Find Minimum in Rotated Sorted Array II

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.

follow up,和一主要的不同在于有重复数字。当

class Solution(object):
    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        l = 0
        r = len(nums) - 1
        
        while l + 1 < r:
            mid = l + (r - l)/2
            if nums[mid] == nums[r]:  ＃只用移动一位就可以。
                r -= 1
            elif nums[mid] > nums[r]:
                l = mid
            else:
                r = mid    
        return min(nums[l], nums[r])

 

转载于:https://www.cnblogs.com/sherylwang/p/5674910.html