归并排序

 1 #include<stdio.h>

 2 #include<stdlib.h>

 3 #include<limits.h>

 4 void merge(int a[], int p, int q, int r);

 5 void merge_sort(int a[], int p, int r);

 6 int main() {

 7     int a[10] = { 6,5, 4, 3, 1, 2, 0, 7,8,9};

 8     int i = 0;

 9     /*用来测试merge函数的一个实例*

10 

11     int idxa = 0, idxb = 4, idxc = 9;

12 

13     printf("Please input 10 numbers to the array:");

14 

15     printf("Three indexes are %d, %d and %d.\nThe first subarray is:", idxa, idxb, idxc);

16     for (i = idxa; i <= idxb; i++)

17         printf(" %d", a[i]);

18     printf("\nThe second subarray is:");

19     for (i = idxb + 1; i <= idxc; i++)

20         printf(" %d", a[i]);

21     printf("\n");

22 

23     merge(a, idxa, idxb, idxc);

24     printf("The merged array is:");

25     for (i = idxa; i <= idxc; i++)

26         printf(" %d", a[i]);

27     printf("\n");

28     getchar();

29     *测试用例到此处结束*/

30 

31     merge_sort(a, 0, 9);

32     printf("The sorted array is:");

33     for (i = 0; i < 10; i++)

34         printf(" %d", a[i]);

35     printf("\n");

36     getchar();

37     return 0;

38 }

39 void merge_sort(int a[], int p, int r){

40     int q = 0;

41     if (p < r) {

42         q = (p + r) / 2;    //由于整数精度限制，自动对q进行了下取整

43         merge_sort(a, p, q);

44         merge_sort(a, q + 1, r);

45         merge(a, p, q, r);

46     }

47 }

48 void merge(int a[], int p, int q, int r) {

49     /*无聊的声明，用于循环等操作*/

50     int i = 0, j = 0, k = 0;

51     /*声明需要的存储空间来存放数据*/

52     int n1 = q - p + 1;               //设置子数组A[p...q],    此时需要q-p+1的大小的数组

53     int n2 = r - q;                   //设置子数组A[q+1...r],    此时需要r - (q+1) + 1的大小的数组

54     /*令L[n1 + 1]和R[n2 + 1]成为新的数组，其中 +1 是为了放置哨兵牌（一个代表极值的数）*/

55     int *left  = (int *)malloc(sizeof(int) * (n1 + 1));    //申请一段存放左侧数组数据的空间

56     int *right = (int *)malloc(sizeof(int) * (n2 + 1));    //申请一段存放左侧数组数据的空间

57     /*初始化Left 和 Right*/

58     for (i = 0; i < n1; i++) {

59         *(left + i) = a[p + i];

60     }

61     for (j = 0; j < n2; j++) {

62         *(right + j) = a[q + j + 1];

63     }

64     *(left    + n1) = INT_MAX;        //将最后一个设置为哨兵牌,因为是从小到大排序，所以将哨兵牌设为整数最大值

65     *(right + n2) = INT_MAX;

66 

67     i = 0;            //无聊的初始化无聊变量，用于循环

68     j = 0;

69     /*从Left和Right中选小的一个合并到一起*/

70     for (k = p; k <= r; k++) {

71         if (*(left + i) <= *(right + j)) {

72             a[k] = *(left + i);

73             i++;

74         }

75         else {

76             a[k] = *(right + j);

77             j++;

78         }

79     }

80     free(left);

81     free(right);

82 }