题目要求：Search in Rotated Sorted Array

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.

代码如下：

class Solution {

public:

    int search(int A[], int n, int target) {

        int index = -1;

        //题目简化成找到数组A中的元素位置……

        for(int i = 0; i < n; i++){

            if(A[i] == target){

                index = i;

                break;

            }

        }

        return index;

    }

};