LeetCode81：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?

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LeetCode：链接

在LeetCode33：Search in Rotated Sorted Array基础上扩展。

当 nums[mid] = nums[left] 时，这时由于很难判断 target 会落在哪，那么只能采取 left++

当 nums[mid] > nums[left] 时，这时可以分为两种情况，判断左半部比较简单

当 nums[mid] < nums[left] 时，这时可以分为两种情况，判断右半部比较简单

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