题目

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

坑：太久没写链表了，竟然犯了在函数内部定义节点的错误，ListNode head = ListNode(0); ×
应该使用new！！！ 临时变量无法取址，报错 ‘’error: taking address of temporary[-fpermissive]‘’

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* head = new ListNode(0);
        ListNode* current = head;
        int carry = 0;
        int sum = 0;
        while ((l1 != NULL) || (l2 != NULL) || carry)
        {
            sum = carry;
            carry = 0;
            if (l1 != NULL) {
                sum += l1->val;
                l1 = l1->next;
            }
            if (l2 != NULL)
            {
                sum += l2->val;
                l2 = l2->next;
            }
            if (sum > 9)
            {
                carry = 1;
                sum -= 10;
            }
            current->next = new ListNode(sum);
            current = current->next;
        }
        return head->next;
    }
};