# 19. Remove Nth Node From End of List

#### Medium

***

Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/10/03/remove_ex1.jpg)

<pre><code>Input: head = [1,2,3,4,5], n = 2
<strong>Output:
</strong> [1,2,3,5]
</code></pre>

**Example 2:**

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

**Example 3:**

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

&#x20;

**Constraints:**

* The number of nodes in the list is `sz`.
* `1 <= sz <= 30`
* `0 <= Node.val <= 100`
* `1 <= n <= sz`

&#x20;

**Follow up:** Could you do this in one pass?

```python
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        slow, fast = head, head
        for _ in range(n):
            fast = fast.next
        if not fast:
            return slow.next
        while fast.next:
            slow = slow.next
            fast = fast.next
        slow.next = slow.next.next
        return head
```