[LeetCode January Challange] Day 12 - 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 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.

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Solution

Time complexity : O(max(l1+l2))
Space complexity : O(max(l1+l2))

/**
 * 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* cur = dummy;
        short carry = 0;
        while (l1 || l2 || carry) {
            short a = l1 ? l1->val : 0;
            short b = l2 ? l2->val : 0;
            short val = a + b + carry;
            carry = val / 10;
            cur->next = new ListNode(val%10);
            cur = cur->next;
            l1 = l1 ? l1->next : nullptr;
            l2 = l2 ? l2->next : nullptr;
        }
        
        return dummy->next;
    }
};