29. Reverse Linked List

Q: Recursive or iterative?

Iterative is preferred — O(1) space vs O(n) recursion stack. The recursive form is elegant but doesn't survive 10K+ length lists in JS.

Q: Why save next BEFORE reassigning?

curr.next = prev overwrites curr.next, so if you didn't save it first, you've lost your way to the rest of the list.

easyAsked at Datadog

Reverse a singly linked list. Datadog uses this as the bedrock linked-list question — pointer manipulation here is required for harder problems like reverse-in-groups or merge-k-sorted.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-20

Source citations

Public interview reports confirming this problem appears in Datadog loops.

Glassdoor (2026-Q1)— Datadog phone screen — used to filter on pointer manipulation.
Blind (2025-12)— Recurring Datadog warmup.

Problem

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

Constraints

The number of nodes in the list is the range [0, 5000].
-5000 <= Node.val <= 5000

Examples

Example 1

Input

head = [1,2,3,4,5]

Output

[5,4,3,2,1]

Example 2

Input

head = [1,2]

Output

[2,1]

Example 3

Input

head = []

Output

[]

Approaches

1. Collect to array + rebuild

Dump values to array, build a new list from the end.

Time: O(n)
Space: O(n)

function reverseList(head) {
  const vals = [];
  while (head) { vals.push(head.val); head = head.next; }
  const dummy = { val: 0, next: null };
  let tail = dummy;
  for (let i = vals.length - 1; i >= 0; i--) {
    tail.next = { val: vals[i], next: null };
    tail = tail.next;
  }
  return dummy.next;
}

Tradeoff: O(n) extra memory + allocation. Datadog will fail this for not doing in-place pointer reversal.

2. Iterative three-pointer (optimal)

Walk forward with prev, curr, next. At each step: save next, flip curr.next to prev, advance.

Time: O(n)
Space: O(1)

function reverseList(head) {
  let prev = null;
  let curr = head;
  while (curr) {
    const next = curr.next;
    curr.next = prev;
    prev = curr;
    curr = next;
  }
  return prev;
}

Tradeoff: O(1) extra space, single pass. The save-flip-advance idiom is the foundation for harder Datadog linked-list problems.

Datadog-specific tips

Datadog interviewers will follow up with: 'Now reverse in groups of K' or 'Now reverse a sublist between m and n.' Show that the three-pointer idiom generalizes — every harder linked-list question builds on it. Mention the recursive version too; both should be in your toolkit.

Common mistakes

Forgetting to save curr.next BEFORE reassigning it — loses the rest of the list.
Returning head at the end — head is now the tail; return prev (the new head).
Recursive version with O(n) stack — works but blows up on long lists in JS.

Follow-up questions

An interviewer at Datadog may pivot to one of these next:

Reverse Linked List II (LC 92) — reverse between positions m and n.
Reverse Nodes in K-Group (LC 25) — generalization.
Palindrome Linked List (LC 234) — uses reverse-second-half as a subroutine.

Solve it now

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Output

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FAQ

Recursive or iterative?

Iterative is preferred — O(1) space vs O(n) recursion stack. The recursive form is elegant but doesn't survive 10K+ length lists in JS.

Why save next BEFORE reassigning?

curr.next = prev overwrites curr.next, so if you didn't save it first, you've lost your way to the rest of the list.

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