题目描述

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.

解题思路

利用二分查找查找分界点。

代码

class Solution:
    def findMin(self, nums: List[int]) -> int:
        n = len(nums)
        start = 0
        last = n - 1
        pos = 0
        while start <= last:
            mid = (start + last) >> 1
            if nums[mid] < nums[mid - 1] and (mid == n - 1 or nums[mid] < nums[mid + 1]):
                pos = mid
                break
            elif nums[mid] > nums[last]:
                start = mid + 1
            else:
                last = mid - 1
        
        return min(nums[0], nums[pos])