Add Two Numbers.cpp

/* 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 = [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.
*/

/*
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

//Time complexity: O(max(M,N)). Space complexity: O(max(M,N)).
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode *res = new ListNode(-1);
        ListNode *cur = res;
        int carry = 0;
        
        while(l1 || l2){
            int n1 = l1 ? l1->val : 0;
            int n2 = l2 ? l2->val : 0;               
            int sum = n1 + n2 + carry;            
            carry = sum / 10;        
            cur->next = new ListNode(sum % 10);
            cur = cur->next;
                
            if(l1) 
                l1 = l1->next;            
            if(l2)
                l2 = l2->next;
            
        }
        if(carry)
            cur->next = new ListNode(1);        
        
        return res->next;
    }
};