81. Search in Rotated Sorted Array II

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Follow up:

• This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
• Would this affect the run-time complexity? How and why?

Code：

class Solution:
    def search(self, nums, target):
        if not nums:
            return False
        low = 0
        high = len(nums) - 1
        while low <= high:
            while low < high and nums[low] == nums[high]:
            #这样的目的是为了能准确判断mid位置，所以算法的最坏时间复杂度为O(n)
                low += 1                  
            mid = (low+high)//2
            if target == nums[mid]:
                return True         
            elif nums[mid] >= nums[low]: #高区
                if nums[low] <= target < nums[mid]:  
                    high = mid - 1
                else:
                    low = mid + 1
            elif nums[mid] <= nums[high]:  #低区
                if nums[mid] < target <= nums[high]:
                    low = mid + 1
                else:
                    high = mid - 1
        return False