旋转数组的二分查找

问题描述：已知有序数组a[N]， 从中间某个位置k（k未知，k=-1表示整个数组有序）分开，然后将前后两部分互换，得到新的数组，在该新数组的查找元素x。如：a[]={1,2,5,7,9,10,15}，从k=4分开，得到新数组a={9,10,15, 1,2,5,7}。

#include "stdafx.h"

int search(int *a, int key, int low, int high)
{
    if (low > high)
        return -1;

    int mid = (low + high)/2;
    if (key == a[mid])  
        return mid;   
    
    if (a[mid] <= a[high])       //右边有序。
    {
        if(a[mid] < key && key <= a[high])      //在右边有序部分寻找。        
            return search(a, key, mid + 1, high);    
        else                                                //排除右边有序部分。          
            return search(a, key, low, mid - 1);   
    }
    else                        //左边有序   
    {
        if(key < a[mid] && a[low] <= key)     //在左边有序部分寻找。              
            return search(a, key, low , mid - 1);
        else                                                //排除左边有序部分。      
            return search(a, key, mid + 1, high);      
    }   
}

int find(int *a, int size, int key)
{    
    return search(a, key, 0, size - 1);
}

int _tmain(int argc, _TCHAR* argv[])
{
    int a[] = {9, 10, 15, 1, 2, 5, 7};
    int b[] = {1,0,1,1,1};
    printf("%d
",find(b, 5, 1));
    printf("%d
",find(b, 5, 0));
    printf("%d
",find(b, 5, 1));
    printf("%d
",find(b, 5, 1));
    printf("%d
",find(b, 5, 1));

    printf("
");
    printf("%d
",find(a, 7, 9));
    printf("%d
",find(a, 7, 10));
    printf("%d
",find(a, 7, 15));
    printf("%d
",find(a, 7, 1));
    printf("%d
",find(a, 7, 2));
    printf("%d
",find(a, 7, 5));
    printf("%d
",find(a, 7, 7));

    return 0;
}

参考了这个博客：

http://www.jobcoding.com/datastructure-and-algorithm/array/one-sorted-array/rotate_array/