Leetcode Find Minimum in Rotated Sorted Array I and II
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.
You may assume no duplicate exists in the array.
在任何一个sublist中,如果头元素大于尾元素,那么这个minimum一定在这个sublist中间的某一个位置。可以用二分法找到这个元素,复杂度是O(logN)。本题O(N)的算法也可以通过OJ,思路就是最简答的从头元素往后比,直到出现 nums[i] > nums[i+1]的情况,那么nums[i+1]就是Minimum element。
class Solution(object):
def findMin(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
n = len(nums)
left = 0
right = n-1
while left < right:
mid = (left+right)//2
if nums[mid] > nums[right]:
left = mid + 1
else:
right = mid
return nums[left]
以下引用来自: http://bangbingsyb.blogspot.com/2014/11/leecode-find-minimum-in-rotated-sorted.html
3 3 3 3 1 2 3
class Solution(object):
def findMin(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
n = len(nums)
left = 0
right = n-1
while left < right:
mid = (left+right)//2
if nums[mid] > nums[right]:
left = mid + 1
elif nums[mid] < nums[right]:
right = mid
else:
right -= 1 return nums[left]
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