[LeetCode] 2. Add Two Numbers (C++)

タイトルの説明:
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 contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

解題の構想:チェーンテーブルを遍歴して、簡単な数学、1つの数で得た値の各位を一時的に保存して、1つは進位を一時的に保存します

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        if (!l1 || !l2)
            return nullptr;
        ListNode* resList = new ListNode(0);
        ListNode* Now = resList;
        int carry = 0;
        int sum = 0;
        
        ListNode* l = l1;
        ListNode* r = l2;
        
        while (l != nullptr && r != nullptr) {
            sum = (l->val + r->val + carry) % 10;
            carry = (l->val + r->val + carry) / 10;
            ListNode* temp = new ListNode(sum);
            Now->next = temp;
            Now = temp;
            l = l->next;
            r = r->next;
        }
        
        while (l != nullptr) {
            sum = (l->val + carry) % 10;
            carry = (l->val + carry) / 10;
            ListNode* temp = new ListNode(sum);
            Now->next = temp;
            Now = temp;
            l = l->next;
        }
        
        while (r != nullptr) { 
            sum = (r->val + carry) % 10;
            carry = (r->val + carry) / 10;
            ListNode* temp = new ListNode(sum);
            Now->next = temp;
            Now = temp;
            r = r->next;
        }
        
        if (carry == 0) {
            Now -> next = nullptr; 
        } else {
            Now->next = new ListNode(carry);
        }
            
        return resList->next;
        
    }
};