Subsets II

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],  []]

解题思路：采用递归，分别得到对应长度的子数组的值，注意子数组要求不重复，可以用哈希来做；

#include<iostream>#include<vector>#include<set>#include<algorithm>using namespace std;set<vector<int> >CompareSet;//先输出SubsetLen长度的子串，然后确定索引为SubsetLen+1的值void SubsetsRec(vector<int> &S,vector<int>tmp ,vector<vector<int>>&ResultSubset, int SubsetLen = 0, int Pos = 0){vector<int> v;for (int i = 0; i < SubsetLen;++i)v.push_back(tmp[i]);if (!CompareSet.count(v))//判别当前子数组是否重复{ResultSubset.push_back(v);CompareSet.insert(v);}for (int i = Pos; i < S.size();++i){tmp[SubsetLen] = S[i];//确定子数组中索引号为SubsetLen的值SubsetsRec(S, tmp, ResultSubset, SubsetLen + 1, i + 1);}}vector<vector<int> > subsetsWithDup(vector<int> &S) {vector<vector<int>>ResultSubset;vector<int>SubsetTmp(S.size());sort(S.begin(), S.end());SubsetsRec(S,SubsetTmp ,ResultSubset);return ResultSubset;}