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31. Add Two Numbers

mediumAsked at Salesforce

Add two numbers represented as linked lists (digits in reverse order). Salesforce asks this to test carry propagation and dummy-head pattern application together.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Salesforce loops.

  • Glassdoor (2026-Q1)— Salesforce billing-platform team uses this exact pattern for arbitrary-precision arithmetic.
  • Blind (2025-11)— Recurring on Salesforce backend onsites.

Problem

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list. You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Constraints

  • The number of nodes in each linked list is in the range [1, 100].
  • 0 <= Node.val <= 9
  • It is guaranteed that the list represents a number that does not have leading zeros.

Examples

Example 1

Input
l1 = [2,4,3], l2 = [5,6,4]
Output
[7,0,8]

Explanation: 342 + 465 = 807.

Example 2

Input
l1 = [0], l2 = [0]
Output
[0]

Example 3

Input
l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output
[8,9,9,9,0,0,0,1]

Approaches

1. Convert to numbers and add

Parse each list as a number, add, build the result list.

Time
O(n)
Space
O(n)
function addTwoNumbers(l1, l2) {
  const toNum = (l) => { let s = ''; while (l) { s = l.val + s; l = l.next; } return BigInt(s); };
  let sum = toNum(l1) + toNum(l2);
  if (sum === 0n) return { val: 0, next: null };
  const dummy = { next: null };
  let cur = dummy;
  while (sum > 0n) { cur.next = { val: Number(sum % 10n), next: null }; cur = cur.next; sum /= 10n; }
  return dummy.next;
}

Tradeoff: Works with BigInt but bypasses the actual exercise. Salesforce will dock you for not doing column addition.

2. Single pass with carry and dummy head

Walk both lists in parallel; sum corresponding digits plus carry; create a new node with sum % 10; carry = sum / 10. Use a dummy head.

Time
O(max(m, n))
Space
O(max(m, n))
function addTwoNumbers(l1, l2) {
  const dummy = { next: null };
  let cur = dummy, carry = 0;
  while (l1 || l2 || carry) {
    const sum = (l1?.val || 0) + (l2?.val || 0) + carry;
    carry = Math.floor(sum / 10);
    cur.next = { val: sum % 10, next: null };
    cur = cur.next;
    l1 = l1?.next;
    l2 = l2?.next;
  }
  return dummy.next;
}

Tradeoff: Single pass with the trifecta: dummy head, optional chaining, and 'while carry' to flush the trailing carry digit. Salesforce's preferred answer.

Salesforce-specific tips

Salesforce billing-platform team uses this exact pattern for currency arithmetic across very large amounts. They grade on whether you include `carry` in the while condition (so '99 + 1' produces '100' not '00'). Bonus signal: mention that the same algorithm extends to bignum arithmetic — Salesforce's BigDecimal Apex code uses identical logic.

Common mistakes

  • Forgetting carry in the loop condition — drops the final carry digit on inputs like [5] + [5].
  • Not using a dummy head — special-casing the first node bloats the code.
  • Computing sum % 10 and sum / 10 in wrong order or with bad division (use Math.floor).

Follow-up questions

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

  • Add Two Numbers II (LC 445) — digits in forward order.
  • Multiply Strings (LC 43).
  • What if the lists could have negative numbers?

Solve it now

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Output

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FAQ

Why does the loop continue while carry is non-zero?

After both lists are exhausted, a leftover carry needs a new node. Example: [9] + [9] = [8, 1]. Without `|| carry`, the trailing 1 would be lost.

Why use a dummy head?

Saves a special case for the first node. With dummy, every iteration uniformly appends to cur.next.

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