<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>> out(1);
        size_t ok = 0;
        for (size_t ki = 0; ki < k; ki++) {
            size_t on = out.size();
            for (size_t oi = ok; oi < on; oi++) {
                for (size_t i = 0; i < n; i++) {
                    if (!out[oi].empty() && i < out[oi].back()) continue;
                    out.push_back(out[oi]);
                    out.back().push_back(i+1);
                }
            }
            ok = on;
            //out.erase(out.begin(), out.begin()+on);
        }

        return vector<vector<int>>(out.begin()+ok, out.end());
    }
};
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