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],
  []
]

S先排序下，使全集能够按顺序输出。

class Solution {
public:
    vector<vector<int> >re;
    vector<vector<int> > subsets(vector<int> &S) {
        sort(S.begin(),S.end());
        for(int i = 0 ; i <= S.size() ; i++)
        {
            find(S , i);
        }
        return re;
        
    }
    void find(vector<int> ve,int k)
    {
        vector<int> result;
        if(k == 0)
        {
            re.push_back(result);
            return;
        }
        if(k == ve.size())
        {
            re.push_back(ve);
            return;
        }
        get(ve,-1,k,result);
        
    }
    void get(vector<int> ve,int begin,int k,vector<int> result)
    {
        if(begin >= 0)
        {
            result.push_back(ve[begin]);
        }
        if(k == 0)
        {
            sort(result.begin(),result.end());
            re.push_back(result);
            return;
        }
        for(int i = begin+1; i <= ve.size() - k  ; i++)
        {
            get(ve,i,k-1,result);
        }
    }
};

　　有更简单的做法，维护一个index。当index到尾部是则将集合返回。

看到一个更漂亮的做法，忍不住贴过来。源自http://www.cppblog.com/Uriel/articles/205467.html

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