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],
  []
]

　　DFS 的简单应用 ： 求组合

class Solution {
public:
  void  DFS(vector<int> &S, vector<int> &temp,int n, int size,int start)
    {

        if(n == size)
        {
            result.push_back(temp);
            return ;
        }
        if(n > size)
            return ;

        for(int i = start; i< len ;i++)
        {
            if(flag[i] == false)
            {
                flag[i] = true;
                temp.push_back(S[i]);
                DFS(S, temp, n+, size,i+);
                temp.pop_back();
                flag[i] = false;
            }
        }

    }
    vector<vector<int> > subsets(vector<int> &S) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        result.clear();
        len = S.size();
        flag.resize(len,false);
        vector<int> temp;
        result.push_back(temp) ;
        sort(S.begin(), S.end());

        for(int i = ; i <= len ; i++)
             DFS(S, temp,, i,);

        return result;
    }
private:
    vector<vector<int> > result ;
    vector<bool> flag;
    int len;
};

解释下start,因为组合和排列不同，组合不考虑排序，所以必须给元素进入temp指定一个次序，这个规则定义就是通过start,这样保证temp是有序的，也就保证result中没有重复

重写后的代码：

class Solution {
public:
    void DFS(vector<int> &S, int currentSize, int length, int currentPos, vector<int> &ans)
    {

        if(length == currentSize){

            res.push_back(ans);
            return;
        }
        for(int i = currentPos; i < S.size(); i++)
        {
            ans.push_back(S[i]);
            DFS(S, currentSize, length + , i+, ans);
            ans.pop_back();
        }
    }
    vector<vector<int> > subsets(vector<int> &S) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sort(S.begin(), S.end());
        res.clear();
        vector<int> empt;
        res.push_back(empt);
        for(int i = ; i <= S.size(); i++)
        {
            vector<int> ans;
            DFS(S, i, , , ans);
        }

        return res;
    }
private:
    vector<vector<int>> res;
};