链表逆向加和

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) 
    {
        ListNode *res = new ListNode(-1);   //添加虚拟头结点，简化边界情况的判断
        ListNode *cur = res;
        int carry = 0;  //表示进位
        while (l1 || l2) 
        {
            int n1 = l1 ? l1->val : 0;
            int n2 = l2 ? l2->val : 0;
            int sum = n1 + n2 + 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); //如果最高位有进位，则需在最前面补1.
        return res->next;   //返回真正的头结点
    }
};