android / platform / libcore / gingerbread / . / luni / src / main / java / java / util / TimSort.java

/* | |

* 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 java.util; | |

/** | |

* 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 final T[] a; | |

/** | |

* The comparator for this sort. | |

*/ | |

private final Comparator<? super T> c; | |

/** | |

* When we get into galloping mode, we stay there until both runs win less | |

* often than MIN_GALLOP consecutive times. | |

*/ | |

private static final int MIN_GALLOP = 7; | |

/** | |

* This controls when we get *into* galloping mode. It is initialized | |

* to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for | |

* random data, and lower for highly structured data. | |

*/ | |

private int minGallop = MIN_GALLOP; | |

/** | |

* Maximum initial size of tmp array, which is used for merging. The array | |

* can grow to accommodate demand. | |

* | |

* Unlike Tim's original C version, we do not allocate this much storage | |

* when sorting smaller arrays. This change was required for performance. | |

*/ | |

private static final int INITIAL_TMP_STORAGE_LENGTH = 256; | |

/** | |

* Temp storage for merges. | |

*/ | |

private T[] tmp; // Actual runtime type will be Object[], regardless of T | |

/** | |

* A stack of pending runs yet to be merged. Run i starts at | |

* address base[i] and extends for len[i] elements. It's always | |

* true (so long as the indices are in bounds) that: | |

* | |

* runBase[i] + runLen[i] == runBase[i + 1] | |

* | |

* so we could cut the storage for this, but it's a minor amount, | |

* and keeping all the info explicit simplifies the code. | |

*/ | |

private int stackSize = 0; // Number of pending runs on stack | |

private final int[] runBase; | |

private final int[] runLen; | |

/** | |

* Asserts have been placed in if-statements for performace. To enable them, | |

* set this field to true and enable them in VM with a command line flag. | |

* If you modify this class, please do test the asserts! | |

*/ | |

private static final boolean DEBUG = false; | |

/** | |

* 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; | |

@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) | |

T[] newArray = (T[]) new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ? | |

len >>> 1 : INITIAL_TMP_STORAGE_LENGTH]; | |

tmp = newArray; | |

/* | |

* Allocate runs-to-be-merged stack (which cannot be expanded). The | |

* stack length requirements are described in listsort.txt. The C | |

* version always uses the same stack length (85), but this was | |

* measured to be too expensive when sorting "mid-sized" arrays (e.g., | |

* 100 elements) in Java. Therefore, we use smaller (but sufficiently | |

* large) stack lengths for smaller arrays. The "magic numbers" in the | |

* computation below must be changed if MIN_MERGE is decreased. See | |

* the MIN_MERGE declaration above for more information. | |

*/ | |

int stackLen = (len < 120 ? 5 : | |

len < 1542 ? 10 : | |

len < 119151 ? 19 : 40); | |

runBase = new int[stackLen]; | |

runLen = new int[stackLen]; | |

} | |

/* | |

* The next two methods (which are package private and static) constitute | |

* the entire API of this class. Each of these methods obeys the contract | |

* of the public method with the same signature in java.util.Arrays. | |

*/ | |

static <T> void sort(T[] a, Comparator<? super T> c) { | |

sort(a, 0, a.length, c); | |

} | |

static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c) { | |

if (c == null) { | |

Arrays.sort(a, lo, hi); | |

return; | |

} | |

rangeCheck(a.length, lo, hi); | |

int nRemaining = hi - lo; | |

if (nRemaining < 2) | |

return; // Arrays of size 0 and 1 are always sorted | |

// If array is small, do a "mini-TimSort" with no merges | |

if (nRemaining < MIN_MERGE) { | |

int initRunLen = countRunAndMakeAscending(a, lo, hi, c); | |

binarySort(a, lo, hi, lo + initRunLen, c); | |

return; | |

} | |

/** | |

* March over the array once, left to right, finding natural runs, | |

* extending short natural runs to minRun elements, and merging runs | |

* to maintain stack invariant. | |

*/ | |

TimSort<T> ts = new TimSort<T>(a, c); | |

int minRun = minRunLength(nRemaining); | |

do { | |

// Identify next run | |

int runLen = countRunAndMakeAscending(a, lo, hi, c); | |

// If run is short, extend to min(minRun, nRemaining) | |

if (runLen < minRun) { | |

int force = nRemaining <= minRun ? nRemaining : minRun; | |

binarySort(a, lo, lo + force, lo + runLen, c); | |

runLen = force; | |

} | |

// Push run onto pending-run stack, and maybe merge | |

ts.pushRun(lo, runLen); | |

ts.mergeCollapse(); | |

// Advance to find next run | |

lo += runLen; | |

nRemaining -= runLen; | |

} while (nRemaining != 0); | |

// Merge all remaining runs to complete sort | |

if (DEBUG) assert lo == hi; | |

ts.mergeForceCollapse(); | |

if (DEBUG) assert ts.stackSize == 1; | |

} | |

/** | |

* Sorts the specified portion of the specified array using a binary | |

* insertion sort. This is the best method for sorting small numbers | |

* of elements. It requires O(n log n) compares, but O(n^2) data | |

* movement (worst case). | |

* | |

* If the initial part of the specified range is already sorted, | |

* this method can take advantage of it: the method assumes that the | |

* elements from index {@code lo}, inclusive, to {@code start}, | |

* exclusive are already sorted. | |

* | |

* @param a the array in which a range is to be sorted | |

* @param lo the index of the first element in the range to be sorted | |

* @param hi the index after the last element in the range to be sorted | |

* @param start the index of the first element in the range that is | |

* not already known to be sorted (@code lo <= start <= hi} | |

* @param c comparator to used for the sort | |

*/ | |

@SuppressWarnings("fallthrough") | |

private static <T> void binarySort(T[] a, int lo, int hi, int start, | |

Comparator<? super T> c) { | |

if (DEBUG) assert lo <= start && start <= hi; | |

if (start == lo) | |

start++; | |

for ( ; start < hi; start++) { | |

T pivot = a[start]; | |

// Set left (and right) to the index where a[start] (pivot) belongs | |

int left = lo; | |

int right = start; | |

if (DEBUG) assert left <= right; | |

/* | |

* Invariants: | |

* pivot >= all in [lo, left). | |

* pivot < all in [right, start). | |

*/ | |

while (left < right) { | |

int mid = (left + right) >>> 1; | |

if (c.compare(pivot, a[mid]) < 0) | |

right = mid; | |

else | |

left = mid + 1; | |

} | |

if (DEBUG) assert left == right; | |

/* | |

* The invariants still hold: pivot >= all in [lo, left) and | |

* pivot < all in [left, start), so pivot belongs at left. Note | |

* that if there are elements equal to pivot, left points to the | |

* first slot after them -- that's why this sort is stable. | |

* Slide elements over to make room to make room for pivot. | |

*/ | |

int n = start - left; // The number of elements to move | |

// Switch is just an optimization for arraycopy in default case | |

switch(n) { | |

case 2: a[left + 2] = a[left + 1]; | |

case 1: a[left + 1] = a[left]; | |

break; | |

default: System.arraycopy(a, left, a, left + 1, n); | |

} | |

a[left] = pivot; | |

} | |

} | |

/** | |

* Returns the length of the run beginning at the specified position in | |

* the specified array and reverses the run if it is descending (ensuring | |

* that the run will always be ascending when the method returns). | |

* | |

* A run is the longest ascending sequence with: | |

* | |

* a[lo] <= a[lo + 1] <= a[lo + 2] <= ... | |

* | |

* or the longest descending sequence with: | |

* | |

* a[lo] > a[lo + 1] > a[lo + 2] > ... | |

* | |

* For its intended use in a stable mergesort, the strictness of the | |

* definition of "descending" is needed so that the call can safely | |

* reverse a descending sequence without violating stability. | |

* | |

* @param a the array in which a run is to be counted and possibly reversed | |

* @param lo index of the first element in the run | |

* @param hi index after the last element that may be contained in the run. | |

It is required that @code{lo < hi}. | |

* @param c the comparator to used for the sort | |

* @return the length of the run beginning at the specified position in | |

* the specified array | |

*/ | |

private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi, | |

Comparator<? super T> c) { | |

if (DEBUG) assert lo < hi; | |

int runHi = lo + 1; | |

if (runHi == hi) | |

return 1; | |

// Find end of run, and reverse range if descending | |

if (c.compare(a[runHi++], a[lo]) < 0) { // Descending | |

while(runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0) | |

runHi++; | |

reverseRange(a, lo, runHi); | |

} else { // Ascending | |

while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0) | |

runHi++; | |

} | |

return runHi - lo; | |

} | |

/** | |

* Reverse the specified range of the specified array. | |

* | |

* @param a the array in which a range is to be reversed | |

* @param lo the index of the first element in the range to be reversed | |

* @param hi the index after the last element in the range to be reversed | |

*/ | |

private static void reverseRange(Object[] a, int lo, int hi) { | |

hi--; | |

while (lo < hi) { | |

Object t = a[lo]; | |

a[lo++] = a[hi]; | |

a[hi--] = t; | |

} | |

} | |

/** | |

* Returns the minimum acceptable run length for an array of the specified | |

* length. Natural runs shorter than this will be extended with | |

* {@link #binarySort}. | |

* | |

* Roughly speaking, the computation is: | |

* | |

* If n < MIN_MERGE, return n (it's too small to bother with fancy stuff). | |

* Else if n is an exact power of 2, return MIN_MERGE/2. | |

* Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k | |

* is close to, but strictly less than, an exact power of 2. | |

* | |

* For the rationale, see listsort.txt. | |

* | |

* @param n the length of the array to be sorted | |

* @return the length of the minimum run to be merged | |

*/ | |

private static int minRunLength(int n) { | |

if (DEBUG) assert n >= 0; | |

int r = 0; // Becomes 1 if any 1 bits are shifted off | |

while (n >= MIN_MERGE) { | |

r |= (n & 1); | |

n >>= 1; | |

} | |

return n + r; | |

} | |

/** | |

* Pushes the specified run onto the pending-run stack. | |

* | |

* @param runBase index of the first element in the run | |

* @param runLen the number of elements in the run | |

*/ | |

private void pushRun(int runBase, int runLen) { | |

this.runBase[stackSize] = runBase; | |

this.runLen[stackSize] = runLen; | |

stackSize++; | |

} | |

/** | |

* Examines the stack of runs waiting to be merged and merges adjacent runs | |

* until the stack invariants are reestablished: | |

* | |

* 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] | |

* 2. runLen[i - 2] > runLen[i - 1] | |

* | |

* This method is called each time a new run is pushed onto the stack, | |

* so the invariants are guaranteed to hold for i < stackSize upon | |

* entry to the method. | |

*/ | |

private void mergeCollapse() { | |

while (stackSize > 1) { | |

int n = stackSize - 2; | |

if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) { | |

if (runLen[n - 1] < runLen[n + 1]) | |

n--; | |

mergeAt(n); | |

} else if (runLen[n] <= runLen[n + 1]) { | |

mergeAt(n); | |

} else { | |

break; // Invariant is established | |

} | |

} | |

} | |

/** | |

* Merges all runs on the stack until only one remains. This method is | |

* called once, to complete the sort. | |

*/ | |

private void mergeForceCollapse() { | |

while (stackSize > 1) { | |

int n = stackSize - 2; | |

if (n > 0 && runLen[n - 1] < runLen[n + 1]) | |

n--; | |

mergeAt(n); | |

} | |

} | |

/** | |

* Merges the two runs at stack indices i and i+1. Run i must be | |

* the penultimate or antepenultimate run on the stack. In other words, | |

* i must be equal to stackSize-2 or stackSize-3. | |

* | |

* @param i stack index of the first of the two runs to merge | |

*/ | |

private void mergeAt(int i) { | |

if (DEBUG) assert stackSize >= 2; | |

if (DEBUG) assert i >= 0; | |

if (DEBUG) assert i == stackSize - 2 || i == stackSize - 3; | |

int base1 = runBase[i]; | |

int len1 = runLen[i]; | |

int base2 = runBase[i + 1]; | |

int len2 = runLen[i + 1]; | |

if (DEBUG) assert len1 > 0 && len2 > 0; | |

if (DEBUG) assert base1 + len1 == base2; | |

/* | |

* Record the length of the combined runs; if i is the 3rd-last | |

* run now, also slide over the last run (which isn't involved | |

* in this merge). The current run (i+1) goes away in any case. | |

*/ | |

runLen[i] = len1 + len2; | |

if (i == stackSize - 3) { | |

runBase[i + 1] = runBase[i + 2]; | |

runLen[i + 1] = runLen[i + 2]; | |

} | |

stackSize--; | |

/* | |

* Find where the first element of run2 goes in run1. Prior elements | |

* in run1 can be ignored (because they're already in place). | |

*/ | |

int k = gallopRight(a[base2], a, base1, len1, 0, c); | |

if (DEBUG) assert k >= 0; | |

base1 += k; | |

len1 -= k; | |

if (len1 == 0) | |

return; | |

/* | |

* Find where the last element of run1 goes in run2. Subsequent elements | |

* in run2 can be ignored (because they're already in place). | |

*/ | |

len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c); | |

if (DEBUG) assert len2 >= 0; | |

if (len2 == 0) | |

return; | |

// Merge remaining runs, using tmp array with min(len1, len2) elements | |

if (len1 <= len2) | |

mergeLo(base1, len1, base2, len2); | |

else | |

mergeHi(base1, len1, base2, len2); | |

} | |

/** | |

* Locates the position at which to insert the specified key into the | |

* specified sorted range; if the range contains an element equal to key, | |

* returns the index of the leftmost equal element. | |

* | |

* @param key the key whose insertion point to search for | |

* @param a the array in which to search | |

* @param base the index of the first element in the range | |

* @param len the length of the range; must be > 0 | |

* @param hint the index at which to begin the search, 0 <= hint < n. | |

* The closer hint is to the result, the faster this method will run. | |

* @param c the comparator used to order the range, and to search | |

* @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k], | |

* pretending that a[b - 1] is minus infinity and a[b + n] is infinity. | |

* In other words, key belongs at index b + k; or in other words, | |

* the first k elements of a should precede key, and the last n - k | |

* should follow it. | |

*/ | |

private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint, | |

Comparator<? super T> c) { | |

if (DEBUG) assert len > 0 && hint >= 0 && hint < len; | |

int lastOfs = 0; | |

int ofs = 1; | |

if (c.compare(key, a[base + hint]) > 0) { | |

// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs] | |

int maxOfs = len - hint; | |

while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) { | |

lastOfs = ofs; | |

ofs = (ofs << 1) + 1; | |

if (ofs <= 0) // int overflow | |

ofs = maxOfs; | |

} | |

if (ofs > maxOfs) | |

ofs = maxOfs; | |

// Make offsets relative to base | |

lastOfs += hint; | |

ofs += hint; | |

} else { // key <= a[base + hint] | |

// Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs] | |

final int maxOfs = hint + 1; | |

while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) { | |

lastOfs = ofs; | |

ofs = (ofs << 1) + 1; | |

if (ofs <= 0) // int overflow | |

ofs = maxOfs; | |

} | |

if (ofs > maxOfs) | |

ofs = maxOfs; | |

// Make offsets relative to base | |

int tmp = lastOfs; | |

lastOfs = hint - ofs; | |

ofs = hint - tmp; | |

} | |

if (DEBUG) assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; | |

/* | |

* Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere | |

* to the right of lastOfs but no farther right than ofs. Do a binary | |

* search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs]. | |

*/ | |

lastOfs++; | |

while (lastOfs < ofs) { | |

int m = lastOfs + ((ofs - lastOfs) >>> 1); | |

if (c.compare(key, a[base + m]) > 0) | |

lastOfs = m + 1; // a[base + m] < key | |

else | |

ofs = m; // key <= a[base + m] | |

} | |

if (DEBUG) assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs] | |

return ofs; | |

} | |

/** | |

* Like gallopLeft, except that if the range contains an element equal to | |

* key, gallopRight returns the index after the rightmost equal element. | |

* | |

* @param key the key whose insertion point to search for | |

* @param a the array in which to search | |

* @param base the index of the first element in the range | |

* @param len the length of the range; must be > 0 | |

* @param hint the index at which to begin the search, 0 <= hint < n. | |

* The closer hint is to the result, the faster this method will run. | |

* @param c the comparator used to order the range, and to search | |

* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k] | |

*/ | |

private static <T> int gallopRight(T key, T[] a, int base, int len, | |

int hint, Comparator<? super T> c) { | |

if (DEBUG) assert len > 0 && hint >= 0 && hint < len; | |

int ofs = 1; | |

int lastOfs = 0; | |

if (c.compare(key, a[base + hint]) < 0) { | |

// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs] | |

int maxOfs = hint + 1; | |

while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) { | |

lastOfs = ofs; | |

ofs = (ofs << 1) + 1; | |

if (ofs <= 0) // int overflow | |

ofs = maxOfs; | |

} | |

if (ofs > maxOfs) | |

ofs = maxOfs; | |

// Make offsets relative to b | |

int tmp = lastOfs; | |

lastOfs = hint - ofs; | |

ofs = hint - tmp; | |

} else { // a[b + hint] <= key | |

// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs] | |

int maxOfs = len - hint; | |

while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) { | |

lastOfs = ofs; | |

ofs = (ofs << 1) + 1; | |

if (ofs <= 0) // int overflow | |

ofs = maxOfs; | |

} | |

if (ofs > maxOfs) | |

ofs = maxOfs; | |

// Make offsets relative to b | |

lastOfs += hint; | |

ofs += hint; | |

} | |

if (DEBUG) assert -1 <= lastOfs && lastOfs < ofs && ofs <= len; | |

/* | |

* Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to | |

* the right of lastOfs but no farther right than ofs. Do a binary | |

* search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs]. | |

*/ | |

lastOfs++; | |

while (lastOfs < ofs) { | |

int m = lastOfs + ((ofs - lastOfs) >>> 1); | |

if (c.compare(key, a[base + m]) < 0) | |

ofs = m; // key < a[b + m] | |

else | |

lastOfs = m + 1; // a[b + m] <= key | |

} | |

if (DEBUG) assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs] | |

return ofs; | |

} | |

/** | |

* Merges two adjacent runs in place, in a stable fashion. The first | |

* element of the first run must be greater than the first element of the | |

* second run (a[base1] > a[base2]), and the last element of the first run | |

* (a[base1 + len1-1]) must be greater than all elements of the second run. | |

* | |

* For performance, this method should be called only when len1 <= len2; | |

* its twin, mergeHi should be called if len1 >= len2. (Either method | |

* may be called if len1 == len2.) | |

* | |

* @param base1 index of first element in first run to be merged | |

* @param len1 length of first run to be merged (must be > 0) | |

* @param base2 index of first element in second run to be merged | |

* (must be aBase + aLen) | |

* @param len2 length of second run to be merged (must be > 0) | |

*/ | |

private void mergeLo(int base1, int len1, int base2, int len2) { | |

if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2; | |

// Copy first run into temp array | |

T[] a = this.a; // For performance | |

T[] tmp = ensureCapacity(len1); | |

System.arraycopy(a, base1, tmp, 0, len1); | |

int cursor1 = 0; // Indexes into tmp array | |

int cursor2 = base2; // Indexes int a | |

int dest = base1; // Indexes int a | |

// Move first element of second run and deal with degenerate cases | |

a[dest++] = a[cursor2++]; | |

if (--len2 == 0) { | |

System.arraycopy(tmp, cursor1, a, dest, len1); | |

return; | |

} | |

if (len1 == 1) { | |

System.arraycopy(a, cursor2, a, dest, len2); | |

a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge | |

return; | |

} | |

Comparator<? super T> c = this.c; // Use local variable for performance | |

int minGallop = this.minGallop; // " " " " " | |

outer: | |

while (true) { | |

int count1 = 0; // Number of times in a row that first run won | |

int count2 = 0; // Number of times in a row that second run won | |

/* | |

* Do the straightforward thing until (if ever) one run starts | |

* winning consistently. | |

*/ | |

do { | |

if (DEBUG) assert len1 > 1 && len2 > 0; | |

if (c.compare(a[cursor2], tmp[cursor1]) < 0) { | |

a[dest++] = a[cursor2++]; | |

count2++; | |

count1 = 0; | |

if (--len2 == 0) | |

break outer; | |

} else { | |

a[dest++] = tmp[cursor1++]; | |

count1++; | |

count2 = 0; | |

if (--len1 == 1) | |

break outer; | |

} | |

} while ((count1 | count2) < minGallop); | |

/* | |

* One run is winning so consistently that galloping may be a | |

* huge win. So try that, and continue galloping until (if ever) | |

* neither run appears to be winning consistently anymore. | |

*/ | |

do { | |

if (DEBUG) assert len1 > 1 && len2 > 0; | |

count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c); | |

if (count1 != 0) { | |

System.arraycopy(tmp, cursor1, a, dest, count1); | |

dest += count1; | |

cursor1 += count1; | |

len1 -= count1; | |

if (len1 <= 1) // len1 == 1 || len1 == 0 | |

break outer; | |

} | |

a[dest++] = a[cursor2++]; | |

if (--len2 == 0) | |

break outer; | |

count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c); | |

if (count2 != 0) { | |

System.arraycopy(a, cursor2, a, dest, count2); | |

dest += count2; | |

cursor2 += count2; | |

len2 -= count2; | |

if (len2 == 0) | |

break outer; | |

} | |

a[dest++] = tmp[cursor1++]; | |

if (--len1 == 1) | |

break outer; | |

minGallop--; | |

} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); | |

if (minGallop < 0) | |

minGallop = 0; | |

minGallop += 2; // Penalize for leaving gallop mode | |

} // End of "outer" loop | |

this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field | |

if (len1 == 1) { | |

if (DEBUG) assert len2 > 0; | |

System.arraycopy(a, cursor2, a, dest, len2); | |

a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge | |

} else if (len1 == 0) { | |

throw new IllegalArgumentException( | |

"Comparison method violates its general contract!"); | |

} else { | |

if (DEBUG) assert len2 == 0; | |

if (DEBUG) assert len1 > 1; | |

System.arraycopy(tmp, cursor1, a, dest, len1); | |

} | |

} | |

/** | |

* Like mergeLo, except that this method should be called only if | |

* len1 >= len2; mergeLo should be called if len1 <= len2. (Either method | |

* may be called if len1 == len2.) | |

* | |

* @param base1 index of first element in first run to be merged | |

* @param len1 length of first run to be merged (must be > 0) | |

* @param base2 index of first element in second run to be merged | |

* (must be aBase + aLen) | |

* @param len2 length of second run to be merged (must be > 0) | |

*/ | |

private void mergeHi(int base1, int len1, int base2, int len2) { | |

if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2; | |

// Copy second run into temp array | |

T[] a = this.a; // For performance | |

T[] tmp = ensureCapacity(len2); | |

System.arraycopy(a, base2, tmp, 0, len2); | |

int cursor1 = base1 + len1 - 1; // Indexes into a | |

int cursor2 = len2 - 1; // Indexes into tmp array | |

int dest = base2 + len2 - 1; // Indexes into a | |

// Move last element of first run and deal with degenerate cases | |

a[dest--] = a[cursor1--]; | |

if (--len1 == 0) { | |

System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2); | |

return; | |

} | |

if (len2 == 1) { | |

dest -= len1; | |

cursor1 -= len1; | |

System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); | |

a[dest] = tmp[cursor2]; | |

return; | |

} | |

Comparator<? super T> c = this.c; // Use local variable for performance | |

int minGallop = this.minGallop; // " " " " " | |

outer: | |

while (true) { | |

int count1 = 0; // Number of times in a row that first run won | |

int count2 = 0; // Number of times in a row that second run won | |

/* | |

* Do the straightforward thing until (if ever) one run | |

* appears to win consistently. | |

*/ | |

do { | |

if (DEBUG) assert len1 > 0 && len2 > 1; | |

if (c.compare(tmp[cursor2], a[cursor1]) < 0) { | |

a[dest--] = a[cursor1--]; | |

count1++; | |

count2 = 0; | |

if (--len1 == 0) | |

break outer; | |

} else { | |

a[dest--] = tmp[cursor2--]; | |

count2++; | |

count1 = 0; | |

if (--len2 == 1) | |

break outer; | |

} | |

} while ((count1 | count2) < minGallop); | |

/* | |

* One run is winning so consistently that galloping may be a | |

* huge win. So try that, and continue galloping until (if ever) | |

* neither run appears to be winning consistently anymore. | |

*/ | |

do { | |

if (DEBUG) assert len1 > 0 && len2 > 1; | |

count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c); | |

if (count1 != 0) { | |

dest -= count1; | |

cursor1 -= count1; | |

len1 -= count1; | |

System.arraycopy(a, cursor1 + 1, a, dest + 1, count1); | |

if (len1 == 0) | |

break outer; | |

} | |

a[dest--] = tmp[cursor2--]; | |

if (--len2 == 1) | |

break outer; | |

count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c); | |

if (count2 != 0) { | |

dest -= count2; | |

cursor2 -= count2; | |

len2 -= count2; | |

System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2); | |

if (len2 <= 1) // len2 == 1 || len2 == 0 | |

break outer; | |

} | |

a[dest--] = a[cursor1--]; | |

if (--len1 == 0) | |

break outer; | |

minGallop--; | |

} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP); | |

if (minGallop < 0) | |

minGallop = 0; | |

minGallop += 2; // Penalize for leaving gallop mode | |

} // End of "outer" loop | |

this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field | |

if (len2 == 1) { | |

if (DEBUG) assert len1 > 0; | |

dest -= len1; | |

cursor1 -= len1; | |

System.arraycopy(a, cursor1 + 1, a, dest + 1, len1); | |

a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge | |

} else if (len2 == 0) { | |

throw new IllegalArgumentException( | |

"Comparison method violates its general contract!"); | |

} else { | |

if (DEBUG) assert len1 == 0; | |

if (DEBUG) assert len2 > 0; | |

System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2); | |

} | |

} | |

/** | |

* Ensures that the external array tmp has at least the specified | |

* number of elements, increasing its size if necessary. The size | |

* increases exponentially to ensure amortized linear time complexity. | |

* | |

* @param minCapacity the minimum required capacity of the tmp array | |

* @return tmp, whether or not it grew | |

*/ | |

private T[] ensureCapacity(int minCapacity) { | |

if (tmp.length < minCapacity) { | |

// Compute smallest power of 2 > minCapacity | |

int newSize = minCapacity; | |

newSize |= newSize >> 1; | |

newSize |= newSize >> 2; | |

newSize |= newSize >> 4; | |

newSize |= newSize >> 8; | |

newSize |= newSize >> 16; | |

newSize++; | |

if (newSize < 0) // Not bloody likely! | |

newSize = minCapacity; | |

else | |

newSize = Math.min(newSize, a.length >>> 1); | |

@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"}) | |

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); | |

} | |

} |