Rotate array minimum number

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

There are three situations to consider ：

array[mid] > array[high]

What happens is array similar [3,4,5,6,0,1,2], At this point, the minimum number must be in mid To the right of

low = mid + 1

array[mid] < array[high]

What happens is array similar [2,2,3,4,5,6,6], The minimum number must be array[mid] Or in mid Left side . Because the right side must be increasing

high = mid

array[mid] == array[high]

What happens is array similar [1,0,1,1,1] perhaps [1,1,1,0,1], At this time, the minimum number is difficult to judge mid Left or right , I have to try one by one

high = high - 1

public class Solution {
    public int minNumberInRotateArray(int [] array) {
        int low = 0;
        int high = array.length - 1;
        int mid = (low + high) / 2;
        while(low < high) {
        	//  At this point, the minimum number must be in  mid  To the right of 
            if(array[mid] > array[high]) {
                low = mid + 1;
            //  The minimum number must be  array[mid]  Or in  mid  Left side 
            } else if(array[mid] < array[high]) {
                high = mid;
            //  Cannot judge in  mid  Left or right , At this time, we have to try one by one 
            } else {
                high--;
            }
            mid = (low + high) / 2;
        }
        return array[low];
    }
}