【LEETCODE】81-Search in Rotated Sorted Array II [Python]

Follow up for "Search in Rotated Sorted Array":

What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

题意：

接Search in Rotated Sorted Array

如果允许重复

这会影响时间复杂度么，how and why

给定target，判断它是否在数组中

思路：

有重复的话，多了一个判断条件就是三点相等时，左右端点同时变化

影响就是，如果在重复中间截断逆转，之后再用 nums[start]<=target<nums[mid] 去判断，就找不到这个target

Python：

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: bool
        """
        
        start=0
        end=len(nums)-1
        
        while start<=end:
            
            mid=(start+end)/2
            
            if nums[mid]==target:
                return True
            
            if nums[mid]==nums[start]==nums[end]:
                start+=1
                end-=1
            elif nums[start]<=nums[mid]:
                if nums[start]<=target<nums[mid]:
                    end=mid-1
                else:
                    start=mid+1
            else:
                if nums[mid]<=target<=nums[end]:
                    start=mid+1
                else:
                    end=mid-1
        return False