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40 lines
1000 B
HTML
40 lines
1000 B
HTML
<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>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg" style="width: 542px; height: 222px;" />
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<pre>
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<strong>输入：</strong>head = [1,2,3,4,5], n = 2
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<strong>输出：</strong>[1,2,3,5]
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>head = [1], n = 1
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<strong>输出：</strong>[]
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</pre>
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<p><strong>示例 3：</strong></p>
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<pre>
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<strong>输入：</strong>head = [1,2], n = 1
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<strong>输出：</strong>[1]
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li>链表中结点的数目为 <code>sz</code></li>
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<li><code>1 <= sz <= 30</code></li>
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<li><code>0 <= Node.val <= 100</code></li>
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<li><code>1 <= n <= sz</code></li>
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</ul>
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<p> </p>
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<p><strong>进阶：</strong>你能尝试使用一趟扫描实现吗？</p>
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