Arrays排序原理

Java Arrays排序原理

计数排序源码

(short为例)

​

for (int i = left - 1; ++i <= right;count[a[i] - Short.MIN_VALUE]++);

​

for (int i = NUM_SHORT_VALUES, k = right + 1; k > left; ) {

while (count[--i] == 0);

short value = (short) (i + Short.MIN_VALUE);

int s = count[i];

​

do {

a[--k] = value;

} while (--s > 0);

}

插入排序源码

(int为例)

/*传统插排(无哨兵Sentinel)

* 遍历

* 循环向左比较(

*/

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 {

//如果一开始就是排好序的——直接返回

do {

if (left >= right) {

return;

}

} while (a[++left] >= a[left - 1]);

​

//优化插排(以两个为单位遍历，大的元素充当哨兵，以减少小的元素循环向左比较的范围)

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;

}

//确保最后一个元素被排序

int last = a[right];

​

while (last < a[--right]) {

a[right + 1] = a[right];

}

a[right + 1] = last;

}

return;

快排源码

(int为例)

// 快排阈值是286 其7分之一小于等于1/8+1/64+1

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

​

// 获取分成7份的五个中间点

int e3 = (left + right) >>> 1; // The midpoint

int e2 = e3 - seventh;

int e1 = e2 - seventh;

int e4 = e3 + seventh;

int e5 = e4 + seventh;

​

// 保证中间点的元素从小到大排序

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; }

}

}

}

​

// 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

​

//点彼此不相等——分三段快排，否则分两段

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];

​

while (a[++less] < pivot1);

while (a[--great] > pivot2);

​

/*

* Partitioning:

*

* left part center part right part

* +--------------------------------------------------------------+

* | < pivot1 | pivot1 <= && <= pivot2 | ? | > pivot2 |

* +--------------------------------------------------------------+

* ^ ^ ^

* | | |

* less k great

*/

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);

}

归并排序前置处理

//判断结构是否适合归并排序

int[] run = new int[MAX_RUN_COUNT + 1];

int count = 0; run[0] = left;

​

// 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 {

//连续MAX_RUN_LENGTH(33)个相等元素，使用快排

for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {

if (--m == 0) {

sort(a, left, right, true);

return;

}

}

}

​

//count达到MAX_RUN_LENGTH，使用快排

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;

}

归并排序源码

(int为例)

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;

}