leetcode153. Find Minimum in Rotated Sorted Array

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, a1, a2, ..., an-1] 1 time results in the array a\[n-1, a0, a1, a2, ..., an-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.

https://leetcode.cn/problems/find-minimum-in-rotated-sorted-array/description/

思路：简洁二分法

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left, right = -1, len(nums) - 1  # 开区间 (-1, n-1)
        while left + 1 < right:  # 开区间不为空
            mid = (left + right) // 2
            if nums[mid] < nums[-1]:
                right = mid
            else:
                left = mid
        return nums[right]