归并排序算法

    归并排序，它采取分而治之（Divide-and-Conquer）的策略，时间复杂度是Θ(nlgn)。归并排序的步骤如下：
1. Divide: 把长度为n的输入序列分成两个长度为n/2的子序列。
2. Conquer: 对这两个子序列分别采用归并排序。
3. Combine: 将两个排序好的子序列合并成一个最终的排序序列。
在描述归并排序的步骤时又调用了归并排序本身，可见这是一个递归的过程。

#include

#define LEN 8
int a[LEN] = { 5, 2, 4, 7, 1, 3, 2, 6 };

void merge(int start, int mid, int end)
{
    int n1 = mid - start 1;
    int n2 = end - mid;
    int left[n1], right[n2];
    int i, j, k;

    for (i = 0; i < n1; i++) /* left holds a[start..mid] */
    left[i] = a[start+i];

    for (j = 0; j < n2; j++) /* right holds a[mid 1..end] */
    right[j] = a[mid+1+j];

    i = j = 0;
    k = start;

    while (i < n1 && j < n2)
        if (left[i] < right[j])
            a[k++] = left[i++];
        else
            a[k++] = right[j++];

    while (i < n1) /* left[] is not exhausted */
        a[k++] = left[i++];
    while (j < n2) /* right[] is not exhausted */
        a[k++] = right[j++];
}

void sort(int start, int end)
{
    int mid;

    if (start < end)
    {
        mid = (start+end) / 2;
        printf("sort (%d-%d, %d-%d) %d %d %d %d %d %d %d %d\n", start, mid, mid+1,
                end, a[0], a[1], a[2], a[3], a[4], a[5], a[6], a[7]);
        sort(start, mid);
        sort(mid+1, end);
        merge(start, mid, end);
        printf("merge (%d-%d, %d-%d) to %d %d %d %d %d %d %d %d\n", start, mid,
               mid+1, end, a[0], a[1], a[2], a[3], a[4], a[5], a[6], a[7]);
    }
}

int main(void)
{
    sort(0, LEN-1);
    return 0;
}

执行结果：

sort (0-3, 4-7) 5 2 4 7 1 3 2 6

sort (0-1, 2-3) 5 2 4 7 1 3 2 6

sort (0-0, 1-1) 5 2 4 7 1 3 2 6

merge (0-0, 1-1) to 2 5 4 7 1 3 2 6

sort (2-2, 3-3) 2 5 4 7 1 3 2 6

merge (2-2, 3-3) to 2 5 4 7 1 3 2 6

merge 0-1, 2-3) to 2 4 5 7 1 3 2 6

sort (4-5, 6-7) 2 4 5 7 1 3 2 6

sort (4-4, 5-5) 2 4 5 7 1 3 2 6

merge (4-4, 5-5) to 2 4 5 7 1 3 2 6

sort (6-6, 7-7) 2 4 5 7 1 3 2 6

merge (6-6, 7-7) to 2 4 5 7 1 3 2 6

merge (4-5, 6-7) to 2 4 5 7 1 2 3 6

merge (0-3, 4-7) to 1 2 2 3 4 5 6 7