Swap Nodes in Pairs
<p>Given a&nbsp;linked list, swap every two adjacent nodes and return its head. You must solve the problem without&nbsp;modifying the values in the list&#39;s nodes (i.e., only nodes themselves may be changed.)</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">head = [1,2,3,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">[2,1,4,3]</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0024.Swap%20Nodes%20in%20Pairs/images/swap_ex1.jpg" style="width: 422px; height: 222px;" /></p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">head = []</span></p>

<p><strong>Output:</strong> <span class="example-io">[]</span></p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">head = [1]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1]</span></p>
</div>

<p><strong class="example">Example 4:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">head = [1,2,3]</span></p>

<p><strong>Output:</strong> <span class="example-io">[2,1,3]</span></p>
</div>

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

<ul>
	<li>The number of nodes in the&nbsp;list&nbsp;is in the range <code>[0, 100]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 100</code></li>
</ul>
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 swap_pairs(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut dummy = Some(Box::new(ListNode { val: 0, next: head }));
        let mut cur = dummy.as_mut().unwrap();
        while cur.next.is_some() && cur.next.as_ref().unwrap().next.is_some() {
            cur.next = {
                let mut b = cur.next.as_mut().unwrap().next.take();
                cur.next.as_mut().unwrap().next = b.as_mut().unwrap().next.take();
                let a = cur.next.take();
                b.as_mut().unwrap().next = a;
                b
            };
            cur = cur.next.as_mut().unwrap().next.as_mut().unwrap();
        }
        dummy.unwrap().next
    }
}
