LeetCode: Combinations

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

For example,
If n = 4 and k = 2, a solution is:

[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

#include<iostream>
#include<vector>
#include<map>
#include <algorithm>
using namespace::std;


#include <algorithm>
#include <algorithm>

class Solution {
public:
	void findCombination(int n, int k, vector<vector<int> >& result, vector<int> path, int start)
	{
		if(n <= 0 || k <= 0)
			return;
		if(k > n)
			return;
		//cout<<path.size()<<endl;
		if(path.size() == k)
		{
			//cout<<"here"<<endl;
			result.push_back(path);
			
		}
		if(path.size() < k)
		{
			
			//vector<int> tmp_vec(path);
			for(int i = start; i <= n; i++)
			{
				path.push_back(i);
				findCombination(n, k, result, path, path.back() + 1);
				path.pop_back();
			}

		}
	}
    vector<vector<int> > combine(int n, int k) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
		vector<vector<int> > result;
		vector<int> path;
        findCombination(n, k, result, path, 1);
		
		return result;
    }
};


int main()
{
	vector<int> input;
	input.push_back(1);
	input.push_back(2);
	//input.push_back(10);
	Solution ss;
	
	vector<vector<int> > result = ss.combine(1, 1);
	for(int i = 0; i < result.size(); i++)
	{
		for(int j = 0; j < result[i].size(); j++)
		{
			cout<<result[i][j];

		}
	cout<<endl;
	}
}

Round 2:

class Solution {
public:
    vector<vector<int> > combine(int n, int k) {
        vector<vector<int> > result;
        if(n == 0 || k == 0)
            return result;
        vector<int> cur;
        dfs(k, n, result, 1, cur);
        return result;
    }
private:
    void dfs(int k, int n, vector<vector<int> > &result, int index, vector<int> &cur)
    {
        
        if(cur.size() == k)
        {
            result.push_back(cur);
            return;
        }
        if(index > n)
            return;
        dfs(k, n, result, index+1, cur);
        cur.push_back(index);
        dfs(k, n, result, index+1, cur);
        cur.pop_back();
    }
};