diff -r 8d0be6a9a809 -r d382dacfd73f rt/emul/compact/src/main/java/java/util/ComparableTimSort.java
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/rt/emul/compact/src/main/java/java/util/ComparableTimSort.java	Tue Feb 26 16:54:16 2013 +0100
@@ -0,0 +1,896 @@
+/*
+ * Copyright 2009 Google Inc.  All Rights Reserved.
+ * 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,
+ * 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.
+ */
+
+package java.util;
+
+
+/**
+ * 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.
+     */
+    private static final int MIN_MERGE = 32;
+
+    /**
+     * 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.
+     */
+    private static final int  MIN_GALLOP = 7;
+
+    /**
+     * This controls when we get *into* galloping mode.  It is initialized
+     * to MIN_GALLOP.  The mergeLo and mergeHi methods nudge it higher for
+     * random data, and lower for highly structured data.
+     */
+    private int minGallop = MIN_GALLOP;
+
+    /**
+     * Maximum initial size of tmp array, which is used for merging.  The array
+     * can grow to accommodate demand.
+     *
+     * Unlike Tim's original C version, we do not allocate this much storage
+     * when sorting smaller arrays.  This change was required for performance.
+     */
+    private static final int INITIAL_TMP_STORAGE_LENGTH = 256;
+
+    /**
+     * Temp storage for merges.
+     */
+    private Object[] tmp;
+
+    /**
+     * A stack of pending runs yet to be merged.  Run i starts at
+     * address base[i] and extends for len[i] elements.  It's always
+     * true (so long as the indices are in bounds) that:
+     *
+     *     runBase[i] + runLen[i] == runBase[i + 1]
+     *
+     * so we could cut the storage for this, but it's a minor amount,
+     * and keeping all the info explicit simplifies the code.
+     */
+    private int stackSize = 0;  // Number of pending runs on stack
+    private final int[] runBase;
+    private final int[] runLen;
+
+    /**
+     * Creates a TimSort instance to maintain the state of an ongoing sort.
+     *
+     * @param a the array to be sorted
+     */
+    private ComparableTimSort(Object[] a) {
+        this.a = a;
+
+        // Allocate temp storage (which may be increased later if necessary)
+        int len = a.length;
+        @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
+        Object[] newArray = 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];
+    }
+
+    /*
+     * 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 void sort(Object[] a) {
+          sort(a, 0, a.length);
+    }
+
+    static void sort(Object[] a, int lo, int hi) {
+        rangeCheck(a.length, lo, hi);
+        int nRemaining  = hi - lo;
+        if (nRemaining < 2)
+            return;  // Arrays of size 0 and 1 are always sorted
+
+        // If array is small, do a "mini-TimSort" with no merges
+        if (nRemaining < MIN_MERGE) {
+            int initRunLen = countRunAndMakeAscending(a, lo, hi);
+            binarySort(a, lo, hi, lo + initRunLen);
+            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.
+         */
+        ComparableTimSort ts = new ComparableTimSort(a);
+        int minRun = minRunLength(nRemaining);
+        do {
+            // Identify next run
+            int runLen = countRunAndMakeAscending(a, lo, hi);
+
+            // 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);
+                runLen = force;
+            }
+
+            // Push run onto pending-run stack, and maybe merge
+            ts.pushRun(lo, runLen);
+            ts.mergeCollapse();
+
+            // Advance to find next run
+            lo += runLen;
+            nRemaining -= runLen;
+        } while (nRemaining != 0);
+
+        // Merge all remaining runs to complete sort
+        assert lo == hi;
+        ts.mergeForceCollapse();
+        assert ts.stackSize == 1;
+    }
+
+    /**
+     * Sorts the specified portion of the specified array using a binary
+     * insertion sort.  This is the best method for sorting small numbers
+     * of elements.  It requires O(n log n) compares, but O(n^2) data
+     * movement (worst case).
+     *
+     * If the initial part of the specified range is already sorted,
+     * this method can take advantage of it: the method assumes that the
+     * elements from index {@code lo}, inclusive, to {@code start},
+     * exclusive are already sorted.
+     *
+     * @param a the array in which a range is to be sorted
+     * @param lo the index of the first element in the range to be sorted
+     * @param hi the index after the last element in the range to be sorted
+     * @param start the index of the first element in the range that is
+     *        not already known to be sorted ({@code lo <= start <= hi})
+     */
+    @SuppressWarnings("fallthrough")
+    private static void binarySort(Object[] a, int lo, int hi, int start) {
+        assert lo <= start && start <= hi;
+        if (start == lo)
+            start++;
+        for ( ; start < hi; start++) {
+            @SuppressWarnings("unchecked")
+            Comparable<Object> pivot = (Comparable) 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 (pivot.compareTo(a[mid]) < 0)
+                    right = mid;
+                else
+                    left = mid + 1;
+            }
+            assert left == right;
+
+            /*
+             * The invariants still hold: pivot >= all in [lo, left) and
+             * pivot < all in [left, start), so pivot belongs at left.  Note
+             * that if there are elements equal to pivot, left points to the
+             * first slot after them -- that's why this sort is stable.
+             * Slide elements over to make room for pivot.
+             */
+            int n = start - left;  // The number of elements to move
+            // Switch is just an optimization for arraycopy in default case
+            switch (n) {
+                case 2:  a[left + 2] = a[left + 1];
+                case 1:  a[left + 1] = a[left];
+                         break;
+                default: System.arraycopy(a, left, a, left + 1, n);
+            }
+            a[left] = pivot;
+        }
+    }
+
+    /**
+     * Returns the length of the run beginning at the specified position in
+     * the specified array and reverses the run if it is descending (ensuring
+     * that the run will always be ascending when the method returns).
+     *
+     * A run is the longest ascending sequence with:
+     *
+     *    a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
+     *
+     * or the longest descending sequence with:
+     *
+     *    a[lo] >  a[lo + 1] >  a[lo + 2] >  ...
+     *
+     * For its intended use in a stable mergesort, the strictness of the
+     * definition of "descending" is needed so that the call can safely
+     * reverse a descending sequence without violating stability.
+     *
+     * @param a the array in which a run is to be counted and possibly reversed
+     * @param lo index of the first element in the run
+     * @param hi index after the last element that may be contained in the run.
+              It is required that {@code lo < hi}.
+     * @return  the length of the run beginning at the specified position in
+     *          the specified array
+     */
+    @SuppressWarnings("unchecked")
+    private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
+        assert lo < hi;
+        int runHi = lo + 1;
+        if (runHi == hi)
+            return 1;
+
+        // Find end of run, and reverse range if descending
+        if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending
+            while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0)
+                runHi++;
+            reverseRange(a, lo, runHi);
+        } else {                              // Ascending
+            while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0)
+                runHi++;
+        }
+
+        return runHi - lo;
+    }
+
+    /**
+     * Reverse the specified range of the specified array.
+     *
+     * @param a the array in which a range is to be reversed
+     * @param lo the index of the first element in the range to be reversed
+     * @param hi the index after the last element in the range to be reversed
+     */
+    private static void reverseRange(Object[] a, int lo, int hi) {
+        hi--;
+        while (lo < hi) {
+            Object t = a[lo];
+            a[lo++] = a[hi];
+            a[hi--] = t;
+        }
+    }
+
+    /**
+     * Returns the minimum acceptable run length for an array of the specified
+     * length. Natural runs shorter than this will be extended with
+     * {@link #binarySort}.
+     *
+     * Roughly speaking, the computation is:
+     *
+     *  If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
+     *  Else if n is an exact power of 2, return MIN_MERGE/2.
+     *  Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
+     *   is close to, but strictly less than, an exact power of 2.
+     *
+     * For the rationale, see listsort.txt.
+     *
+     * @param n the length of the array to be sorted
+     * @return the length of the minimum run to be merged
+     */
+    private static int minRunLength(int n) {
+        assert n >= 0;
+        int r = 0;      // Becomes 1 if any 1 bits are shifted off
+        while (n >= MIN_MERGE) {
+            r |= (n & 1);
+            n >>= 1;
+        }
+        return n + r;
+    }
+
+    /**
+     * Pushes the specified run onto the pending-run stack.
+     *
+     * @param runBase index of the first element in the run
+     * @param runLen  the number of elements in the run
+     */
+    private void pushRun(int runBase, int runLen) {
+        this.runBase[stackSize] = runBase;
+        this.runLen[stackSize] = runLen;
+        stackSize++;
+    }
+
+    /**
+     * Examines the stack of runs waiting to be merged and merges adjacent runs
+     * until the stack invariants are reestablished:
+     *
+     *     1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
+     *     2. runLen[i - 2] > runLen[i - 1]
+     *
+     * This method is called each time a new run is pushed onto the stack,
+     * so the invariants are guaranteed to hold for i < stackSize upon
+     * entry to the method.
+     */
+    private void mergeCollapse() {
+        while (stackSize > 1) {
+            int n = stackSize - 2;
+            if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
+                if (runLen[n - 1] < runLen[n + 1])
+                    n--;
+                mergeAt(n);
+            } else if (runLen[n] <= runLen[n + 1]) {
+                mergeAt(n);
+            } else {
+                break; // Invariant is established
+            }
+        }
+    }
+
+    /**
+     * Merges all runs on the stack until only one remains.  This method is
+     * called once, to complete the sort.
+     */
+    private void mergeForceCollapse() {
+        while (stackSize > 1) {
+            int n = stackSize - 2;
+            if (n > 0 && runLen[n - 1] < runLen[n + 1])
+                n--;
+            mergeAt(n);
+        }
+    }
+
+    /**
+     * Merges the two runs at stack indices i and i+1.  Run i must be
+     * the penultimate or antepenultimate run on the stack.  In other words,
+     * i must be equal to stackSize-2 or stackSize-3.
+     *
+     * @param i stack index of the first of the two runs to merge
+     */
+    @SuppressWarnings("unchecked")
+    private void mergeAt(int i) {
+        assert stackSize >= 2;
+        assert i >= 0;
+        assert i == stackSize - 2 || i == stackSize - 3;
+
+        int base1 = runBase[i];
+        int len1 = runLen[i];
+        int base2 = runBase[i + 1];
+        int len2 = runLen[i + 1];
+        assert len1 > 0 && len2 > 0;
+        assert base1 + len1 == base2;
+
+        /*
+         * Record the length of the combined runs; if i is the 3rd-last
+         * run now, also slide over the last run (which isn't involved
+         * in this merge).  The current run (i+1) goes away in any case.
+         */
+        runLen[i] = len1 + len2;
+        if (i == stackSize - 3) {
+            runBase[i + 1] = runBase[i + 2];
+            runLen[i + 1] = runLen[i + 2];
+        }
+        stackSize--;
+
+        /*
+         * Find where the first element of run2 goes in run1. Prior elements
+         * in run1 can be ignored (because they're already in place).
+         */
+        int k = gallopRight((Comparable<Object>) a[base2], a, base1, len1, 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((Comparable<Object>) a[base1 + len1 - 1], a,
+                base2, len2, len2 - 1);
+        assert len2 >= 0;
+        if (len2 == 0)
+            return;
+
+        // Merge remaining runs, using tmp array with min(len1, len2) elements
+        if (len1 <= len2)
+            mergeLo(base1, len1, base2, len2);
+        else
+            mergeHi(base1, len1, base2, len2);
+    }
+
+    /**
+     * Locates the position at which to insert the specified key into the
+     * specified sorted range; if the range contains an element equal to key,
+     * returns the index of the leftmost equal element.
+     *
+     * @param key the key whose insertion point to search for
+     * @param a the array in which to search
+     * @param base the index of the first element in the range
+     * @param len the length of the range; must be > 0
+     * @param hint the index at which to begin the search, 0 <= hint < n.
+     *     The closer hint is to the result, the faster this method will run.
+     * @return the int k,  0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
+     *    pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
+     *    In other words, key belongs at index b + k; or in other words,
+     *    the first k elements of a should precede key, and the last n - k
+     *    should follow it.
+     */
+    private static int gallopLeft(Comparable<Object> key, Object[] a,
+            int base, int len, int hint) {
+        assert len > 0 && hint >= 0 && hint < len;
+
+        int lastOfs = 0;
+        int ofs = 1;
+        if (key.compareTo(a[base + hint]) > 0) {
+            // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
+            int maxOfs = len - hint;
+            while (ofs < maxOfs && key.compareTo(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 && key.compareTo(a[base + hint - ofs]) <= 0) {
+                lastOfs = ofs;
+                ofs = (ofs << 1) + 1;
+                if (ofs <= 0)   // int overflow
+                    ofs = maxOfs;
+            }
+            if (ofs > maxOfs)
+                ofs = maxOfs;
+
+            // Make offsets relative to base
+            int tmp = lastOfs;
+            lastOfs = hint - ofs;
+            ofs = hint - tmp;
+        }
+        assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
+
+        /*
+         * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
+         * to the right of lastOfs but no farther right than ofs.  Do a binary
+         * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
+         */
+        lastOfs++;
+        while (lastOfs < ofs) {
+            int m = lastOfs + ((ofs - lastOfs) >>> 1);
+
+            if (key.compareTo(a[base + m]) > 0)
+                lastOfs = m + 1;  // a[base + m] < key
+            else
+                ofs = m;          // key <= a[base + m]
+        }
+        assert lastOfs == ofs;    // so a[base + ofs - 1] < key <= a[base + ofs]
+        return ofs;
+    }
+
+    /**
+     * Like gallopLeft, except that if the range contains an element equal to
+     * key, gallopRight returns the index after the rightmost equal element.
+     *
+     * @param key the key whose insertion point to search for
+     * @param a the array in which to search
+     * @param base the index of the first element in the range
+     * @param len the length of the range; must be > 0
+     * @param hint the index at which to begin the search, 0 <= hint < n.
+     *     The closer hint is to the result, the faster this method will run.
+     * @return the int k,  0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
+     */
+    private static int gallopRight(Comparable<Object> key, Object[] a,
+            int base, int len, int hint) {
+        assert len > 0 && hint >= 0 && hint < len;
+
+        int ofs = 1;
+        int lastOfs = 0;
+        if (key.compareTo(a[base + hint]) < 0) {
+            // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
+            int maxOfs = hint + 1;
+            while (ofs < maxOfs && key.compareTo(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 && key.compareTo(a[base + hint + ofs]) >= 0) {
+                lastOfs = ofs;
+                ofs = (ofs << 1) + 1;
+                if (ofs <= 0)   // int overflow
+                    ofs = maxOfs;
+            }
+            if (ofs > maxOfs)
+                ofs = maxOfs;
+
+            // Make offsets relative to b
+            lastOfs += hint;
+            ofs += hint;
+        }
+        assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
+
+        /*
+         * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
+         * the right of lastOfs but no farther right than ofs.  Do a binary
+         * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
+         */
+        lastOfs++;
+        while (lastOfs < ofs) {
+            int m = lastOfs + ((ofs - lastOfs) >>> 1);
+
+            if (key.compareTo(a[base + m]) < 0)
+                ofs = m;          // key < a[b + m]
+            else
+                lastOfs = m + 1;  // a[b + m] <= key
+        }
+        assert lastOfs == ofs;    // so a[b + ofs - 1] <= key < a[b + ofs]
+        return ofs;
+    }
+
+    /**
+     * Merges two adjacent runs in place, in a stable fashion.  The first
+     * element of the first run must be greater than the first element of the
+     * second run (a[base1] > a[base2]), and the last element of the first run
+     * (a[base1 + len1-1]) must be greater than all elements of the second run.
+     *
+     * For performance, this method should be called only when len1 <= len2;
+     * its twin, mergeHi should be called if len1 >= len2.  (Either method
+     * may be called if len1 == len2.)
+     *
+     * @param base1 index of first element in first run to be merged
+     * @param len1  length of first run to be merged (must be > 0)
+     * @param base2 index of first element in second run to be merged
+     *        (must be aBase + aLen)
+     * @param len2  length of second run to be merged (must be > 0)
+     */
+    @SuppressWarnings("unchecked")
+    private void mergeLo(int base1, int len1, int base2, int len2) {
+        assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
+
+        // Copy first run into temp array
+        Object[] a = this.a; // For performance
+        Object[] 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;
+        }
+
+        int minGallop = this.minGallop;  // Use local variable for performance
+    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 (((Comparable) a[cursor2]).compareTo(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((Comparable) a[cursor2], tmp, cursor1, len1, 0);
+                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((Comparable) tmp[cursor1], a, cursor2, len2, 0);
+                if (count2 != 0) {
+                    System.arraycopy(a, cursor2, a, dest, count2);
+                    dest += count2;
+                    cursor2 += count2;
+                    len2 -= count2;
+                    if (len2 == 0)
+                        break outer;
+                }
+                a[dest++] = tmp[cursor1++];
+                if (--len1 == 1)
+                    break outer;
+                minGallop--;
+            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
+            if (minGallop < 0)
+                minGallop = 0;
+            minGallop += 2;  // Penalize for leaving gallop mode
+        }  // End of "outer" loop
+        this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field
+
+        if (len1 == 1) {
+            assert len2 > 0;
+            System.arraycopy(a, cursor2, a, dest, len2);
+            a[dest + len2] = tmp[cursor1]; //  Last elt of run 1 to end of merge
+        } else if (len1 == 0) {
+            throw new IllegalArgumentException(
+                "Comparison method violates its general contract!");
+        } else {
+            assert len2 == 0;
+            assert len1 > 1;
+            System.arraycopy(tmp, cursor1, a, dest, len1);
+        }
+    }
+
+    /**
+     * Like mergeLo, except that this method should be called only if
+     * len1 >= len2; mergeLo should be called if len1 <= len2.  (Either method
+     * may be called if len1 == len2.)
+     *
+     * @param base1 index of first element in first run to be merged
+     * @param len1  length of first run to be merged (must be > 0)
+     * @param base2 index of first element in second run to be merged
+     *        (must be aBase + aLen)
+     * @param len2  length of second run to be merged (must be > 0)
+     */
+    @SuppressWarnings("unchecked")
+    private void mergeHi(int base1, int len1, int base2, int len2) {
+        assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
+
+        // Copy second run into temp array
+        Object[] a = this.a; // For performance
+        Object[] 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;
+        }
+
+        int minGallop = this.minGallop;  // Use local variable for performance
+    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 (((Comparable) tmp[cursor2]).compareTo(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((Comparable) tmp[cursor2], a, base1, len1, len1 - 1);
+                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((Comparable) a[cursor1], tmp, 0, len2, len2 - 1);
+                if (count2 != 0) {
+                    dest -= count2;
+                    cursor2 -= count2;
+                    len2 -= count2;
+                    System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
+                    if (len2 <= 1)
+                        break outer; // len2 == 1 || len2 == 0
+                }
+                a[dest--] = a[cursor1--];
+                if (--len1 == 0)
+                    break outer;
+                minGallop--;
+            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
+            if (minGallop < 0)
+                minGallop = 0;
+            minGallop += 2;  // Penalize for leaving gallop mode
+        }  // End of "outer" loop
+        this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field
+
+        if (len2 == 1) {
+            assert len1 > 0;
+            dest -= len1;
+            cursor1 -= len1;
+            System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
+            a[dest] = tmp[cursor2];  // Move first elt of run2 to front of merge
+        } else if (len2 == 0) {
+            throw new IllegalArgumentException(
+                "Comparison method violates its general contract!");
+        } else {
+            assert len1 == 0;
+            assert len2 > 0;
+            System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
+        }
+    }
+
+    /**
+     * Ensures that the external array tmp has at least the specified
+     * number of elements, increasing its size if necessary.  The size
+     * increases exponentially to ensure amortized linear time complexity.
+     *
+     * @param minCapacity the minimum required capacity of the tmp array
+     * @return tmp, whether or not it grew
+     */
+    private Object[]  ensureCapacity(int minCapacity) {
+        if (tmp.length < minCapacity) {
+            // Compute smallest power of 2 > minCapacity
+            int newSize = minCapacity;
+            newSize |= newSize >> 1;
+            newSize |= newSize >> 2;
+            newSize |= newSize >> 4;
+            newSize |= newSize >> 8;
+            newSize |= newSize >> 16;
+            newSize++;
+
+            if (newSize < 0) // Not bloody likely!
+                newSize = minCapacity;
+            else
+                newSize = Math.min(newSize, a.length >>> 1);
+
+            @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
+            Object[] newArray = new Object[newSize];
+            tmp = newArray;
+        }
+        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
+     */
+    private static void rangeCheck(int arrayLen, int fromIndex, int 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);
+    }
+}