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;

    }

};