leetcode 【 Subsets II 】python 实现

题目：

Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.

For example,
If S = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

代码：oj测试通过 Runtime: 78 ms

 1 class Solution:
 2     # @param num, a list of integer
 3     # @return a list of lists of integer
 4     def dfs(self, start, S, result, father_elements):
 5         if father_elements in result:
 6             return
 7         result.append(father_elements)
 8         for i in range(start, len(S)):
 9             self.dfs(i+1, S, result, father_elements+[S[i]])
10     
11     def subsetsWithDup(self, S):
12         # none case
13         if S is None:
14             return []
15         # deep first search
16         result = []
17         first_depth = {}
18         self.dfs(0, sorted(S), result, [])
19         return result

思路：

大体思路跟Subsets差不多，详情见：

http://www.cnblogs.com/xbf9xbf/p/4253208.html

只需要每次向result中添加子集时注意判断一下这个子集是否已经存在。如果存在那么就直接返回，不做处理。