## TimSort

Some of you many know I have a bit of an obesstion with sorting algorithms ^_^

Anyway their is a revolutionary algorithm come to greater light, developed in 2002 by Tim Peters and called the TimSort ðŸ˜› which is a adaptive, stable, natural mergesort. It has supernatural performance on many kinds of partially ordered arrays (less than lg(N!) comparisons needed, andas few as N-1), yet as fast as Python’s previous highly tuned samplesort hybrid on random arrays.

and for us Java nerds it will be replacing java collections sort() as of Java 7

Anyway its sexy as hell check it out:

/*

* 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. Sun designates this

* particular file as subject to the “Classpath” exception as provided

* by Sun 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 Sun Microsystems, Inc., 4150 Network Circle, Santa Clara,

* CA 95054 USA or visit www.sun.com if you need additional information or

* have any questions.

*/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.

*

* @author Josh Bloch

*/

class TimSort {

/**

* 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 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;/**

* 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 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 void sort(T[] a, Comparator c) {

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

}static void sort(T[] a, int lo, int hi, Comparator 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 ts = new TimSort(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;

}/**

* 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 void binarySort(T[] a, int lo, int hi, int start,

Comparator 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 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 int countRunAndMakeAscending(T[] a, int lo, int hi,

Comparator 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;

}/**

* 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) { 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) { 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); } /** * 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 int gallopLeft(T key, T[] a, int base, int len, int hint,

Comparator 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) // 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;

}

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 int gallopRight(T key, T[] a, int base, int len,

int hint, Comparator 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) // 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;

}

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; } /** * 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 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) // 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) { 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);

}

}/**

* 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 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) // 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) { 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);

}

}/**

* 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);

}

}