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Find Minimum in Rotated Sorted Array II

时间:2015-09-07 00:31:50      阅读:195      评论:0      收藏:0      [点我收藏+]

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Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

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

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

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

Find the minimum element.

The array may contain duplicates.

class Solution(object):
    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) <= 0:
            return 0
        if len(nums) == 1:
            return nums[0]
        size=len(nums)
        left=0
        right=size-1
        while (left<right) and (nums[left] >= nums[right]):
            mid = (left+right)/2
            if nums[mid] > nums[right]:
                left=mid+1
            elif nums[mid] < nums[left]:
                right=mid
            else:
                left+=1
        return nums[left]

 

Find Minimum in Rotated Sorted Array II

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原文地址:http://www.cnblogs.com/allenhaozi/p/4787706.html

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