 void reverseRange(Buffer a, int lo, int hi) {
hi--;
while (lo < hi) {
s.swap(a, lo, hi);
lo++;
hi--;
}
}
/**
* 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 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;
}
private class SortState {
/**
* The Buffer being sorted.
*/
private final Buffer a;
/**
* Length of the sort Buffer.
*/
private final int aLength;
/**
* The comparator for this sort.
*/
private final Comparator<? super K> 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 Buffer tmp; // Actual runtime type will be Object[], regardless of T
/**
* Length of the temp storage.
*/
private int tmpLength = 0;
/**
* 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;
/**
* 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 SortState(Buffer a, Comparator<? super K> c, int len) {
this.aLength = len;
this.a = a;
this.c = c;
// Allocate temp storage (which may be increased later if necessary)
tmpLength = len < 2 * INITIAL_TMP_STORAGE_LENGTH ? len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
tmp = s.allocate(tmpLength);
/*
* 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.
* The maximum value of 49 allows for an array up to length
* Integer.MAX_VALUE-4, if array is filled by the worst case stack size
* increasing scenario. More explanations are given in section 4 of:
* http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf
*/
int stackLen = (len < 120 ? 5 :
len < 1542 ? 10 :
len < 119151 ? 24 : 49);
runBase = new int[stackLen];
runLen = new int[stackLen];
}
/**
* 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.
*
* Thanks to Stijn de Gouw, Jurriaan Rot, Frank S. de Boer,
* Richard Bubel and Reiner Hahnle, this is fixed with respect to
* the analysis in "On the Worst-Case Complexity of TimSort" by
* Nicolas Auger, Vincent Jug, Cyril Nicaud, and Carine Pivoteau.
*/
private void mergeCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1] ||
n > 1 && runLen[n-2] <= runLen[n] + runLen[n-1]) {
if (runLen[n - 1] < runLen[n + 1])
n--;
} else if (n < 0 || 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) {
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--;
K key0 = s.newKey();
/*
* 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(s.getKey(a, base2, key0), 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(s.getKey(a, base1 + len1 - 1, key0), 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);
}
/**
* 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 int gallopLeft(K key, Buffer a, int base, int len, int hint, Comparator<? super K> c) {
assert len > 0 && hint >= 0 && hint < len;
int lastOfs = 0;
int ofs = 1;
K key0 = s.newKey();
if (c.compare(key, s.getKey(a, base + hint, key0)) > 0) {
// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint + ofs, key0)) > 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, s.getKey(a, base + hint - ofs, key0)) <= 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;
}
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, s.getKey(a, base + m, key0)) > 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 int gallopRight(K key, Buffer a, int base, int len, int hint, Comparator<? super K> c) {
assert len > 0 && hint >= 0 && hint < len;
int ofs = 1;
int lastOfs = 0;
K key1 = s.newKey();
if (c.compare(key, s.getKey(a, base + hint, key1)) < 0) {
// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
int maxOfs = hint + 1;
while (ofs < maxOfs && c.compare(key, s.getKey(a, base + hint - ofs, key1)) < 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, s.getKey(a, base + hint + ofs, key1)) >= 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;
}
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, s.getKey(a, base + m, key1)) < 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;
}
/**
* 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
Buffer a = this.a; // For performance
Buffer tmp = ensureCapacity(len1);
s.copyRange(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
s.copyElement(a, cursor2++, a, dest++);
if (--len2 == 0) {
s.copyRange(tmp, cursor1, a, dest, len1);
return;
}
if (len1 == 1) {
s.copyRange(a, cursor2, a, dest, len2);
s.copyElement(tmp, cursor1, a, dest + len2); // Last elt of run 1 to end of merge
return;
}
K key0 = s.newKey();
K key1 = s.newKey();
Comparator<? super K> 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(s.getKey(a, cursor2, key0), s.getKey(tmp, cursor1, key1)) < 0) {
s.copyElement(a, cursor2++, a, dest++);
count2++;
count1 = 0;
if (--len2 == 0)
break outer;
} else {
s.copyElement(tmp, cursor1++, a, dest++);
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(s.getKey(a, cursor2, key0), tmp, cursor1, len1, 0, c);
if (count1 != 0) {
s.copyRange(tmp, cursor1, a, dest, count1);
dest += count1;
cursor1 += count1;
len1 -= count1;
if (len1 <= 1) // len1 == 1 || len1 == 0
break outer;
}
s.copyElement(a, cursor2++, a, dest++);
if (--len2 == 0)
break outer;
count2 = gallopLeft(s.getKey(tmp, cursor1, key0), a, cursor2, len2, 0, c);
if (count2 != 0) {
s.copyRange(a, cursor2, a, dest, count2);
dest += count2;
cursor2 += count2;
len2 -= count2;
if (len2 == 0)
break outer;
}
s.copyElement(tmp, cursor1++, a, dest++);
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;
s.copyRange(a, cursor2, a, dest, len2);
s.copyElement(tmp, cursor1, a, dest + len2); // 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;
s.copyRange(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) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy second run into temp array
Buffer a = this.a; // For performance
Buffer tmp = ensureCapacity(len2);
s.copyRange(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
K key0 = s.newKey();
K key1 = s.newKey();
// Move last element of first run and deal with degenerate cases
s.copyElement(a, cursor1--, a, dest--);
if (--len1 == 0) {
s.copyRange(tmp, 0, a, dest - (len2 - 1), len2);
return;
}
if (len2 == 1) {
dest -= len1;
cursor1 -= len1;
s.copyRange(a, cursor1 + 1, a, dest + 1, len1);
s.copyElement(tmp, cursor2, a, dest);
return;
}
Comparator<? super K> 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(s.getKey(tmp, cursor2, key0), s.getKey(a, cursor1, key1)) < 0) {
s.copyElement(a, cursor1--, a, dest--);
count1++;
count2 = 0;
if (--len1 == 0)
break outer;
} else {
s.copyElement(tmp, cursor2--, a, dest--);
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(s.getKey(tmp, cursor2, key0), a, base1, len1, len1 - 1, c);
if (count1 != 0) {
dest -= count1;
cursor1 -= count1;
len1 -= count1;
s.copyRange(a, cursor1 + 1, a, dest + 1, count1);
if (len1 == 0)
break outer;
}
s.copyElement(tmp, cursor2--, a, dest--);
if (--len2 == 1)
break outer;
count2 = len2 - gallopLeft(s.getKey(a, cursor1, key0), tmp, 0, len2, len2 - 1, c);
if (count2 != 0) {
dest -= count2;
cursor2 -= count2;
len2 -= count2;
s.copyRange(tmp, cursor2 + 1, a, dest + 1, count2);
if (len2 <= 1) // len2 == 1 || len2 == 0
break outer;
}
s.copyElement(a, cursor1--, a, dest--);
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;
s.copyRange(a, cursor1 + 1, a, dest + 1, len1);
s.copyElement(tmp, cursor2, a, dest); // 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;
s.copyRange(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 Buffer ensureCapacity(int minCapacity) {
if (tmpLength < 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, aLength >>> 1);
tmp = s.allocate(newSize);
tmpLength = newSize;
}
return tmp;
}
}
}
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