Remove Nth Node From End - Leetcode 19 | JavaScript Two‑Pointer
Complete LeetCode 19 solution in JavaScript: Remove the nth node from the end of a singly linked list using the two‑pointer technique with a dummy node.
Given head and n, we create dummy = new Node(0, head), then use prev = dummy and end = head. First, move end forward by n steps to create a gap of n between prev and end. Then move both one step at a time while end is not null. When end reaches the end, prev.next is the node to delete. Set prev.next = prev.next.next and return dummy.next.
Example: 1→2→3→4→5, n=2 → 1→2→3→5.
What you’ll learn:
Problem: Remove the nth node from the end and return the head.
Why use a dummy node to handle removing the head easily.
Two‑pointer gap pattern: end moves n steps, then both move together.
JavaScript implementation with clear pointer updates.
Edge cases: removing first node, removing last node, single node list.
Complexity: O(n) time, O(1) extra space.
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