题目
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
首先判断，如果没有旋转，返回第一个数
进行二分，如果中间数大于下一个数，或者中间数小于前一个数，返回结果。如果中间数大于第一个数，在右边二分，否则在左边二分
代码
class Solution:
def findMin(self, nums: List[int]) -> int:
n = len(nums)
if n == 1:
return nums[0]
if nums[-1] > nums[0]:
return nums[0]
l, r = 0, n - 1
while l <= r:
m = (l + r) // 2
if nums[m] > nums[m + 1]:
return nums[m + 1]
if nums[m] < nums[m - 1]:
return nums[m]
if nums[m] > nums[0]:
l = m + 1
else:
r = m - 1
