Remove Nth Node From End of List

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

Example

Key Pointers

• Two pointers offset by n find "n from the end" in a single pass — no need to know the list's length first.
• A dummy node before head removes the special case where the node to delete is the head itself.
• Advance the fast pointer n + 1 steps first, then move both together; when fast reaches the end, slow sits right before the target node.

Step-by-Step Analysis

1. Create a dummy node pointing to head; set slow = dummy, fast = dummy.
2. Advance fast n + 1 steps ahead of slow.
3. Move slow and fast together, one step at a time, until fast is None.
4. Unlink the target node with slow.next = slow.next.next, then return dummy.next.

Complexity

Time: O(n) — Space: O(1)

Code

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def removeNthFromEnd(head: ListNode, n: int) -> ListNode:
    dummy = ListNode(0, head)
    slow = fast = dummy

    for _ in range(n + 1):
        fast = fast.next

    while fast:
        slow = slow.next
        fast = fast.next

    slow.next = slow.next.next
    return dummy.next

Code Explanation

How the solution above works, piece by piece.

1. dummy = ListNode(0, head) sits before the head, so removing the first node works the same as any other.
2. Move fast n + 1 steps: now there are exactly n nodes between slow and fast.
3. Move both one step at a time until fast is None. The gap is preserved, so slow stops just before the n-th node from the end.
4. slow.next = slow.next.next unlinks it.
5. Return dummy.next (the head may have changed). One pass: O(length) time, O(1) space.

Visual Walkthrough

Running it on head = [1, 2, 3, 4, 5], n = 2:

Video Explanation

NeetCode — Oracle Interview Question, LeetCode 19