File	Doc	Category	Size	Date	Package
TimSort.java	API Doc	Android 1.5 API	36239	Wed May 06 22:41:04 BST 2009	java.util

TimSort

• java.lang.Object

Fields Summary
private static final int	MIN_MERGE 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 final T[]	a The array being sorted.
private final Comparator	c The comparator for this sort.
private static final int	MIN_GALLOP When we get into galloping mode, we stay there until both runs win less often than MIN_GALLOP consecutive times.
private int	minGallop 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 static final int	INITIAL_TMP_STORAGE_LENGTH 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 T[]	tmp Temp storage for merges.
private int	stackSize 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
private static final boolean	DEBUG Asserts have been placed in if-statements for performace. To enable them, set this field to true and enable them in VM with a command line flag. If you modify this class, please do test the asserts!

Constructors Summary

private TimSort(T[] a, Comparator c) 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 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];

Methods Summary
private static void	binarySort(T[] a, int lo, int hi, int start, java.util.Comparator c) 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 if (DEBUG) 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; if (DEBUG) 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; } if (DEBUG) assert left == right; /* * The invariants still hold: pivot >= all in [lo, left) and * pivot < all in [left, start), so pivot belongs at left. Note * that if there are elements equal to pivot, left points to the * first slot after them -- that's why this sort is stable. * Slide elements over to make room to make room for pivot. */ int n = start - left; // The number of elements to move // Switch is just an optimization for arraycopy in default case switch(n) { case 2: a[left + 2] = a[left + 1]; case 1: a[left + 1] = a[left]; break; default: System.arraycopy(a, left, a, left + 1, n); } a[left] = pivot; }
private static int	countRunAndMakeAscending(T[] a, int lo, int hi, java.util.Comparator c) 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 if (DEBUG) 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;
private T[]	ensureCapacity(int minCapacity) 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 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;
private static int	gallopLeft(T key, T[] a, int base, int len, int hint, java.util.Comparator c) 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. if (DEBUG) 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; } if (DEBUG) 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] } if (DEBUG) assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs] return ofs;
private static int	gallopRight(T key, T[] a, int base, int len, int hint, java.util.Comparator c) 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] if (DEBUG) 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; } if (DEBUG) 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 } if (DEBUG) assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs] return ofs;
private void	mergeAt(int i) 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 if (DEBUG) assert stackSize >= 2; if (DEBUG) assert i >= 0; if (DEBUG) 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]; if (DEBUG) assert len1 > 0 && len2 > 0; if (DEBUG) 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); if (DEBUG) 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); if (DEBUG) 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);
private void	mergeCollapse() 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. 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 } }
private void	mergeForceCollapse() Merges all runs on the stack until only one remains. This method is called once, to complete the sort. while (stackSize > 1) { int n = stackSize - 2; if (n > 0 && runLen[n - 1] < runLen[n + 1]) n--; mergeAt(n); }
private void	mergeHi(int base1, int len1, int base2, int len2) 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) if (DEBUG) 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 { if (DEBUG) 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 { if (DEBUG) 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) { if (DEBUG) 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 { if (DEBUG) assert len1 == 0; if (DEBUG) assert len2 > 0; System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2); }
private void	mergeLo(int base1, int len1, int base2, int 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.) 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) if (DEBUG) 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 { if (DEBUG) 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 { if (DEBUG) 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) { if (DEBUG) 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 { if (DEBUG) assert len2 == 0; if (DEBUG) assert len1 > 1; System.arraycopy(tmp, cursor1, a, dest, len1); }
private static int	minRunLength(int n) 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 if (DEBUG) 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 void	pushRun(int runBase, int runLen) 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 this.runBase[stackSize] = runBase; this.runLen[stackSize] = runLen; stackSize++;
private static void	rangeCheck(int arrayLen, int fromIndex, int toIndex) 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 > toIndex) throw new IllegalArgumentException("fromIndex(" + fromIndex + ") > toIndex(" + toIndex+")"); if (fromIndex < 0) throw new ArrayIndexOutOfBoundsException(fromIndex); if (toIndex > arrayLen) throw new ArrayIndexOutOfBoundsException(toIndex);
private static void	reverseRange(java.lang.Object[] a, int lo, int hi) 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--; while (lo < hi) { Object t = a[lo]; a[lo++] = a[hi]; a[hi--] = t; }
static void	sort(T[] a, java.util.Comparator c) sort(a, 0, a.length, c);
static void	sort(T[] a, int lo, int hi, java.util.Comparator c) if (c == null) { Arrays.sort(a, lo, hi); return; } rangeCheck(a.length, lo, hi); int nRemaining = hi - lo; if (nRemaining < 2) return; // Arrays of size 0 and 1 are always sorted // If array is small, do a "mini-TimSort" with no merges if (nRemaining < MIN_MERGE) { int initRunLen = countRunAndMakeAscending(a, lo, nRemaining, 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 if (DEBUG) assert lo == hi; ts.mergeForceCollapse(); if (DEBUG) assert ts.stackSize == 1;