LeetCode 19

Subtitle / Summary The hard part of LeetCode 19 is not deleting a node. It is locating the predecessor of the n-th node from the end in a singly linked list without walking backward. Reading time: 12-15 min Tags: Hot100, linked list, two pointers, dummy node SEO keywords: Remove Nth Node From End of List, one-pass two pointers, dummy node, LeetCode 19, Hot100 Meta description: A derivation-first ACERS explanation of LeetCode 19 with fixed-gap two pointers, dummy node boundary handling, engineering mapping, and runnable multi-language implementations. A - Algorithm (Problem and Algorithm) Problem Restatement Given the head of a linked list, remove the n-th node from the end of the list and return the head of the modified list. ...

Subtitle / Summary The hard part is not deletion itself, but locating the predecessor of the nth node from the end in a singly linked list. This article derives the one-pass two-pointer solution from simpler baselines and explains correctness, boundaries, and engineering transfer. Reading time: 12-15 min Tags: linked list, two pointers, interview high frequency SEO keywords: LeetCode 19, Remove Nth Node From End of List, linked list, fast/slow pointers, dummy node Meta description: A complete ACERS walkthrough for removing the nth node from the end: from brute force to one-pass two pointers, with complexity, pitfalls, engineering scenarios, and Python/C/C++/Go/Rust/JS implementations. Target Readers Beginners building a stable template for linked-list interview problems Developers who know fast/slow pointers but still make boundary mistakes Backend/system engineers who want to transfer problem-solving templates to chain-structured data in production Background / Motivation “Remove the nth node from the end” is a classic medium-level linked-list problem. The challenge is usually not the delete operation itself, but: ...