1.力扣C源码

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


struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
//新链表的头节点，无数据
    struct ListNode* p = (struct ListNode*)malloc(sizeof(struct ListNode));
    struct ListNode* q = p;
    int carry = 0;
    while (l1 || l2) {
        struct ListNode* r = (struct ListNode*)malloc(sizeof(struct ListNode));
        int val1 = l1 ? l1->val : 0;
        int val2 = l2 ? l2->val : 0;
        int val = val1 + val2 + carry;
        if (val >= 10) {
            val -= 10;
            carry = 1;
        }
        else {
            carry = 0;
        }
        r->val = val;
        q->next = r;
        q = q->next;//赋值后，q指向当前有数据节点
        l1 = l1 ? l1->next : l1;
        l2 = l2 ? l2->next : l2;
    }
    if (carry == 1) {
        struct ListNode* r = (struct ListNode*)malloc(sizeof(struct ListNode));
        r->val = 1;
        q->next = r;
        q = q->next;
    }
    q->next = NULL;
    return p->next;
}

2.题解

模拟两数相加的过程。设置一个辅助位carry记录进位（0 or 1）。加法的过程示意图如下：
在这里插入图片描述

1.考虑数字相加的进位问题。用carry来记录（取值为0或者1）
2.最高位进位问题，即在当前链表后再加一个节点，值为1
3.如何处理链表的头节点问题？设置两个节点p，q，p作为头节点（没有实际意义），q往后递增，作为指向每次添加到链表末尾的节点的指针，结束后返回p->next即可

3.VS可运行源码

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<vector>
#include<limits>
#include<algorithm>
#pragma warning(disable:4996)
using namespace std;

//特别注意：
//这行代码不要忘，一定要加上
//q->next = NULL;
struct ListNode {
    int val;
    struct ListNode *next;
};

int n3;
ListNode* List_TailInsert3()//尾插法建立带有头节点的链表
{
    ListNode* head = (ListNode*)malloc(sizeof(ListNode));
    ListNode* rear = head;
    printf("输入链表数据：");
    int num;
    for (int i = 0; i < n3; i++) {
        scanf("%d", &num);
        ListNode* p = (ListNode*)malloc(sizeof(ListNode));
        p->val = num;
        rear->next = p;
        rear = p;
    }
    rear->next = NULL;
    return head;
}
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    //本题的头节点指的是第一个节点(有数据)，和我定义的头节点不一样
    //所以把我所定义的头节点的下一个节点作为本题的头节点
    //复制到力扣上时把以下这两行代码删掉即可，并把下面的head1改为l1,head2改为l2
    ListNode* head1 = l1->next;
    ListNode* head2 = l2->next;

    //新链表的头节点，无数据
    ListNode* p = (ListNode*)malloc(sizeof(ListNode));
    ListNode* q = p;
    int carry = 0;
    while (head1 || head2) {
        ListNode* r = (ListNode*)malloc(sizeof(ListNode));
        int val1 = head1 ? head1->val : 0;
        int val2 = head2 ? head2->val : 0;
        int val3 = val1 + val2 + carry;
        if (val3 >= 10) {
            val3 -= 10;
            carry = 1;
        }
        else {
            carry = 0;
        }
        r->val = val3;
        q->next = r;
        q = q->next;//赋值后，q指向当前有数据节点
        head1 = head1 ? head1->next : head1;
        head2 = head2 ? head2->next : head2;
    }
    if (carry == 1) {
        ListNode* r = (ListNode*)malloc(sizeof(ListNode));
        r->val = 1;
        q->next = r;
        q = q->next;
    }
    //这行代码不要忘，一定要加上
    q->next = NULL;
    return p->next;
}
int main()
{
    printf("输入第一个链表的长度：");
    scanf("%d", &n3);
    ListNode* head1 = List_TailInsert3();
    printf("输入第二个链表的长度：");
    scanf("%d", &n3);
    ListNode* head2 = List_TailInsert3();
    ListNode* head = addTwoNumbers(head1, head2);
    ListNode* p = head;
    /*if (p->next->next->next)printf("1");*/
    while (p) {
        printf("%d", p->val);
        p = p->next;
    }

	system("pause");
	return 0;
}