Minimum number of rotation array

import java.util.ArrayList;
public class Solution {
 public int minNumberInRotateArray(int[] array) {
		if (array.length == 0) {
			return 0;
		}
		int min = array[0];
		int index = 0;
		for (int i = 1; i < array.length; i++) {
			if (min > array[i]) {
				min = array[i];
				index = i;
			}
		}

		return array[index];
	}
}

import java.util.ArrayList;
import java.util.*;
public class Solution {
 public int minNumberInRotateArray(int[] array) {
		Arrays.sort(array);
		return array[0];
	}
}

public static int minNumberInRotateArray(int[] array) {
		if (array.length == 0) {
			return 0;
		}
		int left = 0;
		int right = array.length - 1;
		int middle = left;

		while (array[left] >= array[right]) {
			// left points to the last number of the first increment subarray
			// right points to the first number of the second subarray, which is the minimum number of the array
			if (right - left == 1) {
				middle = right;
				break;
			}

			middle = (left + right) / 2;
			// In special cases, if the three digits pointed to by left,middle,right are equal

			// If the intermediate element is in the preceding increment sub array, it should be greater than or equal to the element pointed to by the first pointer.
			// The smallest element in the array should follow the middle element. We can point the first pointer to the intermediate element,
			// This reduces the scope of the search. The first pointer after the move is still in the preceding increment sub array.
			if (array[middle] >= array[left]) {
				left = middle;
			}

			// If the middle element is in the following increasing sub array, it should be less than or equal to the element pointed to by the second pointer.
			// The smallest element in the array should be in front of the middle element.
			if (array[middle] <= array[right]) {
				right = middle;
			}
		}

		return array[middle];
}