原题地址：https://oj.leetcode.com/problems/subsets-ii/
题意：
Given a collection of integers that might contain duplicates, S, return all possible subsets.
Note:
For
example,
If S = [1,2,2],
a solution is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
解题思路：和上一道题一样，求一个集合的所有子集。和上一道题不一样的一点是集合可能有重复元素。这道题同样使用dfs来解题，只是需要在dfs函数里加一个剪枝的条件，排除掉同样的子集。
代码：
class Solution: # @param num, a list of integer # @return a list of lists of integer def subsetsWithDup(self, S): def dfs(depth, start, valuelist): if valuelist not in res: res.append(valuelist) if depth == len(S): return for i in range(start, len(S)): dfs(depth+1, i+1, valuelist+[S[i]]) S.sort() res = [] dfs(0, 0, []) return res
[leetcode]Subsets II @ Python,布布扣,bubuko.com
原文：http://www.cnblogs.com/zuoyuan/p/3758346.html