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   /*
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    * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
    *
    * This code is free software; you can redistribute it and/or modify it
    * under the terms of the GNU General Public License version 2 only, as
    * published by the Free Software Foundation.  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,
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   *
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  package com.WazaBe.HoloEverywhere.util;
  
A stable, adaptive, iterative mergesort that requires far fewer than n lg(n) comparisons when running on partially sorted arrays, while offering performance comparable to a traditional mergesort when run on random arrays. Like all proper mergesorts, this sort is stable and runs O(n log n) time (worst case). In the worst case, this sort requires temporary storage space for n/2 object references; in the best case, it requires only a small constant amount of space. This implementation was adapted from Tim Peters's list sort for Python, which is described in detail here: http://svn.python.org/projects/python/trunk/Objects/listsort.txt Tim's C code may be found here: http://svn.python.org/projects/python/trunk/Objects/listobject.c The underlying techniques are described in this paper (and may have even earlier origins): "Optimistic Sorting and Information Theoretic Complexity" Peter McIlroy SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms), pp 467-474, Austin, Texas, 25-27 January 1993. While the API to this class consists solely of static methods, it is (privately) instantiable; a TimSort instance holds the state of an ongoing sort, assuming the input array is large enough to warrant the full-blown TimSort. Small arrays are sorted in place, using a binary insertion sort.

Author(s):
Josh Bloch
  
  @SuppressWarnings({ "unchecked" })
  class TimSort<T> {
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;

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 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(int) 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;

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 lo, inclusive, to start, exclusive are already sorted.

Parameters:
a the array in which a range is to be sorted
lo the index of the first element in the range to be sorted
hi the index after the last element in the range to be sorted
start the index of the first element in the range that is not already known to be sorted (lo <= start <= hi)
c comparator to used for the sort
 
 	@SuppressWarnings("fallthrough")
 	private static <T> void binarySort(T[] aint loint hiint start,
 			Comparator<? super T> c) {
 		assert lo <= start && start <= hi;
 		if (start == lo) {
 			start++;
 		}
 		for (; start < histart++) {
 			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(pivota[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(aleftaleft + 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.

Parameters:
a the array in which a run is to be counted and possibly reversed
lo index of the first element in the run
hi index after the last element that may be contained in the run. It is required that lo < hi.
c the comparator to used for the sort
Returns:
the length of the run beginning at the specified position in the specified array
 
 	private static <T> int countRunAndMakeAscending(T[] aint loint 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(alorunHi);
 		} else { // Ascending
 			while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0) {
 				runHi++;
 			}
 		}
 
 		return runHi - lo;
 	}

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.

Parameters:
key the key whose insertion point to search for
a the array in which to search
base the index of the first element in the range
len the length of the range; must be > 0
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.
c the comparator used to order the range, and to search
Returns:
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[] aint baseint len,
 			int hintComparator<? super T> c) {
 		assert len > 0 && hint >= 0 && hint < len;
 		int lastOfs = 0;
 		int ofs = 1;
 		if (c.compare(keya[base + hint]) > 0) {
 			// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
 			int maxOfs = len - hint;
 			while (ofs < maxOfs && c.compare(keya[base + hint + ofs]) > 0) {
 				lastOfs = ofs;
 				ofs = (ofs << 1) + 1;
 				if (ofs <= 0) {
 					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(keya[base + hint - ofs]) <= 0) {
 				lastOfs = ofs;
 				ofs = (ofs << 1) + 1;
 				if (ofs <= 0) {
 					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(keya[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.

Parameters:
key the key whose insertion point to search for
a the array in which to search
base the index of the first element in the range
len the length of the range; must be > 0
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.
c the comparator used to order the range, and to search
Returns:
the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
 
 	private static <T> int gallopRight(T key, T[] aint baseint len,
 			int hintComparator<? super T> c) {
 		assert len > 0 && hint >= 0 && hint < len;
 
 		int ofs = 1;
 		int lastOfs = 0;
 		if (c.compare(keya[base + hint]) < 0) {
 			// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
 			int maxOfs = hint + 1;
 			while (ofs < maxOfs && c.compare(keya[base + hint - ofs]) < 0) {
 				lastOfs = ofs;
 				ofs = (ofs << 1) + 1;
 				if (ofs <= 0) {
 					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(keya[base + hint + ofs]) >= 0) {
 				lastOfs = ofs;
 				ofs = (ofs << 1) + 1;
 				if (ofs <= 0) {
 					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(keya[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;
 	}

Returns the minimum acceptable run length for an array of the specified length. Natural runs shorter than this will be extended with binarySort(java.lang.Object[],int,int,int,java.util.Comparator). 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.

Parameters:
n the length of the array to be sorted
Returns:
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 >= ) {
 			r |= n & 1;
 			n >>= 1;
 		}
 		return n + r;
 	}

Checks that fromIndex and toIndex are in range, and throws an appropriate exception if they aren't.

Parameters:
arrayLen the length of the array
fromIndex the index of the first element of the range
toIndex the index after the last element of the range
Throws:
java.lang.IllegalArgumentException if fromIndex > toIndex
java.lang.ArrayIndexOutOfBoundsException if fromIndex < 0 or toIndex > arrayLen
 
 	private static void rangeCheck(int arrayLenint fromIndexint toIndex) {
 		if (fromIndex > toIndex) {
 			throw new IllegalArgumentException("fromIndex(" + fromIndex
 					+ ") > toIndex(" + toIndex + ")");
 		}
 		if (fromIndex < 0) {
 			throw new ArrayIndexOutOfBoundsException(fromIndex);
 		}
 		if (toIndex > arrayLen) {
 			throw new ArrayIndexOutOfBoundsException(toIndex);
 		}
 	}

Reverse the specified range of the specified array.

Parameters:
a the array in which a range is to be reversed
lo the index of the first element in the range to be reversed
hi the index after the last element in the range to be reversed
 
 	private static void reverseRange(Object[] aint loint hi) {
 		hi--;
 		while (lo < hi) {
 			Object t = a[lo];
 			a[lo++] = a[hi];
 			a[hi--] = t;
 		}
 	}
 
 	static <T> void sort(T[] aComparator<? super T> c) {
 		sort(a, 0, a.lengthc);
 	}
 
 	/*
 	 * 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[] aint loint hiComparator<? super T> c) {
 		if (c == null) {
 			Arrays.sort(alohi);
 			return;
 		}
 
 		rangeCheck(a.lengthlohi);
 		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 < ) {
 			int initRunLen = countRunAndMakeAscending(alohic);
 			binarySort(alohilo + initRunLenc);
 			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>(ac);
 		int minRun = minRunLength(nRemaining);
 		do {
 			// Identify next run
 			int runLen = countRunAndMakeAscending(alohic);
 
 			// If run is short, extend to min(minRun, nRemaining)
 			if (runLen < minRun) {
 				int force = nRemaining <= minRun ? nRemaining : minRun;
 				binarySort(alolo + forcelo + runLenc);
 				runLen = force;
 			}
 
 			// Push run onto pending-run stack, and maybe merge
 			ts.pushRun(lorunLen);
 
 			// Advance to find next run
 			lo += runLen;
 			nRemaining -= runLen;
 		} while (nRemaining != 0);
 
 		// Merge all remaining runs to complete sort
 		assert lo == hi;
 		assert ts.stackSize == 1;
 	}

The array being sorted.
 
 	private final T[] a;

The comparator for this sort.
 
 	private final Comparator<? super T> c;

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 = ;
 
 	private final int[] runBase;
 
 	private final int[] runLen;

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
 
Temp storage for merges.
 
 	private T[] tmp// Actual runtime type will be Object[], regardless of T
 
Creates a TimSort instance to maintain the state of an ongoing sort.

Parameters:
a the array to be sorted
c the comparator to determine the order of the sort
 
 	private TimSort(T[] aComparator<? super T> c) {
 		this. = a;
 		this. = c;
 
 		// Allocate temp storage (which may be increased later if necessary)
 		int len = a.length;
 		T[] newArray = (T[]) new Object[len < 2 *  ? len >>> 1
 		 = 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;
 		 = new int[stackLen];
 		 = new int[stackLen];
 	}

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.

Parameters:
minCapacity the minimum required capacity of the tmp array
Returns:
tmp, whether or not it grew
 
 	private T[] ensureCapacity(int minCapacity) {
 		if (. < 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) {
 				newSize = minCapacity;
 			} else {
 				newSize = Math.min(newSize. >>> 1);
 			}
 
 			T[] newArray = (T[]) new Object[newSize];
 			 = newArray;
 		}
 		return ;
 	}

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.

Parameters:
i stack index of the first of the two runs to merge
 
 	private void mergeAt(int i) {
 		assert  >= 2;
 		assert i >= 0;
 		assert i ==  - 2 || i ==  - 3;
 
 		int base1 = [i];
 		int len1 = [i];
 		int base2 = [i + 1];
 		int len2 = [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.
 		 */
 		[i] = len1 + len2;
 		if (i ==  - 3) {
 			[i + 1] = [i + 2];
 			[i + 1] = [i + 2];
 		}
 
 		/*
 		 * 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([base2], base1len1, 0, );
 		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([base1 + len1 - 1], base2len2len2 - 1, );
 		assert len2 >= 0;
 		if (len2 == 0) {
 			return;
 		}
 
 		// Merge remaining runs, using tmp array with min(len1, len2) elements
 		if (len1 <= len2) {
 			mergeLo(base1len1base2len2);
 		} else {
 			mergeHi(base1len1base2len2);
 		}
 	}

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 ( > 1) {
 			int n =  - 2;
 			if (n > 0 && [n - 1] <= [n] + [n + 1]) {
 				if ([n - 1] < [n + 1]) {
 					n--;
 				}
 				mergeAt(n);
 			} else if ([n] <= [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 ( > 1) {
 			int n =  - 2;
 			if (n > 0 && [n - 1] < [n + 1]) {
 				n--;
 			}
 			mergeAt(n);
 		}
 	}

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.)

Parameters:
base1 index of first element in first run to be merged
len1 length of first run to be merged (must be > 0)
base2 index of first element in second run to be merged (must be aBase + aLen)
len2 length of second run to be merged (must be > 0)
 
 	private void mergeHi(int base1int len1int base2int len2) {
 		assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
 
 		// Copy second run into temp array
 		T[] a = this.// For performance
 		T[] tmp = ensureCapacity(len2);
 		System.arraycopy(abase2tmp, 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, adest - (len2 - 1), len2);
 			return;
 		}
 		if (len2 == 1) {
 			dest -= len1;
 			cursor1 -= len1;
 			System.arraycopy(acursor1 + 1, adest + 1, len1);
 			a[dest] = tmp[cursor2];
 			return;
 		}
 
 		Comparator<? super T> c = this.// Use local variable for performance
 		int minGallop = this.// "    " "     " "
 		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], abase1len1len1 - 1, c);
 				if (count1 != 0) {
 					dest -= count1;
 					cursor1 -= count1;
 					len1 -= count1;
 					System.arraycopy(acursor1 + 1, adest + 1, count1);
 					if (len1 == 0) {
 						break outer;
 					}
 				}
 				a[dest--] = tmp[cursor2--];
 				if (--len2 == 1) {
 					break outer;
 				}
 
 				count2 = len2
 						- gallopLeft(a[cursor1], tmp, 0, len2len2 - 1, c);
 				if (count2 != 0) {
 					dest -= count2;
 					cursor2 -= count2;
 					len2 -= count2;
 					System.arraycopy(tmpcursor2 + 1, adest + 1, count2);
 					if (len2 <= 1) {
 						break outer;
 					}
 				}
 				a[dest--] = a[cursor1--];
 				if (--len1 == 0) {
 					break outer;
 				}
 				minGallop--;
 			} while (count1 >=  | count2 >= );
 			if (minGallop < 0) {
 				minGallop = 0;
 			}
 			minGallop += 2; // Penalize for leaving gallop mode
 		} // End of "outer" loop
 		this. = minGallop < 1 ? 1 : minGallop// Write back to field
 
 		if (len2 == 1) {
 			assert len1 > 0;
 			dest -= len1;
 			cursor1 -= len1;
 			System.arraycopy(acursor1 + 1, adest + 1, len1);
 			a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
 		} else if (len2 == 0) {
 					"Comparison method violates its general contract!");
 		} else {
 			assert len1 == 0;
 			assert len2 > 0;
 			System.arraycopy(tmp, 0, adest - (len2 - 1), len2);
 		}
 	}

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.)

Parameters:
base1 index of first element in first run to be merged
len1 length of first run to be merged (must be > 0)
base2 index of first element in second run to be merged (must be aBase + aLen)
len2 length of second run to be merged (must be > 0)
 
 	private void mergeLo(int base1int len1int base2int len2) {
 		assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
 
 		// Copy first run into temp array
 		T[] a = this.// For performance
 		T[] tmp = ensureCapacity(len1);
 		System.arraycopy(abase1tmp, 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(tmpcursor1adestlen1);
 			return;
 		}
 		if (len1 == 1) {
 			System.arraycopy(acursor2adestlen2);
 			a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
 			return;
 		}
 
 		Comparator<? super T> c = this.// Use local variable for performance
 		int minGallop = this.// "    " "     " "
 		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], tmpcursor1len1, 0, c);
 				if (count1 != 0) {
 					System.arraycopy(tmpcursor1adestcount1);
 					dest += count1;
 					cursor1 += count1;
 					len1 -= count1;
 					if (len1 <= 1) {
 						break outer;
 					}
 				}
 				a[dest++] = a[cursor2++];
 				if (--len2 == 0) {
 					break outer;
 				}
 
 				count2 = gallopLeft(tmp[cursor1], acursor2len2, 0, c);
 				if (count2 != 0) {
 					System.arraycopy(acursor2adestcount2);
 					dest += count2;
 					cursor2 += count2;
 					len2 -= count2;
 					if (len2 == 0) {
 						break outer;
 					}
 				}
 				a[dest++] = tmp[cursor1++];
 				if (--len1 == 1) {
 					break outer;
 				}
 				minGallop--;
 			} while (count1 >=  | count2 >= );
 			if (minGallop < 0) {
 				minGallop = 0;
 			}
 			minGallop += 2; // Penalize for leaving gallop mode
 		} // End of "outer" loop
 		this. = minGallop < 1 ? 1 : minGallop// Write back to field
 
 		if (len1 == 1) {
 			assert len2 > 0;
 			System.arraycopy(acursor2adestlen2);
 			a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
 		} else if (len1 == 0) {
 					"Comparison method violates its general contract!");
 		} else {
 			assert len2 == 0;
 			assert len1 > 1;
 			System.arraycopy(tmpcursor1adestlen1);
 		}
 	}

Pushes the specified run onto the pending-run stack.

Parameters:
runBase index of the first element in the run
runLen the number of elements in the run
	private void pushRun(int runBaseint runLen) {
		this.[] = runBase;
		this.[] = runLen;
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