Reverse Linked List II
<p>Given the <code>head</code> of a singly linked list and two integers <code>left</code> and <code>right</code> where <code>left &lt;= right</code>, reverse the nodes of the list from position <code>left</code> to position <code>right</code>, and return <em>the reversed list</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0092.Reverse%20Linked%20List%20II/images/rev2ex2.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], left = 2, right = 4
<strong>Output:</strong> [1,4,3,2,5]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [5], left = 1, right = 1
<strong>Output:</strong> [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;= n &lt;= 500</code></li>
	<li><code>-500 &lt;= Node.val &lt;= 500</code></li>
	<li><code>1 &lt;= left &lt;= right &lt;= n</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you do it in one pass?
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_between(
        head: Option<Box<ListNode>>,
        left: i32,
        right: i32,
    ) -> Option<Box<ListNode>> {
        let mut dummy = Some(Box::new(ListNode { val: 0, next: head }));
        let mut pre = &mut dummy;
        for _ in 1..left {
            pre = &mut pre.as_mut().unwrap().next;
        }
        let mut cur = pre.as_mut().unwrap().next.take();
        for _ in 0..right - left + 1 {
            let mut next = cur.as_mut().unwrap().next.take();
            cur.as_mut().unwrap().next = pre.as_mut().unwrap().next.take();
            pre.as_mut().unwrap().next = cur.take();
            cur = next;
        }
        for _ in 0..right - left + 1 {
            pre = &mut pre.as_mut().unwrap().next;
        }
        pre.as_mut().unwrap().next = cur;
        dummy.unwrap().next
    }
}
