leetcode 【 Search in Rotated Sorted Array II 】python 实现

题目：

与上一道题几乎相同；不同之处在于array中允许有重复元素；但题目要求也简单了，只要返回true or false

http://www.cnblogs.com/xbf9xbf/p/4254590.html

代码：oj测试通过 Runtime: 73 ms

 1 class Solution:
 2     # @param A a list of integers
 3     # @param target an integer
 4     # @return a boolean
 5     def search(self, A, target):
 6         A=list(set(A))
 7         # none case & zero case
 8         if A is None or len(A)==0 :
 9              return False
10         # binary search
11         start = 0
12         end = len(A)-1
13         while start<=end :
14             # one element left case
15             if start == end :
16                 if A[start]==target :
17                     return True
18                 else:
19                     return False
20             # two elements left case
21             if start+1 == end :
22                 if A[start]==target :
23                     return True
24                 elif A[end]==target :
25                     return True
26                 else:
27                     return False
28             # equal or more than three elements case
29             mid = (start+end)/2
30             if A[mid]==target :
31                 return True
32             elif A[mid]>target:
33                 if A[start]>A[mid] and A[end]<A[mid]:
34                     start = mid+1
35                 elif A[start]<A[mid] and A[end]<A[mid]:
36                     if A[end]>=target:
37                         start = mid+1
38                     else:
39                         end = mid-1
40                 elif A[start]>A[mid] and A[end]>A[mid]:
41                      end = mid-1
42                 else:
43                      end = mid-1
44             else:
45                 if A[start]>A[mid] and A[end]<A[mid]:
46                     end = mid-1
47                 elif A[start]<A[mid] and A[end]<A[mid]:
48                     start = mid+1
49                 elif A[start]>A[mid] and A[end]>A[mid]:
50                     if A[end]>=target :
51                         start = mid+1
52                     else:
53                         end = mid-1
54                 else:
55                     start = mid+1
56         return False

思路：

用了一个trick Python数组去重的办法A=list(set(A))

这样A数组中就没有重复的元素了，可以直接用之前一题的代码。

这样的trick应该不是题目的本意，这道二分查找题目比较经典，应该吃透。