归并排序

#include<stdio.h>
#include<stdlib.h>
#include<limits.h>
void merge(int a[], int p, int q, int r);
void merge_sort(int a[], int p, int r);
int main() {
    int a[10] = { 6,5, 4, 3, 1, 2, 0, 7,8,9};
    int i = 0;
    /*用来测试merge函数的一个实例*

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

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

    printf("Three indexes are %d, %d and %d.
The first subarray is:", idxa, idxb, idxc);
    for (i = idxa; i <= idxb; i++)
        printf(" %d", a[i]);
    printf("
The second subarray is:");
    for (i = idxb + 1; i <= idxc; i++)
        printf(" %d", a[i]);
    printf("
");

    merge(a, idxa, idxb, idxc);
    printf("The merged array is:");
    for (i = idxa; i <= idxc; i++)
        printf(" %d", a[i]);
    printf("
");
    getchar();
    *测试用例到此处结束*/

    merge_sort(a, 0, 9);
    printf("The sorted array is:");
    for (i = 0; i < 10; i++)
        printf(" %d", a[i]);
    printf("
");
    getchar();
    return 0;
}
void merge_sort(int a[], int p, int r){
    int q = 0;
    if (p < r) {
        q = (p + r) / 2;    //由于整数精度限制，自动对q进行了下取整
        merge_sort(a, p, q);
        merge_sort(a, q + 1, r);
        merge(a, p, q, r);
    }
}
void merge(int a[], int p, int q, int r) {
    /*无聊的声明，用于循环等操作*/
    int i = 0, j = 0, k = 0;
    /*声明需要的存储空间来存放数据*/
    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
    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
    /*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）*/
    int *left  = (int *)malloc(sizeof(int) * (n1 + 1));    //申请一段存放左侧数组数据的空间
    int *right = (int *)malloc(sizeof(int) * (n2 + 1));    //申请一段存放左侧数组数据的空间
    /*初始化Left 和 Right*/
    for (i = 0; i < n1; i++) {
        *(left + i) = a[p + i];
    }
    for (j = 0; j < n2; j++) {
        *(right + j) = a[q + j + 1];
    }
    *(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
    *(right + n2) = INT_MAX;

    i = 0;            //无聊的初始化无聊变量，用于循环
    j = 0;
    /*从Left和Right中选小的一个合并到一起*/
    for (k = p; k <= r; k++) {
        if (*(left + i) <= *(right + j)) {
            a[k] = *(left + i);
            i++;
        }
        else {
            a[k] = *(right + j);
            j++;
        }
    }
    free(left);
    free(right);
}