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

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