LeetCode: 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],
[]
]

class Solution {
public:
vector<vector<int> > subsetsWithDup(vector<int> &S) {
vector<vector<int> > result;
vector<int> temp;
vector<int> dup;
result.push_back(temp);
sort(S.begin(), S.end());
for(int i = 0; i < S.size(); i++)
{
dup.clear();
dup.push_back(S[i]);
while(i < S.size()-1 && S[i] == S[i+1])
{
dup.push_back(S[i]);
i++;
}
int size = result.size();
for(int j = 0; j < size; j++)
{
temp = result[j];
int index = 0;
while(!dup.empty() && index < dup.size())
{
temp.push_back(dup[index]);
result.push_back(temp);
index++;
}
}
}
return result;
}
};

Round 2:

class Solution {
public:
vector<vector<int> > subsetsWithDup(vector<int> &S) {
vector<vector<int> > result;
if(S.size() == 0)
return result;
std::sort(S.begin(), S.end());
vector<int> cur;
dfs(S, result, 0, cur);
return result;
}
private:
void dfs(vector<int> &input, vector<vector<int> > &result, int index, vector<int> &cur)
{
if(index >= input.size())
{
result.push_back(cur);
return;
}
int count = 1;
int iter = index;
while(iter+1 < input.size() && input[iter] == input[iter+1])
{
count++;
iter++;
}
dfs(input, result, index+count, cur);
for(int i = 0; i < count; i++)
{
cur.push_back(input[index]);
dfs(input, result, index+count, cur);
}
while(cur.size() > 0 && cur.back() == input[index])
cur.pop_back();

}
};