leetcode 5. 两个链表逐个元素相加 Add Two Numbers

7问题：Add Two Numbers 难度-Medium

You are given two linked lists representing two non-negative numbers. 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.

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

解答：
目标：做一个带进位的加法器（243+564 = 708）
本题需要重点考虑的是–两个链表不等长的情况 和 最后的进位

class Solution {
public:

    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        if (l1 == NULL || l2 == NULL)
           return l1 == NULL ? l2 : l1;    
        ListNode dummy(-1); //虚拟节点，指向链表头
        ListNode* l = &dummy;
        int carry = 0;
        int v1, v2, sum;

        while (l1 || l2) {
            if (l1) { 
                v1 = l1->val;
                l1 = l1->next;
            } else {
                v1 = 0;
            }
            if (l2) {
                v2 = l2->val;
                l2 = l2->next;
            } else {
                v2 = 0;
            }

            sum = v1 + v2 + carry;
            l->next = new ListNode(sum % 10);//后插法，如果是C语言，写一个后插函数
            carry = sum / 10;
            l = l->next;
        }
        //l1,l2均达到尾部时，处理最后的进位
        if (carry) {
            l->next = new ListNode(carry);
        }

        return dummy.next;          
    }
};