## Description

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?

## Solutions

### 1. Fast and slow pointers

```
# Time: O(n)
# Space: O(1)
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
fast, slow = head, head
for _ in range(n):
# if not fast.next:
# return None
fast = fast.next
while fast and fast.next:
fast = fast.next
slow = slow.next
if not fast:
return slow.next
else:
slow.next = slow.next.next
return head
# 208/208 cases passed (32 ms)
# Your runtime beats 58.65 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.7 MB)
```