Type：medium

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 ofO(logn).

Example 1:

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

Example 2:

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

给定一个旋转有序数组，一个target值。在这个数组中寻找target，若能找到，返回它的位置，不能找到则返回-1。

采用二分法寻找target。已知在数组中，若中间值大于最右值，则中间值左面的数组是有序的；反之，若中间值小于最右值，则中间值右面的数组是有序的。通过这半边的有序数组，我们可以判断target值是否在这区间内，若在，则在这半边继续搜索；若不在，则在另半边继续搜索。

class Solution {

public:

    int search(vector<int>& nums, int target) {

        int n = nums.size();

        int left = 0;

        int right = n-1;       

        while(left<=right){

            int mid = left + (right - left)/2;

            if(nums[mid] == target) return mid;

            else if(nums[mid] <= nums[right]){

                if(nums[mid] < target && nums[right] >= target) left = mid + 1;

                else right = mid - 1;

            }else if(nums[mid] > nums[right]){

                if(nums[left] <= target && nums[mid] > target) right = mid - 1;

                else left = mid + 1;

            }

        }

        return -1;

    }

};

//二分法