Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

• [4,5,6,7,0,1,2] if it was rotated 4 times.
• [0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 

Constraints:

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• All the integers of nums are unique.
• nums is sorted and rotated between 1 and n times.

这题跟LeetCode 33. Search in Rotated Sorted Array类似，解题思路也基本一样用二分查找法，把数组分成左右两个区间后判断最小值是落在左半区间还是右半区间，关键点是判断的条件该如何确定呢？

如果把中点值与最左边值比较即nums[mid] > nums[left]，左半区间是递增的，但是最小值可能在左半区间也可能在右半区间。如果把中点值与最右边值比较即nums[mid] < nums[right]，右半区间是递增的，最小值可以确定一定在左半区间(包含nums[mid])。因此本题以nums[mid]与nums[right]比较为判断条件。

如果nums[mid]小于nums[r]，最小值落在左半区间(包含nums[mid])，所以右边界更新为r=mid。那如果nums[mid]不小于nums[r]，最小值就落在了右半区间(不包含nums[mid])，所以左边界更新为l=mid + 1。按此条件一直循环折半查找直到区间只剩一个数，这个数就是最小值。

class Solution:
    def findMin(self, nums: List[int]) -> int:
        l, r = 0, len(nums) - 1
        
        while l < r:
            mid = l + (r - l) // 2
            if nums[mid] < nums[r]:
                r = mid
            else:
                l = mid + 1
            
        return nums[l]