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class TimSort<T> {

class Sorter<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

* {@link #minRunLength} computation.

* 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;

* 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) {

public Sorter(SortDataFormat<K, Buffer> sortDataFormat) {

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

this.s = sortDataFormat;

}

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

void sort(Buffer a, int lo, int hi, Comparator<? super K> c) {

if (c == null) {

assert 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);

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

ts.pushRun(lo, runLen);

sortState.pushRun(lo, runLen);

ts.mergeCollapse();

sortState.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();

sortState.mergeForceCollapse();

assert ts.stackSize == 1;

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

private void binarySort(Buffer a, int lo, int hi, int start, Comparator<? super K> c) {

Comparator<? super T> c) {

assert lo <= start && start <= hi;

if (start == lo)

start++;

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

T pivot = a[start];

Buffer pivotStore = s.allocate(1);

s.copyElement(a, start, pivotStore, 0);

K pivot = s.getKey(pivotStore, 0);

// 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)

if (c.compare(pivot, s.getKey(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 2: s.copyElement(a, left + 1, a, left + 2);

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

case 1: s.copyElement(a, left, a, left + 1);

break;

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

default: s.copyRange(a, left, a, left + 1, n);

}

a[left] = pivot;

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

private int countRunAndMakeAscending(Buffer a, int lo, int hi, Comparator<? super K> c) {

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

if (c.compare(s.getKey(a, runHi++), s.getKey(a, lo)) < 0) { // Descending

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

while (runHi < hi && c.compare(s.getKey(a, runHi), s.getKey(a, runHi - 1)) < 0)

runHi++;

reverseRange(a, lo, runHi);

} else { // Ascending

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

while (runHi < hi && c.compare(s.getKey(a, runHi), s.getKey(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) {

private void reverseRange(Buffer a, int lo, int hi) {

hi--;

while (lo < hi) {

Object t = a[lo];

s.swap(a, lo, hi);

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

lo++;

a[hi--] = t;

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 static int minRunLength(int n) {

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 array being sorted.

* The Buffer being sorted.

private final T[] a;

private final Buffer a;

/**

* Length of the sort Buffer.

private final int aLength;

/**

* The comparator for this sort.

private final Comparator<? super T> c;

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

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 TimSort(T[] a, Comparator<? super T> c) {

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)

int len = a.length;

tmpLength = len < 2 * INITIAL_TMP_STORAGE_LENGTH ? len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;

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

tmp = s.allocate(tmpLength);

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

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

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

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

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

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

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

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

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

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

* the MIN_MERGE declaration above for more information.

* 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];

}

/**

* 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);

int k = gallopRight(s.getKey(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);

len2 = gallopLeft(s.getKey(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 <T> int gallopLeft(T key, T[] a, int base, int len, int hint,

private int gallopLeft(K key, Buffer a, int base, int len, int hint, Comparator<? super K> c) {

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) {

if (c.compare(key, s.getKey(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) {

while (ofs < maxOfs && c.compare(key, s.getKey(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) {

while (ofs < maxOfs && c.compare(key, s.getKey(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)

if (c.compare(key, s.getKey(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,

private int gallopRight(K key, Buffer a, int base, int len, int hint, Comparator<? super K> c) {

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) {

if (c.compare(key, s.getKey(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) {

while (ofs < maxOfs && c.compare(key, s.getKey(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) {

while (ofs < maxOfs && c.compare(key, s.getKey(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)

if (c.compare(key, s.getKey(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

Buffer a = this.a; // For performance

T[] tmp = ensureCapacity(len1);

Buffer tmp = ensureCapacity(len1);

System.arraycopy(a, base1, tmp, 0, 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

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

s.copyElement(a, cursor2++, a, dest++);

if (--len2 == 0) {

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

s.copyRange(tmp, cursor1, a, dest, len1);

return;

}

if (len1 == 1) {

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

s.copyRange(a, cursor2, a, dest, len2);

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

s.copyElement(tmp, cursor1, a, dest + len2); // Last elt of run 1 to end of merge

return;

}

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

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(a[cursor2], tmp[cursor1]) < 0) {

if (c.compare(s.getKey(a, cursor2), s.getKey(tmp, cursor1)) < 0) {

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

s.copyElement(a, cursor2++, a, dest++);

count2++;

count1 = 0;

if (--len2 == 0)

break outer;

} else {

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

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(a[cursor2], tmp, cursor1, len1, 0, c);

count1 = gallopRight(s.getKey(a, cursor2), tmp, cursor1, len1, 0, c);

if (count1 != 0) {

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

s.copyRange(tmp, cursor1, a, dest, count1);

dest += count1;

cursor1 += count1;

len1 -= count1;

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

break outer;

}

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

s.copyElement(a, cursor2++, a, dest++);

if (--len2 == 0)

break outer;

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

count2 = gallopLeft(s.getKey(tmp, cursor1), a, cursor2, len2, 0, c);

if (count2 != 0) {

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

s.copyRange(a, cursor2, a, dest, count2);

dest += count2;

cursor2 += count2;

len2 -= count2;

if (len2 == 0)

break outer;

}

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

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

सेव किए गए Diffs

ऑरिजनल टेक्स्ट

फ़ाइल खोलें

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 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
    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 <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;
  }

/**
   * 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;
  }

/**
     * 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;
  
    /**
     * 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];
    }
  
    /**
     * 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 <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)   // 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 <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)   // 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<? 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) // 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<? 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)  // 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;
    }
}

परिवर्तित टेक्स्ट

फ़ाइल खोलें

class Sorter<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 Sorter(SortDataFormat<K, Buffer> sortDataFormat) {
    this.s = sortDataFormat;
  }

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

/**
     * 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);

// 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++;
    for ( ; start < hi; start++) {
      Buffer pivotStore = s.allocate(1);
      s.copyElement(a, start, pivotStore, 0);
      K pivot = s.getKey(pivotStore, 0);

// 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)) < 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;

// Find end of run, and reverse range if descending
    if (c.compare(s.getKey(a, runHi++), s.getKey(a, lo)) < 0) { // Descending
      while (runHi < hi && c.compare(s.getKey(a, runHi), s.getKey(a, runHi - 1)) < 0)
        runHi++;
      reverseRange(a, lo, runHi);
    } else {                              // Ascending
      while (runHi < hi && c.compare(s.getKey(a, runHi), s.getKey(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 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.
       */
      int stackLen = (len <    120  ?  5 :
                      len <   1542  ? 10 :
                      len < 119151  ? 19 : 40);
      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.
     */
    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(s.getKey(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(s.getKey(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 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;
      if (c.compare(key, s.getKey(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, s.getKey(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, s.getKey(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, s.getKey(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 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;
      if (c.compare(key, s.getKey(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, s.getKey(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, s.getKey(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, s.getKey(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
      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;
      }

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), s.getKey(tmp, cursor1)) < 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), 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), 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

// 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;
      }

/*
         * 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), s.getKey(a, cursor1)) < 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), 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), 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);
      }
      return tmp;
    }
  }
}