---
# System prepended metadata

title: '[2. Add Two Numbers](https://leetcode.com/problems/add-two-numbers/description/)'
tags: [Recursion, leetcode, Math, Link List, Medium]

---

###### tags: `leetcode` `Medium` `Link List` `Math` `Recursion`
# [2. Add Two Numbers](https://leetcode.com/problems/add-two-numbers/description/)

## Desciption

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
### Example 1:

![Example1](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.

## Code

```c=
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    if (!l1)
        return l2;
    if (!l2)
        return l1;
    int carry = 0;
    struct ListNode *tmp, *ans, *last;
    ans = (struct ListNode*)malloc(sizeof(struct ListNode));
    tmp = ans;
    while (l1 != NULL || l2 != NULL || carry != 0) {
        tmp->val =
            ((l1 != NULL ? l1->val : 0) + (l2 != NULL ? l2->val : 0) + carry) % 10;
        carry =
            ((l1 != NULL ? l1->val : 0) + (l2 != NULL ? l2->val : 0) + carry) / 10;
        if (l1 != NULL)
            l1 = l1->next;
        if (l2 != NULL)
            l2 = l2->next;
        last = tmp;
        tmp->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        tmp = tmp->next;
    }
    last->next = NULL;
    return ans;
}
```

## Complexity

|Space            |Time            |
|-                |-               |
|$O(Max(l1, l2))$ |$O(Max(l1, l2))$|

## Result
- Runtime: 20 ms, 61.42 %
- Memory usage: 7.8 MB, 29.34 %