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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).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array
代码：oj测试通过 Runtime: 53 ms
1 class Solution: 2 # @param A, a list of integers 3 # @param target, an integer to be searched 4 # @return an integer 5 def search(self, A, target): 6 # none case & zero case 7 if A is None or len(A)==0 : 8 return -1 9 # binary search 10 start = 0 11 end = len(A)-1 12 while start<=end : 13 # one element left case 14 if start == end : 15 if A[start]==target : 16 return start 17 else: 18 return -1 19 # two elements left case 20 if start+1 == end : 21 if A[start]==target : 22 return start 23 elif A[end]==target : 24 return end 25 else: 26 return -1 27 # equal or more than three elements case 28 mid = (start+end)/2 29 if A[mid]==target : 30 return mid 31 elif A[mid]>target: 32 if A[start]>A[mid] and A[end]<A[mid]: 33 start = mid+1 34 elif A[start]<A[mid] and A[end]<A[mid]: 35 if A[end]>=target: 36 start = mid+1 37 else: 38 end = mid-1 39 elif A[start]>A[mid] and A[end]>A[mid]: 40 end = mid-1 41 else: 42 end = mid-1 43 else: 44 if A[start]>A[mid] and A[end]<A[mid]: 45 end = mid-1 46 elif A[start]<A[mid] and A[end]<A[mid]: 47 start = mid+1 48 elif A[start]>A[mid] and A[end]>A[mid]: 49 if A[end]>=target : 50 start = mid+1 51 else: 52 end = mid-1 53 else: 54 start = mid+1 55 return -1
思路：
这个就是binary search的思路。
个人没想出来什么好的方法，硬着头皮硬写了一个暴力解决方法。
传统的binary search只需要判断A[mid]与target的大小就可以了；但这道题是rotated array，光判断A[mid]是不够的。
还需要判断A[start] A[end]与A[mid]的大小才能判断，target可能落在[start,mid]区间还是[mid,end]区间。
自己的代码实在有些繁琐丑陋，估计有些if else条件可以合并，后续会改进。
leetcode 【 Search in Rotated Sorted Array 】python 实现
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4254590.html
