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], [] ]
禁止重复,就使用set,map等容器,就可以很简单解决了。
class Solution {
public:
vector > subsetsWithDup(vector &S)
{
sort(S.begin(), S.end());
vector > rs(1);
vector temp;
set > us;
us.insert(temp);
for (int i = 0; i < S.size(); i++)
{
for (int j = rs.size()-1; j>=0; j--)
{
temp = rs[j];
temp.push_back(S[i]);
rs.push_back(temp);
us.insert(temp);
}
}
rs.clear();
rs.assign(us.begin(), us.end());
return rs;
}
vector > subsetsWithDup2(vector &S)
{
sort(S.begin(), S.end());
vector > rs(1);
vector temp;
set > us;
us.insert(temp);
for (int i = 0; i < S.size(); i++)
{
for (int j = rs.size()-1; j>=0; j--)
{
temp = rs[j];
temp.push_back(S[i]);
rs.push_back(temp);
us.insert(temp);
}
rs.assign(us.begin(),us.end());
}
return rs;
}
};
//2014-2-14 update
vector > subsetsWithDup(vector &S)
{
sort(S.begin(), S.end());
vector > rs(1);
unordered_set > sv(rs.begin(), rs.end());
for (int i = 0; i < S.size(); i++)
{
for (int j = rs.size() - 1; j >= 0 ; j--)
{
rs.push_back(rs[j]);
rs.back().push_back(S[i]);
sv.insert(rs.back());
}
rs.assign(sv.begin(), sv.end());
}
return rs;
}