旋转数组的最小数字

把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个递增排序的数组的一个旋转，输出旋转数组的最小元素。例如数组{3,4,5,1,2}为{1,2,3,4,5}的一个旋转，该数组的最小值为1.

那么如果查找旋转数组的最小数字呢，主要的思想就是二分查找法，即头尾同时查找。
例子查找过程如下：

代码如下：

#include <stdio.h>

int search_min_rotated_sorted_array(int *a, int length);
int min_in_order(int *a, int index1, int index2);

int main(void)
{
    int a[5] = {3, 4, 5, 1, 2};
    int b[5] = {1, 0, 1, 1, 1};
    int c[5] = {1, 1, 1, 0, 1};

    int min = search_min_rotated_sorted_array(a, 5);
    printf("search in rotate sorted array min=%d.\n", min);
    min = search_min_rotated_sorted_array(b, 5);
    printf("search in rotate sorted array b min=%d.\n", min);
    min = search_min_rotated_sorted_array(c, 5);
    printf("search in rotate sorted array c min=%d.\n", min);
    return 0;
}

int search_min_rotated_sorted_array(int *a, int length)
{
    int index1 = 0;
    int index2 = 0;
    int index_mid = 0;

    if (NULL == a || length <= 0){
        printf("param error\n");
        return -1;
    }

    index1 = 0;
    index2 = length - 1;
    index_mid = index1;

    while (a[index1] >= a[index2]){
        if (index2 - index1 == 1){
            index_mid = index2;
            break;
        }

        index_mid = ( index1 + index2 ) / 2;
        if (a[index1] == a[index2] && a[index_mid] == a[index1])
            return min_in_order(a, index1, index2);

        if (a[index_mid] >= a[index1])
            index1 = index_mid;
        else if (a[index_mid] <= a[index2])
            index2 = index_mid;
    }

    return a[index_mid];
}

int min_in_order(int *a, int index1, int index2)
{
    int result = a[index1];
    int i = 0;
    for (i = index1 + 1; i <= index2; ++i){
        if (result > a[i])
            result = a[i];
    }
    return result;
}