Leetcode - 19. Remove Nth Node From End of List

Medium

Dec 24, 2025

#linked-list

19. Remove Nth Node From End of List

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?

Notes

Intuition

By maintaining a gap of n nodes between two pointers, when the right pointer reaches the end, the left pointer will be just before the node to remove.

Implementation

First, advance the right pointer n steps ahead. Create a dummy node pointing to the head to handle edge cases cleanly, and set the left pointer to the dummy. Move both pointers together until right reaches null. Now left is positioned just before the target node—skip over it by setting left.next = left.next.next.

Edge-cases

The dummy node handles removing the first node (when n equals the list length). Without it, we'd need special case logic for updating the head.

Time: O(n) — single pass through the list
Space: O(1) — only using pointers

Complexity

Solution

from typing import Optional
from lcutils import ListNode

# 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]:
        right = head
        while n > 0:
            right = right.next if right else None
            n -= 1

        dummy = ListNode()
        dummy.next = head
        left = dummy
        while left and right:
            left = left.next
            right = right.next

        if left and left.next:
            left.next = left.next.next
        return dummy.next
        



if __name__ == "__main__":
    # Include one-off tests here or debugging logic that can be run by running this file
    # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
    solution = Solution()