# [LC] 2. Add Two Numbers

TechYou 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 = [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

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)
head = root
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)
return head.next
```

Time/Space Complexity: O(N)**Submission**

Runtime: 88 ms, faster than 10.65% of Python3 online submissions for Add Two Numbers.

Memory Usage: 14.4 MB, less than 12.54% of Python3 online submissions for Add Two Numbers.