Find Minimum in Rotated Sorted Array II

一、问题描述

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

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).

Find the minimum element.

The array may contain duplicates.

二、思路

返回数组的最小元素，思路和 Find Minimum in Rotated Sorted Array类似，唯一不同的是，我们需要将当前翻转位置和最小值进行比较。

三、代码

class Solution {public:    int findMin(vector<int>& nums) {        int minNum = nums[0];        for(int i = 0; i < nums.size() - 1; ++i){            if(nums[i] > nums[i + 1]){                if(nums[i + 1] < minNum){                    minNum = nums[i + 1];                }            }        }        return minNum;    }};