2. add two numbers

description:

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.

my answer:

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* dummy = new ListNode(-1), *cur = dummy;//注意这里是*cur
        int carry = 0;
        while (l1 ||l2){
            int val1 = l1 ? l1 -> val : 0;
            int val2 = l2 ? l2 -> val : 0;
            int sum = val1 + val2 + carry; 
            carry = sum / 10;
            cur -> next = new ListNode(sum % 10);
            cur = cur -> next;
            if (l1) l1 = l1 -> next;
            if (l2) l2 = l2 -> next;
        }
        if (carry){
            cur -> next = new ListNode(1);
        }
        return dummy -> next;
    }
};

relative point get√：

• list->next, list->val
• ListNode *dummy = new ListNode(-1), listnode 定义的是一个指针

hint :

nope