reverse-nodes-in-k-group

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Algorithms

Given the <code>head</code> of a linked list, reverse the nodes of the list <code>k</code> at a time, and return the modified list.

<code>k</code> is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of <code>k</code> then left-out nodes, in the end, should remain as it is.

You may not alter the values in the list&#39;s nodes, only nodes themselves may be changed.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]
</pre>

Example 2:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg" style="width: 542px; height: 222px;" />
<pre>
Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]
</pre>

&nbsp;
Constraints:

<ul>
	<li>The number of nodes in the list is <code>n</code>.</li>
	<li><code>1 &lt;= k &lt;= n &lt;= 5000</code></li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
</ul>

&nbsp;
Follow-up: Can you solve the problem in <code>O(1)</code> extra memory space?

reverseKGroup

Reverse Nodes in k-Group

Swap Nodes in Pairs

swap-nodes-in-pairs

两两交换链表中的节点

Swapping Nodes in a Linked List

swapping-nodes-in-a-linked-list

交换链表中的节点

Reverse Nodes in Even Length Groups

reverse-nodes-in-even-length-groups

反转偶数长度组的节点

Recursion

Linked List

给你链表的头节点 <code>head</code> ，每&nbsp;<code>k</code>&nbsp;个节点一组进行翻转，请你返回修改后的链表。

<code>k</code> 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是&nbsp;<code>k</code>&nbsp;的整数倍，那么请将最后剩余的节点保持原有顺序。

你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。

&nbsp;

示例 1：
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
输入：head = [1,2,3,4,5], k = 2
输出：[2,1,4,3,5]
</pre>

示例 2：

<img alt="" src="https://assets.leetcode.com/uploads/2020/10/03/reverse_ex2.jpg" style="width: 542px; height: 222px;" />

<pre>
输入：head = [1,2,3,4,5], k = 3
输出：[3,2,1,4,5]
</pre>

&nbsp;
提示：

<ul>
	<li>链表中的节点数目为 <code>n</code></li>
	<li><code>1 &lt;= k &lt;= n &lt;= 5000</code></li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
</ul>

&nbsp;

进阶：你可以设计一个只用 <code>O(1)</code> 额外内存空间的算法解决此问题吗？

<ul>
</ul>