Project Notes

Remove Nth Node From End of List

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Description

tab: English
 
<p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>
 
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], n = 2
<strong>Output:</strong> [1,2,3,5]
</pre>
 
<p><strong class="example">Example 2:</strong></p>
 
<pre>
<strong>Input:</strong> head = [1], n = 1
<strong>Output:</strong> []
</pre>
 
<p><strong class="example">Example 3:</strong></p>
 
<pre>
<strong>Input:</strong> head = [1,2], n = 1
<strong>Output:</strong> [1]
</pre>
 
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
 
<ul>
	<li>The number of nodes in the list is <code>sz</code>.</li>
	<li><code>1 &lt;= sz &lt;= 30</code></li>
	<li><code>0 &lt;= Node.val &lt;= 100</code></li>
	<li><code>1 &lt;= n &lt;= sz</code></li>
</ul>
 
<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you do this in one pass?</p>
 
 
 
> [!tip]- Hint 1
> 
> Maintain two pointers and update one with a delay of n steps.
 
 
---
 
[submissions](https://leetcode.com/problems/remove-nth-node-from-end-of-list/submissions/) | [solutions](https://leetcode.com/problems/remove-nth-node-from-end-of-list/solutions/)
 
 
tab: 中文
 
<p>给你一个链表，删除链表的倒数第&nbsp;<code>n</code><em>&nbsp;</em>个结点，并且返回链表的头结点。</p>
 
<p>&nbsp;</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>&nbsp;</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 &lt;= sz &lt;= 30</code></li>
	<li><code>0 &lt;= Node.val &lt;= 100</code></li>
	<li><code>1 &lt;= n &lt;= sz</code></li>
</ul>
 
<p>&nbsp;</p>
 
<p><strong>进阶：</strong>你能尝试使用一趟扫描实现吗？</p>
 
 
 
> [!tip]- 提示 1
> 
> Maintain two pointers and update one with a delay of n steps.
 
 
---
 
[提交记录](https://leetcode.cn/problems/remove-nth-node-from-end-of-list/submissions/) | [题解](https://leetcode.cn/problems/remove-nth-node-from-end-of-list/solution/)

Solutions & Notes

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