<p>Given two integers <code>n</code> and <code>k</code>, return <em>all possible combinations of</em> <code>k</code> <em>numbers chosen from the range</em> <code>[1, n]</code>.</p>

<p>You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 4, k = 2
<strong>Output:</strong> [[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
<strong>Explanation:</strong> There are 4 choose 2 = 6 total combinations.
Note that combinations are unordered, i.e., [1,2] and [2,1] are considered to be the same combination.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1, k = 1
<strong>Output:</strong> [[1]]
<strong>Explanation:</strong> There is 1 choose 1 = 1 total combination.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 20</code></li>
	<li><code>1 &lt;= k &lt;= n</code></li>
</ul>

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



class Solution {
    
public:
    vector<vector<int>> answer;
    void dfs(vector<int>v, int n , int k , int cnt , int idx){
        for(int i = idx ; i<=n ; i++){
            if(cnt == k){
                answer.push_back(v);
                return;
            }else{
                v.push_back(i+1);
                dfs(v,n,k,cnt+1,i+1);
                v.pop_back();
                
            }

        }
    }
    vector<vector<int>> combine(int n, int k) {
        //이건 조합 문제네.. 순서 상관 있고 중복 없고.
        //조합은 idx임
        vector<int>ch(n,0);
        vector<int>temp;
        dfs(temp,n,k,0,0);        
        return answer;
    }
};

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