Add Two Numbers — DSA Visualizer

Problem

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each node contains a single digit. Add the two numbers and return the sum as a linked list.

Example 1

Input: l1 = [2,4,3], l2 = [5,6,4]

Output: [7,0,8]

Explanation: 342 + 465 = 807

Example 2

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]

Output: [8,9,9,9,0,0,0,1]

Explanation: 9999999 + 9999 = 10009998

Example 3

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

Output: [0]

Constraints: 1 ≤ length ≤ 100 | 0 ≤ Node.val ≤ 9 | No leading zeros (except "0" itself)

Addition Visualization

l1 (digits in reverse — least significant first)

l2 (digits in reverse — least significant first)

Result (built left to right)

d1 — + d2 — + carry —

sum — = digit — (sum%10)

new carry — (sum/10)

CARRY

propagates to next column

l1 = —

l2 = —

sum = —

State Variables

d1 (l1.val)

—

d2 (l2.val)

—

carry (in)

sum

—

new digit

—

Step Logic

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Ready

0 / 0

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Algorithm

Init dummy node, curr = dummy, carry = 0

Loop while l1 != null || l2 != null || carry != 0

Read d1 = l1.val (or 0), d2 = l2.val (or 0)

Compute sum = d1 + d2 + carry; new carry = sum / 10

Append node sum % 10 to result; advance pointers

Return dummy.next

Time

O(max(m,n))

Space

O(max(m,n))

Key Insight — Why Reverse Order Helps

Storing digits in reverse order means the least-significant digit comes first. This lets us add digit-by-digit from left to right — exactly how long addition works — without needing to reverse anything. A carry from one column flows naturally to the next node. If the final carry is 1 (e.g. 999 + 1), we append one more node without any special handling.