Reverse Linked List

The problem

Given the head of a singly linked list, reverse the list and return the new head.

Every node already holds the pointer you want to change, but swinging it backwards orphans the rest of the list before you can reach it. The entire mechanical challenge is stashing the forward neighbour in a temporary variable before rewriting next to point behind you.

Full solution

Three-pointer iteration is the one to write: O(n) time, O(1) space, no recursion depth risk. Stack and recursion both work but trade memory for no real benefit here.

from typing import Optional


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


def reverse_list(head: Optional[ListNode]) -> Optional[ListNode]:
    prev = None
    curr = head
    while curr is not None:
        ahead = curr.next  # save forward link before it is overwritten
        curr.next = prev
        prev = curr
        curr = ahead
    return prev

class ListNode {
  constructor(val = 0, next = null) {
    this.val = val;
    this.next = next;
  }
}

function reverseList(head) {
  let prev = null;
  let curr = head;
  while (curr !== null) {
    const ahead = curr.next; // save forward link before it is overwritten
    curr.next = prev;
    prev = curr;
    curr = ahead;
  }
  return prev;
}

The theory behind it

Linked List — the ground this problem stands on. All Linked List problems

What Linked List is

A linked list is a chain of separate cargo cars connected by coupling hooks, scattered anywhere across memory rather than sitting in a tidy contiguous row. Each car, called a node, holds a single piece of data and a pointer directing traffic to the address of the next car in line. Because nodes connect only by directional links, jumping straight to the tenth car is impossible without walking past the first nine.

When to reach for it

Choose a linked list when a problem requires frequent insertions and deletions at known positions without shifting whole blocks of surrounding memory. Problems mentioning pointer splicing, reversing subsequences in place, merging sorted streams, or detecting cycles in linear chains strongly point here. It is ideal when total capacity is unpredictable and memory allocation must happen one individual node at a time.

How the pattern works

Think in terms of pointer rewiring before dereferencing. Keep a dummy head node pointing to the start of the list so modifications to the initial item do not require separate edge logic. Always save references to neighboring nodes into temporary variables before cutting or redirecting forward links. When diagnosing loops or locating middle nodes, advance two references simultaneously at differing velocities so traversal completes without supplementary storage.

What each operation costs

Which roles need this problem

Linked List is a core topic for these 3 roles — if you're targeting one of them, this problem is early in your path, not optional.

Secondary for 5 more roles, including Full-Stack Developer, Android Developer, iOS Developer.