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],
  []
]
思路：求集合S的所有组合，就是求n个不同的整数的长度为m（0<=m<=n)的组合，我们把n个整数分成两部分：第一个整数和剩下的所有整数。如果组合包含第一个整数，则下一步在剩余的整数中选取m-1个整数，如果组合里不包含第一个字符，则下一步在剩余的n-1个字符里选取m个字符。这道题很明显使用递归的方式解决。

class Solution {
public:
    void find_subsets(vector<vector<int> > &result,vector<int> &S,vector<int> data,int index)
    {
        if(index==S.size())
        {
            result.push_back(data);
            return;
        }
        data.push_back(S[index]);
        find_subsets(result,S,data,index+1);
        data.pop_back();
        find_subsets(result,S,data,index+1);
    }
    vector<vector<int> > subsets(vector<int> &S) {
        vector<vector<int> > result;
        result.clear();
        vector<int> data;
        if(S.empty())
            return result;
        sort(S.begin(),S.end());
        find_subsets(result,S,data,0);
        return result;
    }
};

 解法二：

class Solution {
public:
    void find_subsets(vector<vector<int> > &result,vector<int> &S,vector<int> data,int index)
    {
        result.push_back(data);
        for(int i=index;i<S.size();i++)
        {
            data.push_back(S[i]);
            find_subsets(result,S,data,i+1);
            data.pop_back();
        }
    }
    vector<vector<int> > subsets(vector<int> &S) {
        vector<vector<int> > result;
        result.clear();
        vector<int> data;
        if(S.empty())
            return result;
        sort(S.begin(),S.end());
        find_subsets(result,S,data,0);
        return result;
    }
};

 解法三：

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