Sword finger offer 11 Rotate the minimum number of the array

The finger of the sword Offer 11. Minimum number of rotation array

The difficulty is simple 626 Switch to English to receive dynamic feedback
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.
Be careful , Array [a[0], a[1], a[2], …, a[n-1]] Rotate once The result is an array [a[n-1], a[0], a[1], a[2], …, a[n-2]] .

 Example  1：
 Input ：
numbers = 

[3,4,5,1,2] Output ：1
 Example  2：
 Input ：
numbers = 

[2,2,2,0,1] Output ：0

Tips ：

• n == numbers.length
• 1 <= n <= 5000
• -5000 <= numbers[i] <= 5000
• numbers It turns out to be an ascending array , And carried on 1 to n Second rotation

class Solution {
    
public:
    int minArray(vector<int>& numbers) {
    
        int m=5001;
        for(int i=0;i<numbers.size();i++){
    
            if(numbers[i]<m)
                m=numbers[i];
        }
        return m;
    }
};