# 2. Add Two Numbers

#### Medium

***

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.

```python
# 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]:
        carry = 0
        head = new = ListNode()
        while l1 and l2:
            num1 = l1.val
            num2 = l2.val
            sumation = num1 + num2 + carry
            carry = (sumation) // 10
            new_node = ListNode(sumation % 10)
            new.next = new_node
            new = new_node
            l1 = l1.next
            l2 = l2.next
        left_node = l1 or l2
        while left_node:
            num2 = left_node.val
            sumation = num2 + carry
            carry = (sumation) // 10
            new_node = ListNode(sumation % 10)
            new.next = new_node
            new = new_node
            left_node = left_node.next
        if carry:
            new_node = ListNode(carry)
            new.next = new_node
            new = new_node
        return head.next
```