/*
* Copyright 2009 Google Inc. All Rights Reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation. Oracle designates this
* particular file as subject to the "Classpath" exception as provided
* by Oracle in the LICENSE file that accompanied this code.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
* or visit www.oracle.com if you need additional information or have any
* questions.
*/
/**
* This is a near duplicate of {@link TimSort}, modified for use with
* arrays of objects that implement {@link Comparable}, instead of using
* explicit comparators.
*
* <p>If you are using an optimizing VM, you may find that ComparableTimSort
* offers no performance benefit over TimSort in conjunction with a
* comparator that simply returns {@code ((Comparable)first).compareTo(Second)}.
* If this is the case, you are better off deleting ComparableTimSort to
* eliminate the code duplication. (See Arrays.java for details.)
*
* @author Josh Bloch
*/
class ComparableTimSort {
/**
* 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.
*/
/**
* The array being sorted.
*/
private final Object[] a;
/**
* When we get into galloping mode, we stay there until both runs win less
* often than MIN_GALLOP consecutive times.
*/
/**
* 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.
*/
/**
* 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.
*/
/**
* Temp storage for merges.
*/
/**
* 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 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
*/
this.a = a;
// Allocate temp storage (which may be increased later if necessary)
/*
* Allocate runs-to-be-merged stack (which cannot be expanded). The
* stack length requirements are described in listsort.txt. The C
* version always uses the same stack length (85), but this was
* measured to be too expensive when sorting "mid-sized" arrays (e.g.,
* 100 elements) in Java. Therefore, we use smaller (but sufficiently
* large) stack lengths for smaller arrays. The "magic numbers" in the
* computation below must be changed if MIN_MERGE is decreased. See
* the MIN_MERGE declaration above for more information.
*/
}
/*
* The 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.
*/
}
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) {
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.
*/
do {
// Identify next run
// If run is short, extend to min(minRun, nRemaining)
}
// Push run onto pending-run stack, and maybe merge
ts.mergeCollapse();
// Advance to find next run
nRemaining -= runLen;
} while (nRemaining != 0);
// Merge all remaining runs to complete sort
}
/**
* 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})
*/
@SuppressWarnings("fallthrough")
start++;
@SuppressWarnings("unchecked")
// Set left (and right) to the index where a[start] (pivot) belongs
/*
* Invariants:
* pivot >= all in [lo, left).
* pivot < all in [right, start).
*/
else
}
/*
* 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.
*/
// Switch is just an optimization for arraycopy in default case
switch (n) {
break;
}
}
}
/**
* 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}.
* @return the length of the run beginning at the specified position in
* the specified array
*/
@SuppressWarnings("unchecked")
return 1;
// Find end of run, and reverse range if descending
runHi++;
} else { // Ascending
runHi++;
}
}
/**
* 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
*/
hi--;
a[hi--] = t;
}
}
/**
* Returns the minimum acceptable run length for an array of the specified
* length. Natural runs shorter than this will be extended with
* {@link #binarySort}.
*
* Roughly speaking, the computation is:
*
* If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
* Else if n is an exact power of 2, return MIN_MERGE/2.
* Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
* is close to, but strictly less than, an exact power of 2.
*
* For the rationale, see listsort.txt.
*
* @param n the length of the array to be sorted
* @return the length of the minimum run to be merged
*/
private static int minRunLength(int n) {
assert n >= 0;
int r = 0; // Becomes 1 if any 1 bits are shifted off
while (n >= MIN_MERGE) {
r |= (n & 1);
n >>= 1;
}
return n + r;
}
/**
* Pushes the specified run onto the pending-run stack.
*
* @param runBase index of the first element in the run
* @param runLen the number of elements in the run
*/
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;
n--;
mergeAt(n);
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;
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
*/
@SuppressWarnings("unchecked")
private void mergeAt(int i) {
assert stackSize >= 2;
assert i >= 0;
/*
* 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.
*/
if (i == stackSize - 3) {
}
stackSize--;
/*
* Find where the first element of run2 goes in run1. Prior elements
* in run1 can be ignored (because they're already in place).
*/
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).
*/
assert len2 >= 0;
if (len2 == 0)
return;
// Merge remaining runs, using tmp array with min(len1, len2) elements
else
}
/**
* 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.
* @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.
*/
int lastOfs = 0;
int ofs = 1;
// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
}
// Make offsets relative to base
} else { // key <= a[base + hint]
// Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
}
// Make offsets relative to base
}
/*
* 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++;
else
ofs = m; // key <= a[base + m]
}
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.
* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
*/
int ofs = 1;
int lastOfs = 0;
// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
}
// Make offsets relative to b
} else { // a[b + hint] <= key
// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
}
// Make offsets relative to b
}
/*
* 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++;
ofs = m; // key < a[b + m]
else
}
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)
*/
@SuppressWarnings("unchecked")
// Copy first run into temp array
Object[] a = this.a; // For performance
// Move first element of second run and deal with degenerate cases
if (--len2 == 0) {
return;
}
if (len1 == 1) {
return;
}
while (true) {
/*
* Do the straightforward thing until (if ever) one run starts
* winning consistently.
*/
do {
count2++;
count1 = 0;
if (--len2 == 0)
break outer;
} else {
count1++;
count2 = 0;
if (--len1 == 1)
break outer;
}
/*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
do {
if (count1 != 0) {
break outer;
}
if (--len2 == 0)
break outer;
if (count2 != 0) {
if (len2 == 0)
break outer;
}
if (--len1 == 1)
break outer;
minGallop--;
if (minGallop < 0)
minGallop = 0;
} // End of "outer" loop
if (len1 == 1) {
assert len2 > 0;
} else if (len1 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
assert len2 == 0;
assert len1 > 1;
}
}
/**
* 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)
*/
@SuppressWarnings("unchecked")
// Copy second run into temp array
Object[] a = this.a; // For performance
// Move last element of first run and deal with degenerate cases
if (--len1 == 0) {
return;
}
if (len2 == 1) {
return;
}
while (true) {
/*
* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
do {
count1++;
count2 = 0;
if (--len1 == 0)
break outer;
} else {
count2++;
count1 = 0;
if (--len2 == 1)
break outer;
}
/*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
do {
if (count1 != 0) {
if (len1 == 0)
break outer;
}
if (--len2 == 1)
break outer;
if (count2 != 0) {
if (len2 <= 1)
break outer; // len2 == 1 || len2 == 0
}
if (--len1 == 0)
break outer;
minGallop--;
if (minGallop < 0)
minGallop = 0;
} // End of "outer" loop
if (len2 == 1) {
assert len1 > 0;
} else if (len2 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
assert len1 == 0;
assert len2 > 0;
}
}
/**
* 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
*/
// Compute smallest power of 2 > minCapacity
int newSize = minCapacity;
newSize++;
else
}
return tmp;
}
/**
* Checks that fromIndex and toIndex are in range, and throws an
* appropriate exception if they aren't.
*
* @param arrayLen the length of the array
* @param fromIndex the index of the first element of the range
* @param toIndex the index after the last element of the range
* @throws IllegalArgumentException if fromIndex > toIndex
* @throws ArrayIndexOutOfBoundsException if fromIndex < 0
* or toIndex > arrayLen
*/
if (fromIndex < 0)
throw new ArrayIndexOutOfBoundsException(fromIndex);
throw new ArrayIndexOutOfBoundsException(toIndex);
}
}