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.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end
# @param {ListNode} head
# @return {ListNode}
def swap_pairs(head)
    dummy = ListNode.new(0, head)
    pre = dummy
    cur = head
    while !cur.nil? && !cur.next.nil?
        t = cur.next
        cur.next = t.next
        t.next = cur
        pre.next = t
        pre = cur
        cur = cur.next
    end
    dummy.next
end
