//数组声明
dataType[] arrayRefVar;	//首选方法
dataType arrayRefVar[];	//效果相同，但是不推荐

//创建数组
arrayRefVar = new dataType[arraySize];	//初始化默认为dataType的default值
dataType[] arrayRefVar = {value0, value1, ..., valuek};	//创建的同时初始化，为静态定义

//数组长度
int i = arrayRefVar.length	//注意数组的length是一个属性，但是String的length是一个方法

//数组声明
dataType[][] arrayRefVar;

//创建数组和一维数组类似

//数组长度
int i = arrayTest.length;	//获取的是一维数组的数量，即第一维的个数
int j = arrayTest[0].length;	//获取的是第一个一维数组的长度，以dataType[][] arrayTest = new dataType[value0][value1]定义时为value1
/*ps:其实记多维数组只是为了这个多维数组的长度而已。。。。。*/

	/**
     * The minimum array length below which a parallel sorting
     * algorithm will not further partition the sorting task. Using
     * smaller sizes typically results in memory contention across
     * tasks that makes parallel speedups unlikely.
     */
    private static final int MIN_ARRAY_SORT_GRAN = 1 << 13;	//也就是最小的能够使用并行处理的数组长度

	/**
     * Sorts the specified array into ascending numerical order.
     *
     * <p>Implementation note: The sorting algorithm is a Dual-Pivot Quicksort
     * by Vladimir Yaroslavskiy, Jon Bentley, and Joshua Bloch. This algorithm
     * offers O(n log(n)) performance on many data sets that cause other
     * quicksorts to degrade to quadratic performance, and is typically
     * faster than traditional (one-pivot) Quicksort implementations.
     *
     * @param a the array to be sorted
     */
	// 升序排序特定数组
    public static void sort(int[] a) {
        DualPivotQuicksort.sort(a, 0, a.length - 1, null, 0, 0);	//Java排序使用的是双轴快排，而不是普通的快排
    }

    /**
     * Sorts the specified range of the array into ascending order. The range
     * to be sorted extends from the index {@code fromIndex}, inclusive, to
     * the index {@code toIndex}, exclusive. If {@code fromIndex == toIndex},
     * the range to be sorted is empty.
     *
     * <p>Implementation note: The sorting algorithm is a Dual-Pivot Quicksort
     * by Vladimir Yaroslavskiy, Jon Bentley, and Joshua Bloch. This algorithm
     * offers O(n log(n)) performance on many data sets that cause other
     * quicksorts to degrade to quadratic performance, and is typically
     * faster than traditional (one-pivot) Quicksort implementations.
     *
     * @param a the array to be sorted
     * @param fromIndex the index of the first element, inclusive, to be sorted
     * @param toIndex the index of the last element, exclusive, to be sorted
     *
     * @throws IllegalArgumentException if {@code fromIndex > toIndex}
     * @throws ArrayIndexOutOfBoundsException
     *     if {@code fromIndex < 0} or {@code toIndex > a.length}
     */
	// 排序特定范围内[fromIndex, toIndex)的数组
    public static void sort(int[] a, int fromIndex, int toIndex) {
        rangeCheck(a.length, fromIndex, toIndex);
        DualPivotQuicksort.sort(a, fromIndex, toIndex - 1, null, 0, 0);
    }


	//在1.8后，Java支持了并行排序，使用的是ArraysParallelSortHelpers类
	/**
     * Sorts the specified array into ascending numerical order.
     *
     * @implNote The sorting algorithm is a parallel sort-merge that breaks the
     * array into sub-arrays that are themselves sorted and then merged. When
     * the sub-array length reaches a minimum granularity, the sub-array is
     * sorted using the appropriate {@link Arrays#sort(byte[]) Arrays.sort}
     * method. If the length of the specified array is less than the minimum
     * granularity, then it is sorted using the appropriate {@link
     * Arrays#sort(byte[]) Arrays.sort} method. The algorithm requires a
     * working space no greater than the size of the original array. The
     * {@link ForkJoinPool#commonPool() ForkJoin common pool} is used to
     * execute any parallel tasks.
     *
     * @param a the array to be sorted
     *
     * @since 1.8
     */
    public static void parallelSort(byte[] a) {
        int n = a.length, p, g;
        if (n <= MIN_ARRAY_SORT_GRAN ||
            (p = ForkJoinPool.getCommonPoolParallelism()) == 1)
            DualPivotQuicksort.sort(a, 0, n - 1);
        else
            new ArraysParallelSortHelpers.FJByte.Sorter
                (null, a, new byte[n], 0, n, 0,
                 ((g = n / (p << 2)) <= MIN_ARRAY_SORT_GRAN) ?
                 MIN_ARRAY_SORT_GRAN : g).invoke();
    }

    /**
     * Sorts the specified range of the array into ascending numerical order.
     * The range to be sorted extends from the index {@code fromIndex},
     * inclusive, to the index {@code toIndex}, exclusive. If
     * {@code fromIndex == toIndex}, the range to be sorted is empty.
     *
     * @implNote The sorting algorithm is a parallel sort-merge that breaks the
     * array into sub-arrays that are themselves sorted and then merged. When
     * the sub-array length reaches a minimum granularity, the sub-array is
     * sorted using the appropriate {@link Arrays#sort(byte[]) Arrays.sort}
     * method. If the length of the specified array is less than the minimum
     * granularity, then it is sorted using the appropriate {@link
     * Arrays#sort(byte[]) Arrays.sort} method. The algorithm requires a working
     * space no greater than the size of the specified range of the original
     * array. The {@link ForkJoinPool#commonPool() ForkJoin common pool} is
     * used to execute any parallel tasks.
     *
     * @param a the array to be sorted
     * @param fromIndex the index of the first element, inclusive, to be sorted
     * @param toIndex the index of the last element, exclusive, to be sorted
     *
     * @throws IllegalArgumentException if {@code fromIndex > toIndex}
     * @throws ArrayIndexOutOfBoundsException
     *     if {@code fromIndex < 0} or {@code toIndex > a.length}
     *
     * @since 1.8
     */
    public static void parallelSort(byte[] a, int fromIndex, int toIndex) {
        rangeCheck(a.length, fromIndex, toIndex);
        int n = toIndex - fromIndex, p, g;
        if (n <= MIN_ARRAY_SORT_GRAN ||
            (p = ForkJoinPool.getCommonPoolParallelism()) == 1)
            DualPivotQuicksort.sort(a, fromIndex, toIndex - 1);
        else
            new ArraysParallelSortHelpers.FJByte.Sorter
                (null, a, new byte[n], fromIndex, n, 0,
                 ((g = n / (p << 2)) <= MIN_ARRAY_SORT_GRAN) ?
                 MIN_ARRAY_SORT_GRAN : g).invoke();
    }
	
	//并行堆叠，其实就是是一个从前往后做加法的运算
	/**
     * Cumulates, in parallel, each element of the given array in place,
     * using the supplied function. For example if the array initially
     * holds {@code [2, 1, 0, 3]} and the operation performs addition,
     * then upon return the array holds {@code [2, 3, 3, 6]}.
     * Parallel prefix computation is usually more efficient than
     * sequential loops for large arrays.
     *
     * @param <T> the class of the objects in the array
     * @param array the array, which is modified in-place by this method
     * @param op a side-effect-free, associative function to perform the
     * cumulation
     * @throws NullPointerException if the specified array or function is null
     * @since 1.8
     */
    public static <T> void parallelPrefix(T[] array, BinaryOperator<T> op) {
        Objects.requireNonNull(op);
        if (array.length > 0)
            new ArrayPrefixHelpers.CumulateTask<>
                    (null, op, array, 0, array.length).invoke();
    }

    /**
     * Performs {@link #parallelPrefix(Object[], BinaryOperator)}
     * for the given subrange of the array.
     *
     * @param <T> the class of the objects in the array
     * @param array the array
     * @param fromIndex the index of the first element, inclusive
     * @param toIndex the index of the last element, exclusive
     * @param op a side-effect-free, associative function to perform the
     * cumulation
     * @throws IllegalArgumentException if {@code fromIndex > toIndex}
     * @throws ArrayIndexOutOfBoundsException
     *     if {@code fromIndex < 0} or {@code toIndex > array.length}
     * @throws NullPointerException if the specified array or function is null
     * @since 1.8
     */
    public static <T> void parallelPrefix(T[] array, int fromIndex,
                                          int toIndex, BinaryOperator<T> op) {
        Objects.requireNonNull(op);
        rangeCheck(array.length, fromIndex, toIndex);
        if (fromIndex < toIndex)
            new ArrayPrefixHelpers.CumulateTask<>
                    (null, op, array, fromIndex, toIndex).invoke();
    }
	
	//fill，顾名思义，就是用来填充数组的，可以全填充，也可以部分填充，但是需要连续
	/**
     * Assigns the specified Object reference to each element of the specified
     * array of Objects.
     *
     * @param a the array to be filled
     * @param val the value to be stored in all elements of the array
     * @throws ArrayStoreException if the specified value is not of a
     *         runtime type that can be stored in the specified array
     */
    public static void fill(Object[] a, Object val) {
        for (int i = 0, len = a.length; i < len; i++)
            a[i] = val;
    }

    /**
     * Assigns the specified Object reference to each element of the specified
     * range of the specified array of Objects.  The range to be filled
     * extends from index <tt>fromIndex</tt>, inclusive, to index
     * <tt>toIndex</tt>, exclusive.  (If <tt>fromIndex==toIndex</tt>, the
     * range to be filled is empty.)
     *
     * @param a the array to be filled
     * @param fromIndex the index of the first element (inclusive) to be
     *        filled with the specified value
     * @param toIndex the index of the last element (exclusive) to be
     *        filled with the specified value
     * @param val the value to be stored in all elements of the array
     * @throws IllegalArgumentException if <tt>fromIndex &gt; toIndex</tt>
     * @throws ArrayIndexOutOfBoundsException if <tt>fromIndex &lt; 0</tt> or
     *         <tt>toIndex &gt; a.length</tt>
     * @throws ArrayStoreException if the specified value is not of a
     *         runtime type that can be stored in the specified array
     */
    public static void fill(Object[] a, int fromIndex, int toIndex, Object val) {
        rangeCheck(a.length, fromIndex, toIndex);
        for (int i = fromIndex; i < toIndex; i++)
            a[i] = val;
    }
	
	//copy，没什么好说的
	/**
     * Copies the specified array, truncating or padding with zeros (if necessary)
     * so the copy has the specified length.  For all indices that are
     * valid in both the original array and the copy, the two arrays will
     * contain identical values.  For any indices that are valid in the
     * copy but not the original, the copy will contain <tt>(byte)0</tt>.
     * Such indices will exist if and only if the specified length
     * is greater than that of the original array.
     *
     * @param original the array to be copied
     * @param newLength the length of the copy to be returned
     * @return a copy of the original array, truncated or padded with zeros
     *     to obtain the specified length
     * @throws NegativeArraySizeException if <tt>newLength</tt> is negative
     * @throws NullPointerException if <tt>original</tt> is null
     * @since 1.6
     */
    public static byte[] copyOf(byte[] original, int newLength) {
        byte[] copy = new byte[newLength];
        System.arraycopy(original, 0, copy, 0,
                         Math.min(original.length, newLength));
        return copy;
    }

	/**
     * Copies the specified range of the specified array into a new array.
     * The initial index of the range (<tt>from</tt>) must lie between zero
     * and <tt>original.length</tt>, inclusive.  The value at
     * <tt>original[from]</tt> is placed into the initial element of the copy
     * (unless <tt>from == original.length</tt> or <tt>from == to</tt>).
     * Values from subsequent elements in the original array are placed into
     * subsequent elements in the copy.  The final index of the range
     * (<tt>to</tt>), which must be greater than or equal to <tt>from</tt>,
     * may be greater than <tt>original.length</tt>, in which case
     * <tt>(byte)0</tt> is placed in all elements of the copy whose index is
     * greater than or equal to <tt>original.length - from</tt>.  The length
     * of the returned array will be <tt>to - from</tt>.
     *
     * @param original the array from which a range is to be copied
     * @param from the initial index of the range to be copied, inclusive
     * @param to the final index of the range to be copied, exclusive.
     *     (This index may lie outside the array.)
     * @return a new array containing the specified range from the original array,
     *     truncated or padded with zeros to obtain the required length
     * @throws ArrayIndexOutOfBoundsException if {@code from < 0}
     *     or {@code from > original.length}
     * @throws IllegalArgumentException if <tt>from &gt; to</tt>
     * @throws NullPointerException if <tt>original</tt> is null
     * @since 1.6
     */
    public static byte[] copyOfRange(byte[] original, int from, int to) {
        int newLength = to - from;
        if (newLength < 0)
            throw new IllegalArgumentException(from + " > " + to);
        byte[] copy = new byte[newLength];
        System.arraycopy(original, from, copy, 0,
                         Math.min(original.length - from, newLength));
        return copy;
    }

	//ArrayList是由数组实现的
	private static class ArrayList<E> extends AbstractList<E>
        implements RandomAccess, java.io.Serializable{...}
    
	//哈希化通过计算数值决定（真的不会出冲突么？？）
     /**
     * Returns a hash code based on the contents of the specified array.  If
     * the array contains other arrays as elements, the hash code is based on
     * their identities rather than their contents.  It is therefore
     * acceptable to invoke this method on an array that contains itself as an
     * element,  either directly or indirectly through one or more levels of
     * arrays.
     *
     * <p>For any two arrays <tt>a</tt> and <tt>b</tt> such that
     * <tt>Arrays.equals(a, b)</tt>, it is also the case that
     * <tt>Arrays.hashCode(a) == Arrays.hashCode(b)</tt>.
     *
     * <p>The value returned by this method is equal to the value that would
     * be returned by <tt>Arrays.asList(a).hashCode()</tt>, unless <tt>a</tt>
     * is <tt>null</tt>, in which case <tt>0</tt> is returned.
     *
     * @param a the array whose content-based hash code to compute
     * @return a content-based hash code for <tt>a</tt>
     * @see #deepHashCode(Object[])
     * @since 1.5
     */
    public static int hashCode(Object a[]) {
        if (a == null)
            return 0;

        int result = 1;

        for (Object element : a)
            result = 31 * result + (element == null ? 0 : element.hashCode());

        return result;
    }

	//deepEquals和deepHashCode只不过适配了Object[]类型的相等比较，其余差别不大

	//Arrays类支持mergeSort和binarySort，用法和sort差不多，此处未列出

	//使用给定空间排序，如果空间足够，则采用merge，左闭右闭
    /**
     * Sorts the specified range of the array using the given
     * workspace array slice if possible for merging
     *
     * @param a the array to be sorted
     * @param left the index of the first element, inclusive, to be sorted
     * @param right the index of the last element, inclusive, to be sorted
     * @param work a workspace array (slice)
     * @param workBase origin of usable space in work array
     * @param workLen usable size of work array
     */
    static void sort(int[] a, int left, int right,
                     int[] work, int workBase, int workLen) {
       	//小数组直接使用快排
        // Use Quicksort on small arrays
        if (right - left < QUICKSORT_THRESHOLD) {
            sort(a, left, right, true);
            return;
        }

        /*
         * Index run[i] is the start of i-th run
         * (ascending or descending sequence).
         */
        int[] run = new int[MAX_RUN_COUNT + 1];
        int count = 0; run[0] = left;

        // 将部分递增序列和递减序列提前排序，并记录趟数和趟数开始地址，用于后续归并排序
        // Check if the array is nearly sorted
        for (int k = left; k < right; run[count] = k) {
            if (a[k] < a[k + 1]) { // ascending
                while (++k <= right && a[k - 1] <= a[k]);
            } else if (a[k] > a[k + 1]) { // descending
                while (++k <= right && a[k - 1] >= a[k]);
                for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
                    int t = a[lo]; a[lo] = a[hi]; a[hi] = t;
                }
            } else { // equal
                for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
                    if (--m == 0) {
                        sort(a, left, right, true);
                        return;
                    }
                }
            }
			
            // 如果数组足够混乱，趟数超过一定数量，就分区进行快排
            /*
             * The array is not highly structured,
             * use Quicksort instead of merge sort.
             */
            if (++count == MAX_RUN_COUNT) {
                sort(a, left, right, true);
                return;
            }
        }

        // Check special cases
        // Implementation note: variable "right" is increased by 1.
        if (run[count] == right++) { // The last run contains one element
            run[++count] = right;
        } else if (count == 1) { // The array is already sorted
            return;
        }
		
        // 后续是使用缓冲区进行归并排序，不进行解释
        // Determine alternation base for merge
        byte odd = 0;
        for (int n = 1; (n <<= 1) < count; odd ^= 1);

        // Use or create temporary array b for merging
        int[] b;                 // temp array; alternates with a
        int ao, bo;              // array offsets from 'left'
        int blen = right - left; // space needed for b
        if (work == null || workLen < blen || workBase + blen > work.length) {
            work = new int[blen];
            workBase = 0;
        }
        if (odd == 0) {
            System.arraycopy(a, left, work, workBase, blen);
            b = a;
            bo = 0;
            a = work;
            ao = workBase - left;
        } else {
            b = work;
            ao = 0;
            bo = workBase - left;
        }

        // Merging
        for (int last; count > 1; count = last) {
            for (int k = (last = 0) + 2; k <= count; k += 2) {
                int hi = run[k], mi = run[k - 1];
                for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
                    if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
                        b[i + bo] = a[p++ + ao];
                    } else {
                        b[i + bo] = a[q++ + ao];
                    }
                }
                run[++last] = hi;
            }
            if ((count & 1) != 0) {
                for (int i = right, lo = run[count - 1]; --i >= lo;
                    b[i + bo] = a[i + ao]
                );
                run[++last] = right;
            }
            int[] t = a; a = b; b = t;
            int o = ao; ao = bo; bo = o;
        }
    }

    /**
     * Sorts the specified range of the array by Dual-Pivot Quicksort.
     *
     * @param a the array to be sorted
     * @param left the index of the first element, inclusive, to be sorted
     * @param right the index of the last element, inclusive, to be sorted
     * @param leftmost indicates if this part is the leftmost in the range
     */
    private static void sort(int[] a, int left, int right, boolean leftmost) {
        int length = right - left + 1;
		
        // 对小数组使用插入排序，源码给的数为47
        // Use insertion sort on tiny arrays
        if (length < INSERTION_SORT_THRESHOLD) {
            if (leftmost) {
                /*
                 * Traditional (without sentinel) insertion sort,
                 * optimized for server VM, is used in case of
                 * the leftmost part.
                 */
                for (int i = left, j = i; i < right; j = ++i) {
                    int ai = a[i + 1];
                    while (ai < a[j]) {
                        a[j + 1] = a[j];
                        if (j-- == left) {
                            break;
                        }
                    }
                    a[j + 1] = ai;
                }
            } else {
                /*
                 * Skip the longest ascending sequence.
                 */
                do {
                    if (left >= right) {
                        return;
                    }
                } while (a[++left] >= a[left - 1]);

                /*
                 * Every element from adjoining part plays the role
                 * of sentinel, therefore this allows us to avoid the
                 * left range check on each iteration. Moreover, we use
                 * the more optimized algorithm, so called pair insertion
                 * sort, which is faster (in the context of Quicksort)
                 * than traditional implementation of insertion sort.
                 */
                for (int k = left; ++left <= right; k = ++left) {
                    int a1 = a[k], a2 = a[left];

                    if (a1 < a2) {
                        a2 = a1; a1 = a[left];
                    }
                    while (a1 < a[--k]) {
                        a[k + 2] = a[k];
                    }
                    a[++k + 1] = a1;

                    while (a2 < a[--k]) {
                        a[k + 1] = a[k];
                    }
                    a[k + 1] = a2;
                }
                int last = a[right];

                while (last < a[--right]) {
                    a[right + 1] = a[right];
                }
                a[right + 1] = last;
            }
            return;
        }

        // Inexpensive approximation of length / 7
        int seventh = (length >> 3) + (length >> 6) + 1;

        /*
         * Sort five evenly spaced elements around (and including) the
         * center element in the range. These elements will be used for
         * pivot selection as described below. The choice for spacing
         * these elements was empirically determined to work well on
         * a wide variety of inputs.
         */
        int e3 = (left + right) >>> 1; // The midpoint
        int e2 = e3 - seventh;
        int e1 = e2 - seventh;
        int e4 = e3 + seventh;
        int e5 = e4 + seventh;

        // Sort these elements using insertion sort
        if (a[e2] < a[e1]) { int t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

        if (a[e3] < a[e2]) { int t = a[e3]; a[e3] = a[e2]; a[e2] = t;
            if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
        }
        if (a[e4] < a[e3]) { int t = a[e4]; a[e4] = a[e3]; a[e3] = t;
            if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
            }
        }
        if (a[e5] < a[e4]) { int t = a[e5]; a[e5] = a[e4]; a[e4] = t;
            if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
                if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
                }
            }
        }

        // Pointers
        int less  = left;  // The index of the first element of center part
        int great = right; // The index before the first element of right part

        if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
            /*
             * Use the second and fourth of the five sorted elements as pivots.
             * These values are inexpensive approximations of the first and
             * second terciles of the array. Note that pivot1 <= pivot2.
             */
            int pivot1 = a[e2];
            int pivot2 = a[e4];

            /*
             * The first and the last elements to be sorted are moved to the
             * locations formerly occupied by the pivots. When partitioning
             * is complete, the pivots are swapped back into their final
             * positions, and excluded from subsequent sorting.
             */
            a[e2] = a[left];
            a[e4] = a[right];

            /*
             * Skip elements, which are less or greater than pivot values.
             */
            while (a[++less] < pivot1);
            while (a[--great] > pivot2);

            /*
             * Partitioning:
             *
             *   left part           center part                   right part
             * +--------------------------------------------------------------+
             * |  < pivot1  |  pivot1 <= && <= pivot2  |    ?    |  > pivot2  |
             * +--------------------------------------------------------------+
             *               ^                          ^       ^
             *               |                          |       |
             *              less                        k     great
             *
             * Invariants:
             *
             *              all in (left, less)   < pivot1
             *    pivot1 <= all in [less, k)     <= pivot2
             *              all in (great, right) > pivot2
             *
             * Pointer k is the first index of ?-part.
             */
            outer:
            for (int k = less - 1; ++k <= great; ) {
                int ak = a[k];
                if (ak < pivot1) { // Move a[k] to left part
                    a[k] = a[less];
                    /*
                     * Here and below we use "a[i] = b; i++;" instead
                     * of "a[i++] = b;" due to performance issue.
                     */
                    a[less] = ak;
                    ++less;
                } else if (ak > pivot2) { // Move a[k] to right part
                    while (a[great] > pivot2) {
                        if (great-- == k) {
                            break outer;
                        }
                    }
                    if (a[great] < pivot1) { // a[great] <= pivot2
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // pivot1 <= a[great] <= pivot2
                        a[k] = a[great];
                    }
                    /*
                     * Here and below we use "a[i] = b; i--;" instead
                     * of "a[i--] = b;" due to performance issue.
                     */
                    a[great] = ak;
                    --great;
                }
            }

            // Swap pivots into their final positions
            a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
            a[right] = a[great + 1]; a[great + 1] = pivot2;

            // Sort left and right parts recursively, excluding known pivots
            sort(a, left, less - 2, leftmost);
            sort(a, great + 2, right, false);

            /*
             * If center part is too large (comprises > 4/7 of the array),
             * swap internal pivot values to ends.
             */
            if (less < e1 && e5 < great) {
                /*
                 * Skip elements, which are equal to pivot values.
                 */
                while (a[less] == pivot1) {
                    ++less;
                }

                while (a[great] == pivot2) {
                    --great;
                }

                /*
                 * Partitioning:
                 *
                 *   left part         center part                  right part
                 * +----------------------------------------------------------+
                 * | == pivot1 |  pivot1 < && < pivot2  |    ?    | == pivot2 |
                 * +----------------------------------------------------------+
                 *              ^                        ^       ^
                 *              |                        |       |
                 *             less                      k     great
                 *
                 * Invariants:
                 *
                 *              all in (*,  less) == pivot1
                 *     pivot1 < all in [less,  k)  < pivot2
                 *              all in (great, *) == pivot2
                 *
                 * Pointer k is the first index of ?-part.
                 */
                outer:
                for (int k = less - 1; ++k <= great; ) {
                    int ak = a[k];
                    if (ak == pivot1) { // Move a[k] to left part
                        a[k] = a[less];
                        a[less] = ak;
                        ++less;
                    } else if (ak == pivot2) { // Move a[k] to right part
                        while (a[great] == pivot2) {
                            if (great-- == k) {
                                break outer;
                            }
                        }
                        if (a[great] == pivot1) { // a[great] < pivot2
                            a[k] = a[less];
                            /*
                             * Even though a[great] equals to pivot1, the
                             * assignment a[less] = pivot1 may be incorrect,
                             * if a[great] and pivot1 are floating-point zeros
                             * of different signs. Therefore in float and
                             * double sorting methods we have to use more
                             * accurate assignment a[less] = a[great].
                             */
                            a[less] = pivot1;
                            ++less;
                        } else { // pivot1 < a[great] < pivot2
                            a[k] = a[great];
                        }
                        a[great] = ak;
                        --great;
                    }
                }
            }

            // Sort center part recursively
            sort(a, less, great, false);

        } else { // Partitioning with one pivot
            /*
             * Use the third of the five sorted elements as pivot.
             * This value is inexpensive approximation of the median.
             */
            int pivot = a[e3];

            /*
             * Partitioning degenerates to the traditional 3-way
             * (or "Dutch National Flag") schema:
             *
             *   left part    center part              right part
             * +-------------------------------------------------+
             * |  < pivot  |   == pivot   |     ?    |  > pivot  |
             * +-------------------------------------------------+
             *              ^              ^        ^
             *              |              |        |
             *             less            k      great
             *
             * Invariants:
             *
             *   all in (left, less)   < pivot
             *   all in [less, k)     == pivot
             *   all in (great, right) > pivot
             *
             * Pointer k is the first index of ?-part.
             */
            for (int k = less; k <= great; ++k) {
                if (a[k] == pivot) {
                    continue;
                }
                int ak = a[k];
                if (ak < pivot) { // Move a[k] to left part
                    a[k] = a[less];
                    a[less] = ak;
                    ++less;
                } else { // a[k] > pivot - Move a[k] to right part
                    while (a[great] > pivot) {
                        --great;
                    }
                    if (a[great] < pivot) { // a[great] <= pivot
                        a[k] = a[less];
                        a[less] = a[great];
                        ++less;
                    } else { // a[great] == pivot
                        /*
                         * Even though a[great] equals to pivot, the
                         * assignment a[k] = pivot may be incorrect,
                         * if a[great] and pivot are floating-point
                         * zeros of different signs. Therefore in float
                         * and double sorting methods we have to use
                         * more accurate assignment a[k] = a[great].
                         */
                        a[k] = pivot;
                    }
                    a[great] = ak;
                    --great;
                }
            }

            /*
             * Sort left and right parts recursively.
             * All elements from center part are equal
             * and, therefore, already sorted.
             */
            sort(a, left, less - 1, leftmost);
            sort(a, great + 1, right, false);
        }
    }