Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

You may assume no duplicate exists in the array.

方法一：遞迴+分治策略。

public class Solution {
    private int min(int[] nums, int from, int to) {
        if (from == to) return nums[from];
        if (nums[from] < nums[to]) return nums[from];
        int m = (from+to)/2;
        if (nums[m] > nums[to]) return min(nums, m + 1, to);
        return min(nums, from, m);
    }
    public int findMin(int[] nums) {
        return min(nums, 0, nums.length-1);
    }
}
 

public class Solution {
    public int findMin(int[] nums) {
        int from = 0, to = nums.length-1;
        while (nums[from]>nums[to]) {
            int mid = (from+to)/2;
            if (nums[mid] > nums[to]) from = mid + 1;
            else to = mid;
        }
        return nums[from];
    }
}