# [LC] 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:

```Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
```

Example 2:

```Input: l1 = , l2 = 
Output: 
```

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

Approach is simple. Just looking all of the list pair of list1 and list2, and calculate sum = (list1.val + list2.val + prev_carry)%10 and carry = (list1.val + list2.val + prev_carry) // 10. Loop both list first, and list1, list2 consequentially. If carry is 1 at the last, we should add it to the last element.

``````# 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: ListNode, l2: ListNode) -> ListNode:
root = ListNode(0)
s,c = 0,0
while l1 and l2:
s,c = (l1.val+l2.val+c)%10, (l1.val+l2.val+c)//10
root.next = ListNode(s)
l1,l2,root = l1.next, l2.next, root.next

while l1:
s,c = (l1.val+c)%10, (l1.val+c)//10
root.next = ListNode(s)
l1, root = l1.next, root.next

while l2:
s,c = (l2.val+c)%10, (l2.val+c)//10
root.next = ListNode(s)
l2, root = l2.next, root.next

if c != 0:
root.next = ListNode(c)