subsets

Q:

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],
  []
]

A: DFS问题。

    void dfs(vector<int>& S, int step,vector<int>& sequence,vector<vector<int> >& result)
    {
        if(step == S.size())
        {
            result.push_back(sequence);
            //sequence.clear();                   //注意这里不要clear，因为在上层做了恢复。
            return;
        }
        dfs(S,step+1,sequence,result);  //not choose S[step]
        sequence.push_back(S[step]);   //choose S[step]
        dfs(S,step+1,sequence,result);
        sequence.pop_back();            //返回时，将sequence恢复原样！！重要
        
    }
    vector<vector<int> > subsets(vector<int> &S) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function>
        vector<int> sequence;
        vector<vector<int> > result;
        sort(S.begin(),S.end());
        dfs(S,0,sequence,result);
        return result;       
    }

　　[second]

    vector<vector<int> > subsets(vector<int> &S) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<vector<int>> res;
        vector<int> path;
        sort(S.begin(),S.end()); //注意，先排序
        construct(0,S,path,res);
        return res;
        
    }
    
    void construct(int pos,vector<int>& S, vector<int>& path,vector<vector<int>>& res)
    {
        if(pos==S.size())
        {
            res.push_back(path);
            return;
        }

        construct(pos+1,S,path,res);
        path.push_back(S[pos]);
        construct(pos+1,S,path,res);
        path.pop_back();
    }