<p>给你一个链表，删除链表的倒数第 <code>n</code><em> </em>个结点，并且返回链表的头结点。</p>

<p><strong>进阶：</strong>你能尝试使用一趟扫描实现吗？</p>

<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>输入：</strong>head = [1,2,3,4,5], n = 2
<strong>输出：</strong>[1,2,3,5]
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>head = [1], n = 1
<strong>输出：</strong>[]
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>head = [1,2], n = 1
<strong>输出：</strong>[1]
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>链表中结点的数目为 <code>sz</code></li>
	<li><code>1 <= sz <= 30</code></li>
	<li><code>0 <= Node.val <= 100</code></li>
	<li><code>1 <= n <= sz</code></li>
</ul>
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