排序算法在JDK中的应用（二）快速排序

/**

* Sorts the  specified range of the array by Dual-Pivot Quicksort.

* 通过双轴快速排序对指定范围内的数据进行排序

* @param a  the array to be sorted 被排序的数组

* @param  left the index of the first element, inclusive, to be sorted 需要排序的第一个元素的位置（包括在内）

* @param  right the index of the last element, inclusive, to be sorted 需要排序的最后一个元素的位置（包括在内）

* @param  leftmost indicates if this part is the leftmost in the range leftmost表示该部分是否是范围内最左的部分

*/

private  static void sort(int[] a, int left, int right, boolean leftmost) {

int  length = right - left + 1;

// Use  insertion sort on tiny arrays

//当数组的长度很小时就是用插入排序，INSERTION_SORT_THRESHOLD=47

if  (length < INSERTION_SORT_THRESHOLD) {

if  (leftmost) {

/*

* Traditional (without sentinel) insertion sort, 传统的插入排序，不使用哨兵元素

* optimized for server VM, is used in case of 针对最左边的部分的情况进行了服务器虚拟机的优化

* 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. 跳过最长的升序情况，提高算法效率

*/

do {

if (left >= right) {

return;

}

}  while (a[++left] >= a[left - 1]);

/*

* Every element from adjoining part plays the role   在这种排序方法中相邻的每个元素都起到了哨兵的作用

* of sentinel, therefore this allows us to avoid the  这种办法可以避免我们每次迭代时都要进行左范围检查。

* left range check on each iteration. Moreover, we use 而且我们还使用了一个效率更好的算法，我们称之为“双插入排序”，

* the more optimized algorithm, so called pair insertion 在快速排序的上下文中（即满足进入sort()方法的数组）他比传统的

* sort, which is faster (in the context of Quicksort) 插入排序更快

* 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;

}

//从这里开始是对待排序的元素进行分组处理

//  Inexpensive approximation of length / 7

// 使用length/7作为近似的加权长度

int  seventh = (length >> 3) + (length >> 6) + 1;

/*

* Sort  five evenly spaced elements around (and including) the 在范围内的中心元素附近找到5个均匀间隔的元素

* center  element in the range. These elements will be used for 这些元素将用于下面代码中的枢轴选择

* pivot  selection as described below. The choice for spacing 根据经验，这些元素的间距能够很好的应对和处理各种各样的输入（待排序的数组）

* 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 使用插入排序对这些元素进行排序

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; }

}

}

}//分组完成

//  Pointers 指针

int  less  = left;  // The index of the first element of center  part 中心部分第一个元素的位置

int great  = right; // The index before the first element of right part 右边第一个元素之前的位置

//五个分位点的值各不相同

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. 使用五个分位点中的第二个和第四个作为枢轴

*  These values are inexpensive approximations of the first and 因为有上面的排序 所以在这里pivot1 <= pivot2

*  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 将要排序的第一个和最后一个元素换到枢轴的位置

*  locations formerly occupied by the pivots. When partitioning 当分区操作完成后，枢轴元素将和这个元素交换回到原来的位置

* is  complete, the pivots are swapped back into their final 并排除到后续排序之外

*  positions, and excluded from subsequent sorting.

*/

a[e2]  = a[left];

a[e4]  = a[right];

/*

*  Skip elements, which are less or greater than pivot values.

* 筛选那些比枢轴元素更大或者更小的元素 以此来确定less和great的位置

*/

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.

* 以下的forless-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。       

*/

//outer标签

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.

* 在这里分开写的原因是因为前者效率更佳

* 想要了解的可以看这里的讨论：https://www.oschina.net/question/3037675_2206753

*/

a[less] = ak;

++less;

}  else if (ak > pivot2) { // Move a[k] to right part

while (a[great] > pivot2) {

if (great-- == k) {

break outer;

}

}

//通过上面已知great<pivot2，但是并不知道great和pivot1的大小关系，

//如果它比pivot1还小，需要移动到到less左边，否则只需要交换到k处。

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

// 将less-1的元素放到对头，great+1的元素放在队尾

//然后将pivot1放在less-1,pivot2放在great+1

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

//递归左右部分进行排序 包括已知的轴心

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.

* 如果中心部分太大(大小超过了整个数组的4/7),就将内部的枢轴值交换到端点

*/

if  (less < e1 && e5 < great) {

/*

* Skip elements, which are equal to pivot values.

* 调整less和great指针的位置，即跳过相等的元素

*/

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

*这下面的排序和上面的区别在于边界条件的判断

* 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];

/*

* 浮点零的相关问题：https://stackoverflow.com/questions/13544342/why-do-floating-point-numbers-have-signed-zeros

* Even though  a[great] equals to pivot1, the 如果a[great]和pivot1是不同符号的浮点零，

* assignment  a[less] = pivot1 may be incorrect, 即使a[great]这个元素等于pivot1，“a[less] = pivot1”这个赋值操作也可能是不正确的

* if a[great]  and pivot1 are floating-point zeros （这里是对单双精度元素排序的一个特例处理，使-0排在+0之前）

* of different signs.  Therefore in float and 因此在单双精度的排序算法中我们必须使用更加精确的赋值即a[less]=a[great]

* 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

// 对中心部分进行递归排序

sort(a, less, great, false);

} else {  // Partitioning with one pivot 采用单轴分区 区别于上面那种情况

/*

*  Use the third of the five sorted elements as pivot. 使用5个排序好的元素中的第三个作为枢轴元素

*  This value is inexpensive approximation of the median. 这个值是数组的中值近似值

*/

int pivot  = a[e3];

/*

*  Partitioning degenerates to the traditional 3-way 分区方式退化为传统的3路形式

*  (or "Dutch National Flag") schema:（或者“荷兰国旗”模式）

* 荷兰国旗问题（Dutch National Flag Problem）:https://www.cnblogs.com/freelancy/p/7940803.html

*   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

* 以下算法与上面算法思路差不多 不再累述

* 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.对左右部分进行递归排序

*  All elements from center part are equal 中间的元素都相等，所以已经排序

*  and, therefore, already sorted.

*/

sort(a, left, less - 1, leftmost);

sort(a, great + 1, right, false);

}

}

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