LeetCode154：Find Minimum in Rotated Sorted Array II

最新推荐文章于 2022-05-26 16:47:57 发布

励志学好数据结构

最新推荐文章于 2022-05-26 16:47:57 发布

阅读量226

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/mengmengdajuanjuan/article/details/84619104

LeetCode 专栏收录该内容

186 篇文章

订阅专栏

本文探讨了在未知旋转点的升序数组中查找最小元素的算法，特别关注含有重复数字的情况，提出了一种改进的二分搜索策略，并通过Python代码实现了解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Suppose an array sorted in ascending order 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.

The array may contain duplicates.

Example 1:

Input: [1,3,5]
Output: 1

Example 2:

Input: [2,2,2,0,1]
Output: 0

Note:

This is a follow up problem to Find Minimum in Rotated Sorted Array.
Would allow duplicates affect the run-time complexity? How and why?

LeetCode：链接

这题和剑指Offer_编程题06：旋转数组的最小数字(指针)一模一样！有重复数字，因为有重复数字，就不能用二分法，就得移动high指针。

class Solution(object):
    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return None
        low = 0
        high = len(nums) - 1
        while low <= high:
            mid = (low + high) // 2
            if nums[mid] > nums[high]:
                low = mid + 1
            elif nums[mid] < nums[high]:
                high = mid
            else:
                high -= 1
        return nums[low]