2. Add Two Numbers

1. Description

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.

2. Example

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]

3. 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.

4. Solutions

Linked List

n = max(len(l1), len(l2))
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode dummy = ListNode();
        ListNode *tail = &dummy;
        int carry = 0;

        while (l1 != nullptr || l2 != nullptr || carry != 0) {
            int sum = carry;
            if (l1 != nullptr) {
                sum += l1->val;
                l1 = l1->next;
            }
            if (l2 != nullptr) {
                sum += l2->val;
                l2 = l2->next;
            }

            carry = sum / 10;
            tail->next = new ListNode(sum % 10);
            tail = tail->next;
        }

        return dummy.next;
    }
};