【LeetCode 33】Search in Rotated Sorted Array

题意：

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).

思路：

想到了二分，但没想到，关键是，可以根据mid和nums[l]和nums[r]的大小关系，判断出一个单调区间，然后根据target和这个区间两端值得大小关系，判断是否在这一边，决定保留哪一边。

代码：

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

来源：CSDN

作者：涛涛酱

链接：https://blog.csdn.net/iCode_girl/article/details/104015949