## 20 Dec 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.

• we need to deal with one situation here:
``````l1     1     3                              1
+      +     +                              +
l2     1     9                              1
-----------------------------------------------------------------------------------
cur    2    12(we need to put 2 here)        2+carry(carry from the previous 12)
12 %10 = 2
carry = 12//10 =1	   ``````

By adding a variable called carry can deal with this situation: align the addends vertically and add the columns, starting from the left-most column. If a column’s sum exceeds nine, the extra digit “carried” add into the next column.

``````l1          1                  3                       1
+           +                  +                       +
l2          1                  9                       1
-----------------------------------------------------------------------------------
carry  0+1+1=3             0+3+9=12              1+1+1 = 3
cur     3%10=3             12%10=2               3%10 = 3
carry  3//10 =0             12//10=1             3//10 = 0 ``````
# 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
result = current = ListNode(0)

while l1 or l2 or carry:
if l1:
carry = carry + l1.val
l1 = l1.next
if l2:
carry = carry +l2.val
l2 = l2.next
current.next = ListNode(carry%10)
current = current.next
carry = carry // 10
return result.next
` `