题目
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.
代码
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class Solution(object):
def search(self, nums, l, r):
if l==r:
return nums[l]
m = l + (r-l)/2
if nums[m] == nums[l]:
return min(self.search(nums,l,m), self.search(nums,m+1,r))
elif nums[m] > nums[l]:
return min(nums[l], self.search(nums, m+1, r))
else:
return self.search(nums, l, m)
def findMin(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if nums==None or len(nums)==0:
return None
return self.search(nums,0,len(nums)-1)
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