LeetCode·33. Search in Rotated Sorted Array

题目：

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

思路：

二分查找，判断最左和中间值的关系分情况讨论。

代码：

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

        return -1;
    }
};

结果：