/*
    Given 2 linked lists, digits stored in reverse order, add them
    Ex. l1 = [2,4,3] l2 = [5,6,4] -> [7,0,8] (342 + 465 = 807)

    Sum digit-by-digit + carry, handle if one list becomes null

    Time: O(max(m, n))
    Space: O(max(m, n))
*/

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode();
        
        ListNode* curr = dummy;
        int carry = 0;
        
        while (l1 != NULL || l2 != NULL) {
            int val1 = (l1 != NULL) ? l1->val : 0;
            int val2 = (l2 != NULL) ? l2->val : 0;
            
            int sum = val1 + val2 + carry;
            carry = sum / 10;
            
            curr->next = new ListNode(sum % 10);
            curr = curr->next;
            
            if (l1 != NULL) {
                l1 = l1->next;
            }
            if (l2 != NULL) {
                l2 = l2->next;
            }
        }
        
        if (carry == 1) {
            curr->next = new ListNode(1);
        }
        
        return dummy->next;
    }
};