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

class Solution:
    # @param num, a list of integer
    # @return a list of lists of integer
    def dfs(self, start, S, result, father_elements):
        if father_elements in result:
            return
        result.append(father_elements)
        for i in range(start, len(S)):
            self.dfs(i+1, S, result, father_elements+[S[i]])
    
    def subsetsWithDup(self, S):
        # none case
        if S is None:
            return []
        # deep first search
        result = []
        first_depth = {}
        self.dfs(0, sorted(S), result, [])
        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。