import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param nums int整型一维数组 
     * @return int整型
     */
    public int minNumberInRotateArray (int[] nums) {
        int left = 0;
        int right = nums.length - 1;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (nums[mid] < nums[right]) right = mid;
            else if (nums[mid] > nums[right]) left = mid + 1;
            else if (nums[mid] < nums[left]) right = mid;
            else {
                if (right - left == 1) break;
                int left_num = minNumberInRotateArray(Arrays.copyOfRange(nums, left, mid));
                int right_num = minNumberInRotateArray(Arrays.copyOfRange(nums, mid, right + 1));
                return Math.min(left_num, right_num);
            }
        }
        return nums[left];
    }
}

解题思路：二分查找

首先，left、mid、right 三个指针对应的元素大小情况划分：

case 1：nums[mid] < nums[right]

    最小值在左侧区间 right = mid

case 2: nums[mid] > nums[right]

    最小值在右侧区间 left = mid + 1

case 3：nums[mid] == nums[right]

    case 3.1: nums[mid] < nums[left]

        最小值在左侧区间

    case 3.2: nums[left] == nums[mid] == nums[right]

        无法区分最小值区间，分割数组，递归

        left_num = minNumberInRotateArray(Arrays.copyOfRange(nums, left, mid))

        right_num = minNumberInRotateArray(Arrays.copyOfRange(nums, mid, right + 1))

        return Math.min(left_num, right_num);