<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>> result(1);
        for (int i = 1; i <= n; ++i ) {
            int stop = result.size();
            int p = 0, cutoff = i+k-n-1;
            for (int j = 0; j < stop; ++j) {
                int n = result[j].size();
                if (n >= k || n < cutoff) continue;
                while (p < j && static_cast<int>(result[p].size()) >= cutoff) ++p;
                if (p < j) {
                    result[p] = result[j];
                    result[p].push_back(i);
                } else {
                    result.push_back(result[j]);
                    result.back().push_back(i);
                }
            }
        }
        vector<vector<int>> ret;
        for (auto& row : result) {
            if (row.size() == k) {
                ret.emplace_back(move(row));
            }
        }
        return ret;
    }
};
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