Search in Rotated Sorted Array

Q: Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

A: 递归+二分查找。

    int bisearch(int A[],int begin,int end,int target)
    {
        if(begin>end)
            return -1;
        int middle = begin+(end-begin)/2;
        if(A[middle] == target)
            return middle;
        if(A[begin]>A[end])
        {
            int pos1 = bisearch(A,begin,middle-1,target);
            int pos2 = bisearch(A,middle+1,end,target);
            return (pos1!=-1?pos1:pos2);
        }else
        {
　　　　　　//这里不能写成return bisearch(A,begin,end,target);---这样相当于搜索的范围永远不会减少。。死循环
            if(A[middle]>target)
                return bisearch(A,begin,middle-1,target);  
            else
                return bisearch(A,middle+1,end,target);
        }
    }
    int search(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return bisearch(A,0,n-1,target);
        
    }