C实现 LeetCode->Add Two Numbers

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

     给定两个链表表示两个非负数。数字逆序存储，每个节点包含一个单一的数字。计算两个数的和，并以链表的形式返还。

//
//  AddTwoNumbers.c
//  Algorithms
//
//  Created by TTc on 15/6/4.
//  Copyright (c) 2015年 TTc. All rights reserved.
//

#include "AddTwoNumbers.h"
#include <stdlib.h>

struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode *
addTwoNumbers(struct ListNode *l1,struct ListNode *l2){
    if(l1 == NULL && l2 == NULL){
        return NULL;
    }
    if(l1 == NULL){
        return l2;
    }
    if(l2 == NULL){
        return l1;
    }
    struct ListNode *resultNode = (struct ListNode *)malloc(sizeof(*resultNode));
    struct ListNode *p1 = resultNode;
    p1->next = NULL;
    struct ListNode *p2;
    
    int buff = 0; int dev = 0;
    while (l1 != NULL || l2 != NULL) {
        if(l1 != NULL && l2 == NULL){
            buff = l1->val + dev;
            p2 = (struct ListNode *)malloc(sizeof(*p2));
            p2->next = NULL;
            p1->next = p2;
            p2->val = buff % 10;
            l1 = l1->next; //遍历指向下一个元素
            p1 = p1->next; //中间变量也指向下一个元素
            dev = buff/10; //计算 这个层级的节点 相加后 个位 上的数字
        }else if(l1 == NULL && l2 != NULL){
            buff = l2->val + dev;
            p2 = (struct ListNode *)malloc(sizeof(*p2));
            p2->next = NULL;
            p1->next = p2;
            p2->val = buff %10;
            
            l2 = l2->next;
            p1 = p1->next;
            dev = buff / 10;
        }else{
            buff = l1->val + l2->val + dev;
            p2 = (struct ListNode *)malloc(sizeof(*p2));
            p2->next = NULL;
            p1->next = p2;
            p2->val = buff % 10;
            l1 = l1->next;
            l2 = l2->next;
            p1 = p1->next;
            dev = buff / 10;
        }
    }
    if(dev){
        p2 = (struct ListNode *)malloc(sizeof(*p2));
        p2->next = NULL;
        p1->next = p2;
        p2->val = dev;
    }
    resultNode = resultNode->next;
    return resultNode;
}