快速排序使用的是分治思想，将原问题分成若干个子问题进行递归解决。选择一个元素作为轴(pivot)，通过一趟排序将要排序的数据分割成独立的两部分，其中一部分的所有数据都比轴元素小，另外一部分的所有数据都比轴元素大，然后再按此方法对这两部分数据分别进行快速排序，整个排序过程可以递归进行，以此达到整个数据变成有序序列。
双轴快排(DualPivotQuicksort)，顾名思义有两个轴元素pivot1，pivot2，且pivot ≤ pivot2，将序列分成三段：x < pivot1、pivot1 ≤ x ≤ pivot2、x > pivot2，然后分别对三段进行递归。这个算法通常会比传统的快排效率更高，也因此被作为Arrays.java中给基本类型的数据排序的具体实现。
下面我们以JDK1.8中Arrays对int型数组的排序为例来介绍其中使用的双轴快排：
Step1.判断数组的长度是否大于286，大于则使用归并排序(merge sort)，否则执行step2。
// Use Quicksort on small arrays
if (right - left < QUICKSORT_THRESHOLD) {
sort(a, left, right, true);
return;
}
// Merge sort
......
Step2.判断数组长度是否小于47，小于则直接采用插入排序(insertion sort)，否则执行step3。
// Use insertion sort on tiny arrays
if (length < INSERTION_SORT_THRESHOLD) {
// Insertion sort
......
}
Step3.用公式length/8+length/64+1近似计算出数组长度的1/7。
// Inexpensive approximation of length / 7
int seventh = (length >> 3) + (length >> 6) + 1;
Step4.取5个根据经验得出的等距点
/*
* 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;
Step5.将这5个元素进行插入排序
// 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; }
}
}
}
Step6.选取a[e2]，a[e4]分别作为pivot1，pivot2。由于步骤5进行了排序，所以必有pivot1 <= pivot2。定义两个指针less和great，less从最左边开始向右遍历，一直找到第一个不小于pivot1的元素，great从右边开始向左遍历，一直找到第一个不大于pivot2的元素。
/*
* 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);
Step7.接着定义指针k从less-1开始向右遍历至great，把小于pivot1的元素移动到less左边，大于pivot2的元素移动到great右边。这里要注意，我们已知great处的元素小于pivot2，但是它于pivot1的大小关系，还需要进行判断，如果比pivot1还小，需要移动到到less左边，否则只需要交换到k处。
/*
* 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;
}
}
Step8.将less-1处的元素移动到队头，great+1处的元素移动到队尾，并把pivot1和pivot2分别放到less-1和great+1处。
// Swap pivots into their final positions
a[left] = a[less - 1]; a[less - 1] = pivot1;
a[right] = a[great + 1]; a[great + 1] = pivot2;
Step9.至此，less左边的元素都小于pivot1，great右边的元素都大于pivot2，分别对两部分进行同样的递归排序。
// Sort left and right parts recursively, excluding known pivots
sort(a, left, less - 2, leftmost);
sort(a, great + 2, right, false);
Step10.对于中间的部分，如果大于4/7的数组长度，很可能是因为重复元素的存在，所以把less向右移动到第一个不等于pivot1的地方，把great向左移动到第一个不等于pivot2的地方，然后再对less和great之间的部分进行递归排序。
/*
* 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;
}
}
......
// Sort center part recursively
sort(a, less, great, false);
总结：与《算法导论》中给出的快排算法相比，Arrays.sort对升序数组、降序数组和重复数组的排序效率有了很大的提升，这里面有几个重大的优化。1.对于小数组来说，插入排序效率更高，每次递归到小于47的大小时，用插入排序代替快排，明显提升了性能。2.双轴快排使用两个pivot，每轮把数组分成3段，在没有明显增加比较次数的情况下巧妙地减少了递归次数。3.pivot的选择上增加了随机性，却没有带来随机数的开销。4.对重复数据进行了优化处理，避免了不必要交换和递归。