leetcode 【 Search in Rotated Sorted Array 】python 实现

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

 class Solution:
     # @param A, a list of integers
     # @param target, an integer to be searched
     # @return an integer
     def search(self, A, target):
         # none case & zero case
         if A is None or len(A)==0 :
             return -1
         # binary search
         start = 0
         end = len(A)-1
         while start<=end :
             # one element left case
             if start == end :
                 if A[start]==target :
                     return start
                 else:
                     return -1
             # two elements left case
             if start+1 == end :
                 if A[start]==target :
                     return start
                 elif A[end]==target :
                     return end
                 else:
                     return -1
             # equal or more than three elements case
             mid = (start+end)/2
             if A[mid]==target :
                 return mid
             elif A[mid]>target:
                 if A[start]>A[mid] and A[end]<A[mid]:
                     start = mid+1
                 elif A[start]<A[mid] and A[end]<A[mid]:
                     if A[end]>=target:
                         start = mid+1
                     else:
                         end = mid-1
                 elif A[start]>A[mid] and A[end]>A[mid]:
                     end = mid-1
                 else:
                     end = mid-1
             else:
                 if A[start]>A[mid] and A[end]<A[mid]:
                     end = mid-1
                 elif A[start]<A[mid] and A[end]<A[mid]:
                     start = mid+1
                 elif A[start]>A[mid] and A[end]>A[mid]:
                     if A[end]>=target :
                         start = mid+1
                     else:
                         end = mid-1
                 else:
                     start = mid+1
         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。