题目链接
https://leetcode.com/problems/search-in-rotated-sorted-array-ii/
题目原文
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.
题目大意
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
解题思路
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
代码
class Solution(object):
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: bool
"""
left = 0
right = len(nums) - 1
while left <= right:
mid = (left+right)//2
if nums[mid]==target:
return True
if nums[mid]==nums[left]==nums[right]:
left += 1
right -= 1
elif nums[left] <= nums[mid]:
if nums[left] <= target and target < nums[mid]:
right = mid -1
else:
left = mid + 1
else:
if nums[mid] <= target and target <= nums[right]:
left = mid + 1
else:
right = mid -1
return False