Subsets I

Elements in a subset must be in non-descending order.

The solution set must not contain duplicate subsets.
For example,
If S = [1,2,3], a solution is:
[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

Solution: 1. Recursive solution.
2. Iterative solution. Contributed by yinlinglin.

 1 class Solution {
 2 public:
 3     
 4     vector<vector<int> > subsets(vector<int> &S) {
 5         vector<vector<int> > res;
 6         sort(S.begin(), S.end());
 7         vector<int> set;
 8         int N = S.size();
 9         for(int i = 0; i <= N; i++) {
10             subsets(S, 0, i, set, res);
11         }
12         return res;
13     }
14     
15     void subsets(vector<int> &S, int start, int length, vector<int> &set, vector<vector<int> > &res) {
16         if(set.size() == length) {
17             res.push_back(set);
18             return;
19         }
20         int N = S.size(), M = set.size();
21         for(int i = start; i < N - (length-M) + 1; i++) {
22             set.push_back(S[i]);
23             subsets(S, i+1, length, set, res);
24             set.pop_back();
25         }
26     }
27 };

10. 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],
[]
]

 1 class Solution {
 2 public:
 3     vector<vector<int> > subsetsWithDup(vector<int> &S) {
 4         vector<vector<int> > res;
 5         vector<int> set;
 6         int N = S.size();
 7         sort(S.begin(), S.end());
 8         res.push_back(vector<int>());
 9         for(int i = 1; i <= N; i++) {
10             subsetsWithDupRe(S, 0, i, set, res);
11         }
12         return res;
13     }
14     
15     void subsetsWithDupRe(vector<int> &S, int start, int length, vector<int> &set, vector<vector<int> > &res)
16     {
17         int N = S.size(), M = set.size();
18         if(M == length) {
19             res.push_back(set);
20             return;
21         }
22         for(int i = start; i < N - (length-M) + 1; i++) {
23             if(i > start && S[i] == S[i-1]) continue;// duplicate
24             set.push_back(S[i]);
25             subsetsWithDupRe(S, i+1, length, set, res);
26             set.pop_back();
27         }
28         
29     }
30 };