[LeetCode]Subsets

Given a set of distinct integers, 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 = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

思考：最后还是按上一题的思路来k=0:n

class Solution {
private:
    vector<vector<int> > res;
    vector<int> ans;
public:
    void DFS(vector<int> S,int n,int start,int dep,int k)
    {
        if(dep==k)
        {
            res.push_back(ans);
            return;
        }
        for(int i=start;i<n-k+1;i++)
        {
            ans.push_back(S[i+dep]);
            DFS(S,n,i,dep+1,k);
            ans.pop_back();
        }
    }
    vector<vector<int> > subsets(vector<int> &S) {
        res.clear();
        ans.clear();
        int n=S.size();
        sort(S.begin(),S.end());
        for(int k=0;k<=n;k++)
        {
            DFS(S,n,0,0,k); //上一题的k
        }
        return res;
    }
};

second time:

class Solution {
public:
    void DFS(vector<vector<int> > &res, vector<int> &ans, vector<int> &S, int dep, int start)
    {
        res.push_back(ans);
        if(dep == S.size()) return;
        for(int i = start; i < S.size(); ++i)
        {
            ans.push_back(S[i]);
            DFS(res, ans, S, dep + 1, i + 1);
            ans.pop_back();
        }
    }
    vector<vector<int> > subsets(vector<int> &S) {
        vector<vector<int> > res;
        vector<int> ans;
        sort(S.begin(), S.end());
        DFS(res, ans, S, 0, 0);
        return res;
    }
};