题目：
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.
代码：oj测试通过 Runtime: 52 ms
1 class Solution: 2 # @param num, a list of integer 3 # @return an integer 4 def findMin(self, num): 5 # none case 6 if num is None: 7 return None 8 # short lenght case 9 if len(num)==1 : 10 return num[0] 11 # binary search 12 start = 0 13 end = len(num)-1 14 while start<=end : 15 if start==end : 16 return num[start] 17 if start+1==end : 18 return min(num[start],num[end]) 19 mid = (start+end)/2 20 if num[mid]>num[start] : 21 if num[mid]>num[end] : 22 start = mid 23 else: 24 return num[start] 25 else: 26 if num[mid]>num[end] : 27 return num[end] 28 else: 29 end = mid
思路：
基本思路还是binary search。
注意修改start或end时的条件：因为mid也有可能是最小值，所以start=mid end=mid，这个跟传统二分查找时start=mid+1以及end=mid-1有所不同。
想明白这些后，代码一次AC。
leetcode 【 Find Minimum in Rotated Sorted Array 】python 实现
原文：http://www.cnblogs.com/xbf9xbf/p/4261334.html