Description
Given the head
of a linked list, remove the nth
node from the end of the list and return its head.
Example 1:

Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5]
Example 2:
Input: head = [1], n = 1 Output: []
Example 3:
Input: head = [1,2], n = 1 Output: [1]
Constraints:
- The number of nodes in the list is
sz
. 1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
Follow up: Could you do this in one pass?
Solution
Python3
# 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]:
res = curr = ListNode(-1, head)
for _ in range(n):
head = head.next
while head and curr:
head = head.next
curr = curr.next
curr.next = curr.next.next
return res.next