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LeetCode 81 Search in Rotated Sorted Array II (Python详解及实现)

2017-08-08 17:39 846 查看
【题目】

Follow up for "Search in RotatedSorted Array":

What if duplicates are allowed?

 

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

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

 

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 01 2).

 

Write a function to determine if a giventarget is in the array.

 

The array may contain duplicates.

对一个给定的有序数组,数组中存在重复数字,以某数字为中心进行旋转,在该数组中寻找目标target是否存在。

 

【思路】

方法一:直接查找

方法二:利用二分查找,因为有旋转的情况,需要注意边界条件

 

【Python实现】

#二分查找

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[left] == nums[right] == nums[mid]:

                left += 1

                right -= 1

           elif nums[left] <= nums[mid]:#升序

                if nums[left] <= target <nums[mid]:

                    right = mid -1

                else:

                    left = mid + 1

           else:#旋转(左大右小)

                if nums[left] > target >=nums[mid]:

                    left = mid + 1

                else:

                    right = mid - 1

       return False

   

if __name__ == '__main__':

    nums = [3, 1]

    target = 1

    S = Solution()

    S.search(nums,target)

                                                          

#直接逐个查找

class Solution1(object):

   def search(self, nums, target):

       """

       :type nums: List[int]

       :type target: int

       :rtype: bool

       """

       for i in nums:

           if i == target:

                return True

       return False       

 

 

if __name__ == '__main__':

     nums= [3, 1]

    target = 1

    S = Solution1()

    S.search(nums,target)

 
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标签:  LeetCode Python