Search in Rotated Sorted Array

题目：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.

思路：

二分法的变形，每次变换一共是三种情况，找出中间值，相等时最理想的情况；

中间值大于最左边的数以及中间值在下面。

对于第二种情况，考虑是不是左边的左边，然后就是right减去1；

对于第三种情况，考虑最右边的右边，然后就是left加上1.

好题目！！！

代码：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left=0,right=nums.size()-1;
        while(left<=right){
            int mid=(left+right)/2;
            if(nums[mid]==target)   return mid;
            if(nums[mid]>=nums[left]){
                //在左边
                if(target>=nums[left]&&target<nums[mid])   
                    right=mid-1;
                else
                    left=mid+1;
            }else{
                //在右边
                if(target<=nums[right]&&nums[mid]<target)
                    left=mid+1;
                else
                    right=mid-1;
            }
        }
        return -1;
    }
};