<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) {
        queue< vector<int> > q;
        for(int i=1; i<=n-k+1; i++){
            q.push(vector<int> (1, i));
        }
        for(int i=n-k+2; i<=n; i++){
            vector<int> vec;
            int qSize;
            qSize = q.size();
            for(int j=0; j<qSize; j++){
                vec = q.front();
                q.pop();
                for(int k=vec.back()+1; k<=i; k++){
                    vector<int> v(vec.begin(), vec.end());
                    v.push_back(k);
                    q.push(v);
                }
            }
        }

        vector< vector<int> > combinations;
        while(!q.empty()){
            combinations.push_back(q.front());
            q.pop();
        }
        return combinations;
    }
};
168462