# Question

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

```
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
```

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

# Solution

```
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def removeNthFromEnd(self, head, n):
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
nth_prev = head
tail = head
for i in range(0, n):
tail = tail.next
if tail is None:
head = head.next
return head
while tail.next is not None:
nth_prev = nth_prev.next
tail = tail.next
nth_prev.next = nth_prev.next.next
return head
```