Arrays.sort() ----- DualPivotQuicksort
DualPivotQuicksort是Arrays.sort()对基本类型的排序算法，它不止使用了双轴快速排序，还使用了TimSort、插入排序、成对插入排序、3-way快速排序。
算法介绍
成对插入排序
具体执行过程：
- 将要插入的数据，第一个值赋值a1,第二个值赋值a2
- 然后判断a1与a2的大小，使a1要大于a2
- 接下来，首先是插入大的数值a1，将a1与k之前的数字一一比较，直到数值小于a1为止，把a1插入到合适的位置，注意：比a1大的值右移2位
- 接下来，插入小的数值a2,将a2与此时k之前的数字一一比较，直到数值小于a2为止，将a2插入到合适的位置，注意：比a2大的值右移1位
- 最后把最后一个没有遍历到的数据插入到合适位置
3-way快速排序
具体执行过程：
- a[k] < pivot 交换a[less]和a[k]，然后less和k都自增1，k继续扫描
- a[k] = pivot k自增1，k接着继续扫描
- a[k] > pivot 交换a[great]和a[k],但是我们不能直接将a[k]与a[great]交换，因为目前a[great]和pivot的关系未知，所以我们这个时候应该从great的位置自右向左扫描。而a[great]与pivot的关系可以继续分为三种情况讨论:
- a[great] > pivot great自减1，great接着继续扫描
- a[great] == pivot 交换a[k]和a[great]，k自增1，great自减1，k继续扫描（注意此时great的扫描就结束了）
- a[great] < pivot: 此时我们注意到a[great] < pivot, a[k] > pivot, a[less] == pivot，那么我们只需要将a[great]放到a[less]上，a[k]放到a[great]上，而a[less]放到a[k]上。然后less和k自增1，great自减1，k继续扫描（注意此时great的扫描就结束了）
- a[great] > pivot great自减1，great接着继续扫描
双轴快速排序
具体执行过程：
- a[k] < pivot1 less先自增，交换a[less]和a[k]，k自增1，k接着继续扫描
- a[k] >= pivot1 && a[k] <= pivot2 k自增1，k接着继续扫描
- a[k] > pivot2: 这个时候显然a[k]应该放到最右端大于pivot2的部分。但此时，我们不能直接将a[k]与a[great]交换,因为目前a[great]和pivot1以及pivot2的关系未知，所以我们这个时候应该从great自右向左扫描。而a[great]与pivot1和pivot2的关系可以继续分为三种情况讨论
- a[great] > pivot2 j接着继续扫描
- a[great] >= pivot1且a[great] <= pivot2 交换a[k]和a[great]，great自减1，k自增1，k继续扫描（注意此时great的扫描就结束了）
- a[great] < pivot1 先将i自增1，此时我们注意到a[great] < pivot1, a[k] > pivot2, pivot1 <= a[less] <=pivot2，那么我们只需要将a[great]放到a[less]上，a[k]放到a[great]上，而a[less]放到a[k]上。k自增1，great自减1，然后k继续扫描（此时great的扫描就结束了）
- a[great] > pivot2 j接着继续扫描
注意
- pivot1和pivot2在始终不参与k，greate扫描过程。
- 扫描结束时，a[less-1]表示了小于pivot1部分的最后一个元素，A[greate+1]表示了大于pivot2的第一个元素，这时我们只需要交换pivot1（即A[L]）和A[less-1]，交换pivot2（即A[R]）与A[greate+1]，同时我们可以确定A[less-1]和A[greate+1]所在的位置在后续的排序过程中不会发生变化（这一步非常重要，否则可能引起无限递归导致的栈溢出），最后我们只需要对A[L, less-2]，A[less, great]，A[great+1, R]这三个部分继续递归上述操作即可。
TimSort
TimSort可以看另一篇文章：Arrays.sort() ----- TimSort
jdk1.8源码
static void sort(int[] a, int left, int right,
int[] work, int workBase, int workLen) {
// Use Quicksort on small arrays
if (right - left < QUICKSORT_THRESHOLD) {
sort(a, left, right, true);
return;
}
//大于阈值使用阉割版的TimSort，去掉了自适应
........
}
private static void sort(int[] a, int left, int right, boolean leftmost) {
int length = right - left + 1;
// 小于一个阈值的使用插入排序
if (length < INSERTION_SORT_THRESHOLD) {
if (leftmost) {//位于整个数组最左边，使用传统插入排序
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 {//成对插入排序
/*
* 跳过左边已排序的部分
*/
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]) {//双轴快速排序
/*
* 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);
//如果中间区域太大，使用双轴的思想，优化中间区域
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 { //使用3-ways快速排序
/*
* 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);
}
}参考资料
https://www.cnblogs.com/nullzx/p/5880191.html
https://www.jianshu.com/p/6d26d525bb96