[leetcode] sword finger offer 11 Rotate the minimum number of the array

subject ：

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

 Example  1：

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

 Input ：[2,2,2,0,1]
 Output ：0

Dichotomy ：

class Solution {
    
    public int minArray(int[] numbers) {
    
        int l = 0,r = numbers.length - 1;
        while(l < r) {
    
            int mid = (l + r) / 2;
            //  If the one on the right is larger than the one in the middle , Then the one on the right is an ordered array , Then narrow the scope , Don't look at the one on the right 
            if(numbers[r] > numbers[mid]){
    
                r = mid;
            }
            //  If the one on the right is smaller than the one in the middle , Then the left is an ordered array , Then narrow the scope , Don't look at the one on the left 
            else if(numbers[r] < numbers[mid]) l = mid + 1; 
            //  duplicate removal 
            else r--;
        }
        return numbers[r];
    }
}