Leetcode the smallest number of the rotation array of the offer of the sword (11)

Title Description

Move the first elements of an array to the end of the array , We call it rotation of arrays .

Give you a chance to exist repeat An array of element values numbers , It turns out to be an ascending array , And a rotation is carried out according to the above situation . Please return the smallest element of the rotation array . for example , Array [3,4,5,1,2] by [1,2,3,4,5] A rotation of , The minimum value of the array is 1.

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/xuan-zhuan-shu-zu-de-zui-xiao-shu-zi-lcof
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Example 1：

 Input ：[3,4,5,1,2]
 Output ：1

Example 2：

 Input ：[2,2,2,0,1]
 Output ：0

Coding ideas ：

Two points search

Python3 Realization

class Solution:
    def minArray(self, numbers: List[int]) -> int:
        low = 0
        high = len(numbers) - 1
        while low < high:
            mid = low + (high - low) // 2
            if numbers[mid] > numbers[high]:
                low = mid + 1
            elif numbers[mid] < numbers[high]:
                high = mid
            else:
                high -= 1
        return numbers[low]

C++ Realization

class Solution {
public:
    int minArray(vector<int>& numbers) {
        int low = 0;
        int high = numbers.size() - 1;
        while(low < high){
            int mid = static_cast<int>(low + (high - low) / 2);
            if(numbers[mid] > numbers[high]){
                low = mid + 1;
            }else if(numbers[mid] < numbers[high]){
                high = mid;
            }else{
                high -= 1;
            }
        }
        return numbers[low];

    }
};