题目描述：

Suppose an array sorted in ascending order 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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

中文理解：

给定一个由递增序列根据某一个下标旋转的序列，找出这个序列是够存在target值，存在的话，返回下标，否则返回-1.

解题思路：

使用双指针，在收尾像中间遍历，若找到target则返回下标，否则返回-1。

代码（java）：

class Solution {
    public int search(int[] nums, int target) {
        if(nums.length==0)return -1;
        int start=0,end=nums.length-1;
        while(start<=end){
            if(nums[start]==target)return start;
            if(nums[end]==target)return end;
            start++;
            end--;
        }
        return -1;
    }
}