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.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn reverse_k_group(head: Option<Box<ListNode>>, k: i32) -> Option<Box<ListNode>> {
        fn reverse(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
            let mut head = head;
            let mut pre = None;
            while let Some(mut node) = head {
                head = node.next.take();
                node.next = pre.take();
                pre = Some(node);
            }
            pre
        }

        let mut dummy = Some(Box::new(ListNode::new(0)));
        let mut pre = &mut dummy;
        let mut cur = head;
        while cur.is_some() {
            let mut q = &mut cur;
            for _ in 0..k - 1 {
                if q.is_none() {
                    break;
                }
                q = &mut q.as_mut().unwrap().next;
            }
            if q.is_none() {
                pre.as_mut().unwrap().next = cur;
                return dummy.unwrap().next;
            }

            let b = q.as_mut().unwrap().next.take();
            pre.as_mut().unwrap().next = reverse(cur);
            while pre.is_some() && pre.as_mut().unwrap().next.is_some() {
                pre = &mut pre.as_mut().unwrap().next;
            }
            cur = b;
        }
        dummy.unwrap().next
    }
}
