Pascal's Triangle II
<p>Given an integer <code>rowIndex</code>, return the <code>rowIndex<sup>th</sup></code> (<strong>0-indexed</strong>) row of the <strong>Pascal&#39;s triangle</strong>.</p>

<p>In <strong>Pascal&#39;s triangle</strong>, each number is the sum of the two numbers directly above it as shown:</p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0119.Pascal%27s%20Triangle%20II/images/PascalTriangleAnimated2.gif" style="height:240px; width:260px" />
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> rowIndex = 3
<strong>Output:</strong> [1,3,3,1]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> rowIndex = 0
<strong>Output:</strong> [1]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> rowIndex = 1
<strong>Output:</strong> [1,1]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= rowIndex &lt;= 33</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you optimize your algorithm to use only <code>O(rowIndex)</code> extra space?</p>
Array; Dynamic Programming
function getRow(rowIndex: number): number[] {
    const f: number[] = Array(rowIndex + 1).fill(1);
    for (let i = 2; i < rowIndex + 1; ++i) {
        for (let j = i - 1; j; --j) {
            f[j] += f[j - 1];
        }
    }
    return f;
}
