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class TimSort<T> {
class Sorter<K, Buffer> {
/**
/**
* This is the minimum sized sequence that will be merged. Shorter
* This is the minimum sized sequence that will be merged. Shorter
* sequences will be lengthened by calling binarySort. If the entire
* sequences will be lengthened by calling binarySort. If the entire
* array is less than this length, no merges will be performed.
* 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
* 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
* implementation, but 32 was empirically determined to work better in
* this implementation. In the unlikely event that you set this constant
* 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
* 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
* If you decrease this constant, you must change the stackLen
* computation in the TimSort constructor, or you risk an
* computation in the TimSort constructor, or you risk an
* ArrayOutOfBounds exception. See listsort.txt for a discussion
* ArrayOutOfBounds exception. See listsort.txt for a discussion
* of the minimum stack length required as a function of the length
* of the minimum stack length required as a function of the length
* of the array being sorted and the minimum merge sequence length.
* of the array being sorted and the minimum merge sequence length.
*/
*/
private static final int MIN_MERGE = 32;
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;
int nRemaining = hi - lo;
if (nRemaining < 2)
if (nRemaining < 2)
return; // Arrays of size 0 and 1 are always sorted
return; // Arrays of size 0 and 1 are always sorted
// If array is small, do a "mini-TimSort" with no merges
// If array is small, do a "mini-TimSort" with no merges
if (nRemaining < MIN_MERGE) {
if (nRemaining < MIN_MERGE) {
int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
binarySort(a, lo, hi, lo + initRunLen, c);
binarySort(a, lo, hi, lo + initRunLen, c);
return;
return;
}
}
/**
/**
* March over the array once, left to right, finding natural runs,
* March over the array once, left to right, finding natural runs,
* extending short natural runs to minRun elements, and merging runs
* extending short natural runs to minRun elements, and merging runs
* to maintain stack invariant.
* to maintain stack invariant.
*/
*/
TimSort<T> ts = new TimSort<T>(a, c);
SortState sortState = new SortState(a, c, hi - lo);
int minRun = minRunLength(nRemaining);
int minRun = minRunLength(nRemaining);
do {
do {
// Identify next run
// Identify next run
int runLen = countRunAndMakeAscending(a, lo, hi, c);
int runLen = countRunAndMakeAscending(a, lo, hi, c);
// If run is short, extend to min(minRun, nRemaining)
// If run is short, extend to min(minRun, nRemaining)
if (runLen < minRun) {
if (runLen < minRun) {
int force = nRemaining <= minRun ? nRemaining : minRun;
int force = nRemaining <= minRun ? nRemaining : minRun;
binarySort(a, lo, lo + force, lo + runLen, c);
binarySort(a, lo, lo + force, lo + runLen, c);
runLen = force;
runLen = force;
}
}
// Push run onto pending-run stack, and maybe merge
// 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
// Advance to find next run
lo += runLen;
lo += runLen;
nRemaining -= runLen;
nRemaining -= runLen;
} while (nRemaining != 0);
} while (nRemaining != 0);
// Merge all remaining runs to complete sort
// Merge all remaining runs to complete sort
assert lo == hi;
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
* Sorts the specified portion of the specified array using a binary
* insertion sort. This is the best method for sorting small numbers
* 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
* of elements. It requires O(n log n) compares, but O(n^2) data
* movement (worst case).
* movement (worst case).
*
*
* If the initial part of the specified range is already sorted,
* If the initial part of the specified range is already sorted,
* this method can take advantage of it: the method assumes that the
* this method can take advantage of it: the method assumes that the
* elements from index {@code lo}, inclusive, to {@code start},
* elements from index {@code lo}, inclusive, to {@code start},
* exclusive are already sorted.
* exclusive are already sorted.
*
*
* @param a the array in which a range is to be 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 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 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
* @param start the index of the first element in the range that is
* not already known to be sorted ({@code lo <= start <= hi})
* not already known to be sorted ({@code lo <= start <= hi})
* @param c comparator to used for the sort
* @param c comparator to used for the sort
*/
*/
@SuppressWarnings("fallthrough")
@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;
assert lo <= start && start <= hi;
if (start == lo)
if (start == lo)
start++;
start++;
for ( ; start < hi; 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
// Set left (and right) to the index where a[start] (pivot) belongs
int left = lo;
int left = lo;
int right = start;
int right = start;
assert left <= right;
assert left <= right;
/*
/*
* Invariants:
* Invariants:
* pivot >= all in [lo, left).
* pivot >= all in [lo, left).
* pivot < all in [right, start).
* pivot < all in [right, start).
*/
*/
while (left < right) {
while (left < right) {
int mid = (left + right) >>> 1;
int mid = (left + right) >>> 1;
if (c.compare(pivot, a[mid]) < 0)
if (c.compare(pivot, s.getKey(a, mid)) < 0)
right = mid;
right = mid;
else
else
left = mid + 1;
left = mid + 1;
}
}
assert left == right;
assert left == right;
/*
/*
* The invariants still hold: pivot >= all in [lo, left) and
* The invariants still hold: pivot >= all in [lo, left) and
* pivot < all in [left, start), so pivot belongs at left. Note
* pivot < all in [left, start), so pivot belongs at left. Note
* that if there are elements equal to pivot, left points to the
* that if there are elements equal to pivot, left points to the
* first slot after them -- that's why this sort is stable.
* first slot after them -- that's why this sort is stable.
* Slide elements over to make room for pivot.
* Slide elements over to make room for pivot.
*/
*/
int n = start - left; // The number of elements to move
int n = start - left; // The number of elements to move
// Switch is just an optimization for arraycopy in default case
// Switch is just an optimization for arraycopy in default case
switch (n) {
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;
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
* Returns the length of the run beginning at the specified position in
* the specified array and reverses the run if it is descending (ensuring
* the specified array and reverses the run if it is descending (ensuring
* that the run will always be ascending when the method returns).
* that the run will always be ascending when the method returns).
*
*
* A run is the longest ascending sequence with:
* A run is the longest ascending sequence with:
*
*
* a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
* a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
*
*
* or the longest descending sequence with:
* or the longest descending sequence with:
*
*
* a[lo] > a[lo + 1] > a[lo + 2] > ...
* a[lo] > a[lo + 1] > a[lo + 2] > ...
*
*
* For its intended use in a stable mergesort, the strictness of the
* For its intended use in a stable mergesort, the strictness of the
* definition of "descending" is needed so that the call can safely
* definition of "descending" is needed so that the call can safely
* reverse a descending sequence without violating stability.
* reverse a descending sequence without violating stability.
*
*
* @param a the array in which a run is to be counted and possibly reversed
* @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 lo index of the first element in the run
* @param hi index after the last element that may be contained in the run.
* @param hi index after the last element that may be contained in the run.
It is required that {@code lo < hi}.
It is required that {@code lo < hi}.
* @param c the comparator to used for the sort
* @param c the comparator to used for the sort
* @return the length of the run beginning at the specified position in
* @return the length of the run beginning at the specified position in
* the specified array
* 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;
assert lo < hi;
int runHi = lo + 1;
int runHi = lo + 1;
if (runHi == hi)
if (runHi == hi)
return 1;
return 1;
// Find end of run, and reverse range if descending
// 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++;
runHi++;
reverseRange(a, lo, runHi);
reverseRange(a, lo, runHi);
} else { // Ascending
} 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++;
runHi++;
}
}
return runHi - lo;
return runHi - lo;
}
}
/**
/**
* Reverse the specified range of the specified array.
* Reverse the specified range of the specified array.
*
*
* @param a the array in which a range is to be reversed
* @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 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
* @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--;
hi--;
while (lo < 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
* Returns the minimum acceptable run length for an array of the specified
* length. Natural runs shorter than this will be extended with
* length. Natural runs shorter than this will be extended with
* {@link #binarySort}.
* {@link #binarySort}.
*
*
* Roughly speaking, the computation is:
* Roughly speaking, the computation is:
*
*
* If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
* 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 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
* 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.
* is close to, but strictly less than, an exact power of 2.
*
*
* For the rationale, see listsort.txt.
* For the rationale, see listsort.txt.
*
*
* @param n the length of the array to be sorted
* @param n the length of the array to be sorted
* @return the length of the minimum run to be merged
* @return the length of the minimum run to be merged
*/
*/
private static int minRunLength(int n) {
private int minRunLength(int n) {
assert n >= 0;
assert n >= 0;
int r = 0; // Becomes 1 if any 1 bits are shifted off
int r = 0; // Becomes 1 if any 1 bits are shifted off
while (n >= MIN_MERGE) {
while (n >= MIN_MERGE) {
r |= (n & 1);
r |= (n & 1);
n >>= 1;
n >>= 1;
}
}
return n + r;
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.
* 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
* When we get into galloping mode, we stay there until both runs win less
* often than MIN_GALLOP consecutive times.
* often than MIN_GALLOP consecutive times.
*/
*/
private static final int MIN_GALLOP = 7;
private static final int MIN_GALLOP = 7;
/**
/**
* This controls when we get *into* galloping mode. It is initialized
* This controls when we get *into* galloping mode. It is initialized
* to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
* to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher for
* random data, and lower for highly structured data.
* random data, and lower for highly structured data.
*/
*/
private int minGallop = MIN_GALLOP;
private int minGallop = MIN_GALLOP;
/**
/**
* Maximum initial size of tmp array, which is used for merging. The array
* Maximum initial size of tmp array, which is used for merging. The array
* can grow to accommodate demand.
* can grow to accommodate demand.
*
*
* Unlike Tim's original C version, we do not allocate this much storage
* Unlike Tim's original C version, we do not allocate this much storage
* when sorting smaller arrays. This change was required for performance.
* when sorting smaller arrays. This change was required for performance.
*/
*/
private static final int INITIAL_TMP_STORAGE_LENGTH = 256;
private static final int INITIAL_TMP_STORAGE_LENGTH = 256;
/**
/**
* Temp storage for merges.
* 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
* 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
* address base[i] and extends for len[i] elements. It's always
* true (so long as the indices are in bounds) that:
* true (so long as the indices are in bounds) that:
*
*
* runBase[i] + runLen[i] == runBase[i + 1]
* runBase[i] + runLen[i] == runBase[i + 1]
*
*
* so we could cut the storage for this, but it's a minor amount,
* so we could cut the storage for this, but it's a minor amount,
* and keeping all the info explicit simplifies the code.
* and keeping all the info explicit simplifies the code.
*/
*/
private int stackSize = 0; // Number of pending runs on stack
private int stackSize = 0; // Number of pending runs on stack
private final int[] runBase;
private final int[] runBase;
private final int[] runLen;
private final int[] runLen;
/**
/**
* Creates a TimSort instance to maintain the state of an ongoing sort.
* Creates a TimSort instance to maintain the state of an ongoing sort.
*
*
* @param a the array to be sorted
* @param a the array to be sorted
* @param c the comparator to determine the order of the sort
* @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.a = a;
this.c = c;
this.c = c;
// Allocate temp storage (which may be increased later if necessary)
// 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 :
int stackLen = (len < 120 ? 5 :
len < 1542 ? 10 :
len < 1542 ? 10 :
len < 119151 ? 19 : 40);
len < 119151 ? 19 : 40);
runBase = new int[stackLen];
runBase = new int[stackLen];
runLen = new int[stackLen];
runLen = new int[stackLen];
}
}
/**
/**
* Pushes the specified run onto the pending-run stack.
* Pushes the specified run onto the pending-run stack.
*
*
* @param runBase index of the first element in the run
* @param runBase index of the first element in the run
* @param runLen the number of elements in the run
* @param runLen the number of elements in the run
*/
*/
private void pushRun(int runBase, int runLen) {
private void pushRun(int runBase, int runLen) {
this.runBase[stackSize] = runBase;
this.runBase[stackSize] = runBase;
this.runLen[stackSize] = runLen;
this.runLen[stackSize] = runLen;
stackSize++;
stackSize++;
}
}
/**
/**
* Examines the stack of runs waiting to be merged and merges adjacent runs
* Examines the stack of runs waiting to be merged and merges adjacent runs
* until the stack invariants are reestablished:
* until the stack invariants are reestablished:
*
*
* 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
* 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
* 2. 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,
* 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
* so the invariants are guaranteed to hold for i < stackSize upon
* entry to the method.
* entry to the method.
*/
*/
private void mergeCollapse() {
private void mergeCollapse() {
while (stackSize > 1) {
while (stackSize > 1) {
int n = stackSize - 2;
int n = stackSize - 2;
if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
if (runLen[n - 1] < runLen[n + 1])
if (runLen[n - 1] < runLen[n + 1])
n--;
n--;
mergeAt(n);
mergeAt(n);
} else if (runLen[n] <= runLen[n + 1]) {
} else if (runLen[n] <= runLen[n + 1]) {
mergeAt(n);
mergeAt(n);
} else {
} else {
break; // Invariant is established
break; // Invariant is established
}
}
}
}
}
}
/**
/**
* Merges all runs on the stack until only one remains. This method is
* Merges all runs on the stack until only one remains. This method is
* called once, to complete the sort.
* called once, to complete the sort.
*/
*/
private void mergeForceCollapse() {
private void mergeForceCollapse() {
while (stackSize > 1) {
while (stackSize > 1) {
int n = stackSize - 2;
int n = stackSize - 2;
if (n > 0 && runLen[n - 1] < runLen[n + 1])
if (n > 0 && runLen[n - 1] < runLen[n + 1])
n--;
n--;
mergeAt(n);
mergeAt(n);
}
}
}
}
/**
/**
* Merges the two runs at stack indices i and i+1. Run i must be
* 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,
* the penultimate or antepenultimate run on the stack. In other words,
* i must be equal to stackSize-2 or stackSize-3.
* i must be equal to stackSize-2 or stackSize-3.
*
*
* @param i stack index of the first of the two runs to merge
* @param i stack index of the first of the two runs to merge
*/
*/
private void mergeAt(int i) {
private void mergeAt(int i) {
assert stackSize >= 2;
assert stackSize >= 2;
assert i >= 0;
assert i >= 0;
assert i == stackSize - 2 || i == stackSize - 3;
assert i == stackSize - 2 || i == stackSize - 3;
int base1 = runBase[i];
int base1 = runBase[i];
int len1 = runLen[i];
int len1 = runLen[i];
int base2 = runBase[i + 1];
int base2 = runBase[i + 1];
int len2 = runLen[i + 1];
int len2 = runLen[i + 1];
assert len1 > 0 && len2 > 0;
assert len1 > 0 && len2 > 0;
assert base1 + len1 == base2;
assert base1 + len1 == base2;
/*
/*
* Record the length of the combined runs; if i is the 3rd-last
* 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
* 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.
* in this merge). The current run (i+1) goes away in any case.
*/
*/
runLen[i] = len1 + len2;
runLen[i] = len1 + len2;
if (i == stackSize - 3) {
if (i == stackSize - 3) {
runBase[i + 1] = runBase[i + 2];
runBase[i + 1] = runBase[i + 2];
runLen[i + 1] = runLen[i + 2];
runLen[i + 1] = runLen[i + 2];
}
}
stackSize--;
stackSize--;
/*
/*
* Find where the first element of run2 goes in run1. Prior elements
* Find where the first element of run2 goes in run1. Prior elements
* in run1 can be ignored (because they're already in place).
* 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;
assert k >= 0;
base1 += k;
base1 += k;
len1 -= k;
len1 -= k;
if (len1 == 0)
if (len1 == 0)
return;
return;
/*
/*
* Find where the last element of run1 goes in run2. Subsequent elements
* Find where the last element of run1 goes in run2. Subsequent elements
* in run2 can be ignored (because they're already in place).
* 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;
assert len2 >= 0;
if (len2 == 0)
if (len2 == 0)
return;
return;
// Merge remaining runs, using tmp array with min(len1, len2) elements
// Merge remaining runs, using tmp array with min(len1, len2) elements
if (len1 <= len2)
if (len1 <= len2)
mergeLo(base1, len1, base2, len2);
mergeLo(base1, len1, base2, len2);
else
else
mergeHi(base1, len1, base2, len2);
mergeHi(base1, len1, base2, len2);
}
}
/**
/**
* Locates the position at which to insert the specified key into the
* Locates the position at which to insert the specified key into the
* specified sorted range; if the range contains an element equal to key,
* specified sorted range; if the range contains an element equal to key,
* returns the index of the leftmost equal element.
* returns the index of the leftmost equal element.
*
*
* @param key the key whose insertion point to search for
* @param key the key whose insertion point to search for
* @param a the array in which to search
* @param a the array in which to search
* @param base the index of the first element in the range
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint < n.
* @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.
* 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
* @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],
* @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.
* 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,
* 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
* the first k elements of a should precede key, and the last n - k
* should follow it.
* 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;
assert len > 0 && hint >= 0 && hint < len;
int lastOfs = 0;
int lastOfs = 0;
int ofs = 1;
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]
// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
int maxOfs = len - hint;
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;
lastOfs = ofs;
ofs = (ofs << 1) + 1;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
if (ofs <= 0) // int overflow
ofs = maxOfs;
ofs = maxOfs;
}
}
if (ofs > maxOfs)
if (ofs > maxOfs)
ofs = maxOfs;
ofs = maxOfs;
// Make offsets relative to base
// Make offsets relative to base
lastOfs += hint;
lastOfs += hint;
ofs += hint;
ofs += hint;
} else { // key <= a[base + hint]
} else { // key <= a[base + hint]
// Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
// Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
final int maxOfs = hint + 1;
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;
lastOfs = ofs;
ofs = (ofs << 1) + 1;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
if (ofs <= 0) // int overflow
ofs = maxOfs;
ofs = maxOfs;
}
}
if (ofs > maxOfs)
if (ofs > maxOfs)
ofs = maxOfs;
ofs = maxOfs;
// Make offsets relative to base
// Make offsets relative to base
int tmp = lastOfs;
int tmp = lastOfs;
lastOfs = hint - ofs;
lastOfs = hint - ofs;
ofs = hint - tmp;
ofs = hint - tmp;
}
}
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
/*
/*
* Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
* 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
* 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].
* search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
*/
*/
lastOfs++;
lastOfs++;
while (lastOfs < ofs) {
while (lastOfs < ofs) {
int m = lastOfs + ((ofs - lastOfs) >>> 1);
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
lastOfs = m + 1; // a[base + m] < key
else
else
ofs = m; // key <= a[base + m]
ofs = m; // key <= a[base + m]
}
}
assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
return ofs;
return ofs;
}
}
/**
/**
* Like gallopLeft, except that if the range contains an element equal to
* Like gallopLeft, except that if the range contains an element equal to
* key, gallopRight returns the index after the rightmost equal element.
* key, gallopRight returns the index after the rightmost equal element.
*
*
* @param key the key whose insertion point to search for
* @param key the key whose insertion point to search for
* @param a the array in which to search
* @param a the array in which to search
* @param base the index of the first element in the range
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint < n.
* @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.
* 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
* @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]
* @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;
assert len > 0 && hint >= 0 && hint < len;
int ofs = 1;
int ofs = 1;
int lastOfs = 0;
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]
// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
int maxOfs = hint + 1;
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;
lastOfs = ofs;
ofs = (ofs << 1) + 1;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
if (ofs <= 0) // int overflow
ofs = maxOfs;
ofs = maxOfs;
}
}
if (ofs > maxOfs)
if (ofs > maxOfs)
ofs = maxOfs;
ofs = maxOfs;
// Make offsets relative to b
// Make offsets relative to b
int tmp = lastOfs;
int tmp = lastOfs;
lastOfs = hint - ofs;
lastOfs = hint - ofs;
ofs = hint - tmp;
ofs = hint - tmp;
} else { // a[b + hint] <= key
} else { // a[b + hint] <= key
// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
int maxOfs = len - hint;
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;
lastOfs = ofs;
ofs = (ofs << 1) + 1;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
if (ofs <= 0) // int overflow
ofs = maxOfs;
ofs = maxOfs;
}
}
if (ofs > maxOfs)
if (ofs > maxOfs)
ofs = maxOfs;
ofs = maxOfs;
// Make offsets relative to b
// Make offsets relative to b
lastOfs += hint;
lastOfs += hint;
ofs += hint;
ofs += hint;
}
}
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
/*
/*
* Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
* 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
* the right of lastOfs but no farther right than ofs. Do a binary
* search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
* search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
*/
*/
lastOfs++;
lastOfs++;
while (lastOfs < ofs) {
while (lastOfs < ofs) {
int m = lastOfs + ((ofs - lastOfs) >>> 1);
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]
ofs = m; // key < a[b + m]
else
else
lastOfs = m + 1; // a[b + m] <= key
lastOfs = m + 1; // a[b + m] <= key
}
}
assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
return ofs;
return ofs;
}
}
/**
/**
* Merges two adjacent runs in place, in a stable fashion. The first
* 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
* 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
* 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.
* (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;
* For performance, this method should be called only when len1 <= len2;
* its twin, mergeHi should be called if len1 >= len2. (Either method
* its twin, mergeHi should be called if len1 >= len2. (Either method
* may be called if len1 == len2.)
* may be called if len1 == len2.)
*
*
* @param base1 index of first element in first run to be merged
* @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 len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
* @param base2 index of first element in second run to be merged
* (must be aBase + aLen)
* (must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
* @param len2 length of second run to be merged (must be > 0)
*/
*/
private void mergeLo(int base1, int len1, int base2, int len2) {
private void mergeLo(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy first run into temp array
// 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 cursor1 = 0; // Indexes into tmp array
int cursor2 = base2; // Indexes int a
int cursor2 = base2; // Indexes int a
int dest = base1; // Indexes int a
int dest = base1; // Indexes int a
// Move first element of second run and deal with degenerate cases
// Move first element of second run and deal with degenerate cases
a[dest++] = a[cursor2++];
s.copyElement(a, cursor2++, a, dest++);
if (--len2 == 0) {
if (--len2 == 0) {
System.arraycopy(tmp, cursor1, a, dest, len1);
s.copyRange(tmp, cursor1, a, dest, len1);
return;
return;
}
}
if (len1 == 1) {
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;
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; // " " " " "
int minGallop = this.minGallop; // " " " " "
outer:
outer:
while (true) {
while (true) {
int count1 = 0; // Number of times in a row that first run won
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
int count2 = 0; // Number of times in a row that second run won
/*
/*
* Do the straightforward thing until (if ever) one run starts
* Do the straightforward thing until (if ever) one run starts
* winning consistently.
* winning consistently.
*/
*/
do {
do {
assert len1 > 1 && len2 > 0;
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++;
count2++;
count1 = 0;
count1 = 0;
if (--len2 == 0)
if (--len2 == 0)
break outer;
break outer;
} else {
} else {
a[dest++] = tmp[cursor1++];
s.copyElement(tmp, cursor1++, a, dest++);
count1++;
count1++;
count2 = 0;
count2 = 0;
if (--len1 == 1)
if (--len1 == 1)
break outer;
break outer;
}
}
} while ((count1 | count2) < minGallop);
} while ((count1 | count2) < minGallop);
/*
/*
* One run is winning so consistently that galloping may be a
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
* neither run appears to be winning consistently anymore.
*/
*/
do {
do {
assert len1 > 1 && len2 > 0;
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) {
if (count1 != 0) {
System.arraycopy(tmp, cursor1, a, dest, count1);
s.copyRange(tmp, cursor1, a, dest, count1);
dest += count1;
dest += count1;
cursor1 += count1;
cursor1 += count1;
len1 -= count1;
len1 -= count1;
if (len1 <= 1) // len1 == 1 || len1 == 0
if (len1 <= 1) // len1 == 1 || len1 == 0
break outer;
break outer;
}
}
a[dest++] = a[cursor2++];
s.copyElement(a, cursor2++, a, dest++);
if (--len2 == 0)
if (--len2 == 0)
break outer;
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) {
if (count2 != 0) {
System.arraycopy(a, cursor2, a, dest, count2);
s.copyRange(a, cursor2, a, dest, count2);
dest += count2;
dest += count2;
cursor2 += count2;
cursor2 += count2;
len2 -= count2;
len2 -= count2;
if (len2 == 0)
if (len2 == 0)
break outer;
break outer;
}
}
a[dest++] = tmp[cursor1++];
s.copyElement(tmp, cursor1++, a, dest++);
if (--len1 == 1)
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