---
description: Linked List
---

# LeetCode 2. Add Two Numbers

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.

**Example 1:**![](https://assets.leetcode.com/uploads/2020/10/02/addtwonumber1.jpg)

```
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.

## Solution:

[English Version in Youtube](https://youtu.be/nrT4oGpOLfk)

[中文版解答Youtube Link](https://youtu.be/LVhNH1BdGGM)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1h54y1W7iR/)

```
class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode *fake_head = new ListNode(0);
        ListNode *head = fake_head;
        int carry = 0;
        
        while(l1 || l2 || carry) {
            int a = (l1 == nullptr? 0 : l1->val);
            int b = (l2 == nullptr? 0 : l2->val);
            int c = a + b + carry;
            carry = c / 10;
            int v = c % 10;
            ListNode *cur = new ListNode(v);
            head->next = cur;
            head = cur;
            if(l1) l1 = l1->next;
            if(l2) l2 = l2->next;
        }

        return fake_head->next;
    }
};
```
