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MEDIUM

Add Two Numbers

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

Input:

l1 = [2,4,3]
l2 = [5,6,4]

Output:

[7,0,8]

Explanation:
342 + 465 = 807.

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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.

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Approach: Iterative with Carry

• Time Complexity: O(max(m, n)), where m and n are the lengths of the two linked lists. Each node is processed once.
• Space Complexity: O(max(m, n)), for the output linked list (if counting the input and output as extra space).

C++

/**
 * 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(0);
        ListNode* cur = dummy;
        int carry = 0;
        while(l1 || l2 || carry) {
            int val = carry;
            if (l1) {
                val += l1->val;
                l1 = l1->next;
            }
            if (l2) {
                val += l2->val;
                l2 = l2->next;
            }
            carry = val/10;
            cur->next = new ListNode(val%10);
            cur = cur->next;
        }
        return dummy->next;
    }
};