题目描述：

给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。

如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外，这两个数都不会以 0 开头。

示例1：

输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[7,0,8]
解释：342 + 465 = 807

示例2：

输入：l1 = [0], l2 = [0]
输出：[0]

示例3：

输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出：[8,9,9,9,0,0,0,1]

提示

• 每个链表中的节点数在范围[1, 100] 内
• 0 <= Node.val <= 9
• 题目数据保证列表表示的数字不含前导零

代码示例：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
    int carry = 0;
    struct ListNode* head = NULL, *tail = NULL;
    while(l1 && l2) {
        if(tail == NULL) {
            tail = (struct ListNode*)malloc(sizeof(struct ListNode));
            head = tail;
        } else {
            tail -> next = (struct ListNode*)malloc(sizeof(struct ListNode));
            tail = tail -> next;
        }
        tail -> val = (l1 -> val + l2 -> val + carry) % 10;
        carry = (l1 -> val + l2 -> val + carry) / 10;
        tail -> next = NULL;
        l1 = l1 -> next;
        l2 = l2 -> next;
    }
    while(l1) {
        if(tail == NULL) {
            tail = (struct ListNode*)malloc(sizeof(struct ListNode));
            head = tail;
        } else {
            tail -> next = (struct ListNode*)malloc(sizeof(struct ListNode));
            tail = tail -> next;
        }
        tail -> val = (l1 -> val + carry) % 10;
        carry = (l1 -> val + carry) / 10;
        tail -> next = NULL;
        l1 = l1 -> next;
    }
    while(l2) {
        if(tail == NULL) {
            tail = (struct ListNode*)malloc(sizeof(struct ListNode));
            head = tail;
        } else {
            tail -> next = (struct ListNode*)malloc(sizeof(struct ListNode));
            tail = tail -> next;
        }
        tail -> val = (l2 -> val + carry) % 10;
        carry = (l2 -> val + carry) / 10;
        tail -> next = NULL;
        l2 = l2 -> next;
    }
    if(carry) {
        tail -> next = (struct ListNode*)malloc(sizeof(struct ListNode));
        tail = tail -> next;
        tail -> val = carry;
        tail -> next = NULL;
    }
    return head;
}