Reverse Linked List

easyTests passed 2026-11-11v1.0.0

TC: O(N)|SC: O(1)

Linked Lists Original Statement (leetcode).md

1. Core Invariant / The Trick

Maintain `prev = None` and `curr = head`. Before reversing `curr.next = prev`, save `nxt = curr.next`, then advance `prev, curr = curr, nxt`. Return `prev`.

2. Complexity Bounds & Proof

Visits each node once redirecting one pointer with constant auxiliary variables.

3. Minimal Idiomatic Code

def reverseList(head: Optional[ListNode]) -> Optional[ListNode]:
    prev, curr = None, head
    while curr:
        nxt = curr.next
        curr.next = prev
        prev = curr
        curr = nxt
    return prev

4. Failure Modes & Interview Follow-ups

Pitfalls:- Dropping `nxt` before mutating `curr.next` loses the rest of the list. - Returning `head` instead of `prev` yields only the old head node with `next = null`.

Follow-ups:- Recursive version uses O(N) call stack space; Reverse Linked List II reverses a slice `[left, right]` with a dummy node.

Personal Notes & Scratchpad (Local-First Yjs CRDT)

Saved instantaneously in IndexedDB. Streams live across peers via Hocuspocus when signed in.

Related Invariants

Reorder List (O(N))Linked List Cycle (O(N))

Browser Scratchpad

Console Output

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