Reverse Linked List

Problem

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

Examples

The brute-force approach

Collect all node values into an array, reverse the array, then build a new list from the reversed values.

values = []
cur = head
while cur:
    values.append(cur.val)
    cur = cur.next

values.reverse()

dummy = ListNode(0)
cur = dummy
for v in values:
    cur.next = ListNode(v)
    cur = cur.next

return dummy.next

This uses O(n) extra space for the values array. You can reverse the list in place using only three pointers and O(1) space, no new nodes needed.

Spotting the pattern

This is a Linked List problem. The key question to ask yourself:

Can I reverse the list in place using prev, curr, and next pointers in a single pass?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.