Remove Nth Node From End of List
<p>Given the <code>head</code> of a linked list, remove the <code>n<sup>th</sup></code> node from the end of the list and return its head.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0019.Remove%20Nth%20Node%20From%20End%20of%20List/images/remove_ex1.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5], n = 2
<strong>Output:</strong> [1,2,3,5]
</pre>

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

<pre>
<strong>Input:</strong> head = [1], n = 1
<strong>Output:</strong> []
</pre>

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

<pre>
<strong>Input:</strong> head = [1,2], n = 1
<strong>Output:</strong> [1]
</pre>

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

<ul>
	<li>The number of nodes in the list is <code>sz</code>.</li>
	<li><code>1 &lt;= sz &lt;= 30</code></li>
	<li><code>0 &lt;= Node.val &lt;= 100</code></li>
	<li><code>1 &lt;= n &lt;= sz</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you do this in one pass?</p>
Linked List; Two Pointers
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {
        let dummy = ListNode(0)
        dummy.next = head
        var fast: ListNode? = dummy
        var slow: ListNode? = dummy
        for _ in 0..<n {
            fast = fast?.next
        }
        while fast?.next != nil {
            fast = fast?.next
            slow = slow?.next
        }
        slow?.next = slow?.next?.next
        return dummy.next
    }
}
