【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