LeetCode 之 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.

仍然是二分查找，和之前的找最小很类似，只不过这次上下界的移动是根据mid和target比较的结果来决定的，情况分析如下：

mid>high,  如果target大于mid或者小于等于high这时候可以确定target位于mid和high 之间，low=mid+1；

mid<high, 如果target大于mid同时小于等于high这时候可以确定target位于mid和high之间，low=mid+1;

代码如下：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int low=0,high=nums.size()-1;
        while(low<high){
            int mid=(low+high)/2;
            if(nums[mid]>nums[high]){
                if(target>nums[mid]||target<=nums[high])
                    low=mid+1;
                else
                    high=mid;
            }else{
                if(target>nums[mid]&&target<=nums[high])
                    low=mid+1;
                else
                    high=mid;
            }
        }
        if(low==high&&target!=nums[low]) return -1;
        return low;
    }
};