Combinations
<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>
Backtracking
impl Solution {
    fn dfs(i: i32, n: i32, k: i32, t: &mut Vec<i32>, ans: &mut Vec<Vec<i32>>) {
        if t.len() == (k as usize) {
            ans.push(t.clone());
            return;
        }
        if i > n {
            return;
        }
        t.push(i);
        Self::dfs(i + 1, n, k, t, ans);
        t.pop();
        Self::dfs(i + 1, n, k, t, ans);
    }

    pub fn combine(n: i32, k: i32) -> Vec<Vec<i32>> {
        let mut ans = vec![];
        Self::dfs(1, n, k, &mut vec![], &mut ans);
        ans
    }
}
