Leetcode Find Minimum in Rotated Sorted Array I and II

标签：find   sans   and   etc   array   ota   before   div   family   

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。

 1 class Solution(object):
 2     def findMin(self, nums):
 3         """
 4         :type nums: List[int]
 5         :rtype: int
 6         """
 7         n = len(nums)
 8         left = 0
 9         right = n-1
10         while left < right:
11             mid = (left+right)//2
12             if nums[mid] > nums[right]:
13                 left = mid + 1
14             else:
15                 right = mid
16         return nums[left]
17         

以下引用来自： http://bangbingsyb.blogspot.com/2014/11/leecode-find-minimum-in-rotated-sorted.html

 1 class Solution(object):
 2     def findMin(self, nums):
 3         """
 4         :type nums: List[int]
 5         :rtype: int
 6         """
 7         n = len(nums)
 8         left = 0
 9         right = n-1
10         while left < right:
11             mid = (left+right)//2
12             if nums[mid] > nums[right]:
13                 left = mid + 1
14             elif nums[mid] < nums[right]:
15                 right = mid
16             else:
17                 right -= 1
18         
19         return nums[left]

Leetcode Find Minimum in Rotated Sorted Array I and II

标签：find   sans   and   etc   array   ota   before   div   family   

原文地址：http://www.cnblogs.com/lettuan/p/6185516.html