Permutations III
<p>Given an integer <code>n</code>, an <strong>alternating permutation</strong> is a permutation of the first <code>n</code> positive integers such that no <strong>two</strong> adjacent elements are <strong>both</strong> odd or <strong>both</strong> even.</p>

<p>Return <em>all such </em><strong>alternating permutations</strong> sorted in lexicographical order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 4</span></p>

<p><strong>Output:</strong> <span class="example-io">[[1,2,3,4],[1,4,3,2],[2,1,4,3],[2,3,4,1],[3,2,1,4],[3,4,1,2],[4,1,2,3],[4,3,2,1]]</span></p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">[[1,2],[2,1]]</span></p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 3</span></p>

<p><strong>Output:</strong> <span class="example-io">[[1,2,3],[3,2,1]]</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10</code></li>
</ul>
Array; Backtracking
function permute(n: number): number[][] {
    const ans: number[][] = [];
    const vis: boolean[] = Array(n).fill(false);
    const t: number[] = Array(n).fill(0);
    const dfs = (i: number) => {
        if (i >= n) {
            ans.push([...t]);
            return;
        }
        for (let j = 1; j <= n; ++j) {
            if (!vis[j] && (i === 0 || t[i - 1] % 2 !== j % 2)) {
                vis[j] = true;
                t[i] = j;
                dfs(i + 1);
                vis[j] = false;
            }
        }
    };
    dfs(0);
    return ans;
}
