remove-nth-node-from-end-of-list

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Algorithms

Given the <code>head</code> of a linked list, remove the <code>nth</code> node from the end of the list and return its head.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]
</pre>

Example 2:

<pre>
Input: head = [1], n = 1
Output: []
</pre>

Example 3:

<pre>
Input: head = [1,2], n = 1
Output: [1]
</pre>

&nbsp;
Constraints:

<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>

&nbsp;
Follow up: Could you do this in one pass?

removeNthFromEnd

Remove Nth Node From End of List

Swapping Nodes in a Linked List

swapping-nodes-in-a-linked-list

交换链表中的节点

Delete N Nodes After M Nodes of a Linked List

delete-n-nodes-after-m-nodes-of-a-linked-list

删除链表 M 个节点之后的 N 个节点

Delete the Middle Node of a Linked List

delete-the-middle-node-of-a-linked-list

删除链表的中间节点

Linked List

Two Pointers

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

&nbsp;

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

示例 2：

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

示例 3：

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

&nbsp;

提示：

<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>

&nbsp;

进阶：你能尝试使用一趟扫描实现吗？