leetcode 153 154 Find Minimum in Rotated Sorted Array I II 二分砍掉一边跨步移动，不找分界线-优快云博客

本文链接：https://blog.youkuaiyun.com/taoqick/article/details/100052403

本文探讨在未知旋转点的升序数组中寻找最小元素的算法，即使数组包含重复元素。通过分析不同情况，如[3,4,5,1,2]和[2,2,2,0,1]，提出了一种改进的二分查找方法，以O(log n)的时间复杂度解决问题。特别关注处理重复元素时的挑战，并提供了一种有效避免无限循环的解决方案。

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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.

You may assume no duplicate exists in the array.

Example 1:

Input: [3,4,5,1,2] 
Output: 1

Example 2:

Input: [4,5,6,7,0,1,2]
Output: 0

Accepted

311,542

Submissions

717,092

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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

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不管重复不重复，都利用（如何写一个正确的二分法代码_taoqick的博客-优快云博客）的二分写法。和传统写法不同的是，由于这里是找最小，所以nums[mid]==nums[r]时可以放心大胆的把nums[r]扔掉（这里也解释了为啥不用nums[l]和nums[mid]，因为nums[l]<nums[mid]的时候，这时候只能说明左边一定没坑，但并不能说明右边一定有坑，这时候丢掉左边或者右边都是不对的，这就是左右的区别！）。同时要注意nums[mid]<nums[r]的时候，nums[mid]有可能是最小(例如[3,1,3])，所以把一边的跨步移动r=mid-1改成r=mid，另外一边仍然跨步移动，就不会陷入死循环。

但是注意这种方式，找到的是最小值，但并不一定是最大和最小的分界线。例如[1,1,3,1]，上来就会把最右边的1给扔掉，因此找分界线更难！！！

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