Reverse Nodes in k-Group
<p>Given the <code>head</code> of a linked list, reverse the nodes of the list <code>k</code> at a time, and return <em>the modified list</em>.</p>

<p><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.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0025.Reverse%20Nodes%20in%20k-Group/images/reverse_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], k = 2
<strong>Output:</strong> [2,1,4,3,5]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0025.Reverse%20Nodes%20in%20k-Group/images/reverse_ex2.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], k = 3
<strong>Output:</strong> [3,2,1,4,5]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<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>

<p>&nbsp;</p>
<p><strong>Follow-up:</strong> Can you solve the problem in <code>O(1)</code> extra memory space?</p>
Recursion; Linked List
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val = 0, ListNode next = null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode ReverseKGroup(ListNode head, int k) {
        var dummy = new ListNode(0);
        dummy.next = head;
        var pre = dummy;

        while (pre != null) {
            var cur = pre;
            for (int i = 0; i < k; i++) {
                if (cur.next == null) {
                    return dummy.next;
                }
                cur = cur.next;
            }

            var node = pre.next;
            var nxt = cur.next;
            cur.next = null;
            pre.next = Reverse(node);
            node.next = nxt;
            pre = node;
        }

        return dummy.next;
    }

    private ListNode Reverse(ListNode head) {
        ListNode prev = null;
        var cur = head;
        while (cur != null) {
            var nxt = cur.next;
            cur.next = prev;
            prev = cur;
            cur = nxt;
        }
        return prev;
    }
}
