Leetcode #2

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;    }};