leetcode/apps/docs/content/(linked-list)/2-add-two-numbers.mdx

---
title: '2. Add Two Numbers'
description: 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
sidebar:
  label: 'Add Two Numbers'
  badge: 'Medium'
---

<Badge variant="accent">Hybrid Structures</Badge>

### 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]`

### 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.

## Approach

```mermaid
flowchart TD
  S(["addTwoNumbers(l1, l2)"]) --> I["dummy = ListNode(), carry = 0, curr = dummy — dummy.next will be the head"]
  I --> W{"l1 or l2 or carry?"}
  W -- no --> E(["return dummy.next"])
  W -- yes --> A["s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry — a missing digit counts as 0"]
  A --> D["carry, val = divmod(s, 10)"]
  D --> N["curr.next = ListNode(val), curr = curr.next"]
  N --> V["l1 = l1.next if l1 else None, l2 = l2.next if l2 else None"]
  V --> W
```

## Solution

```py
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(
        self, l1: Optional[ListNode], l2: Optional[ListNode]
    ) -> Optional[ListNode]:

        dummy = ListNode()
        carry = 0
        curr = dummy
        while l1 or l2 or carry:
            s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry
            carry, val = divmod(s, 10)

            curr.next = ListNode(val)
            curr = curr.next

            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None

        return dummy.next
```

## Explanation

<YouTube id="wgFPrzTjm7s" title="Add Two Numbers - Leetcode 2 - Python" />