<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>

class Solution {
public:
    vector<vector<int>> combine(int n, int k) {
        vector<vector<int>> sol;
        if(k == 1){
            for (int i = 1; i <= n; i++){
                sol.push_back({i});
            }
        }
        else {
            vector<int> cur;
            for (int i = 1; i <= n-k+1; i++){
                cur.push_back(i);
                next(i+1, n, k-1, cur, sol);
                cur.pop_back();
            }
        }
        return sol;
    }

    void next(int p, int n, int k, vector<int> cur, vector<vector<int>> &sol){
        if (k == 0){
            sol.push_back(cur);
            return;
        }
        for (int i = p; i <= n-k+1; i++){
            cur.push_back(i);
            next(i+1, n, k-1, cur, sol);
            cur.pop_back();
        }
        return;

    }
};
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