题解 | #牛牛的双链表求和#

#include <stdio.h>
#include <stdlib.h>
//结构体
typedef struct Node
{
    int num;
    struct Node *next;    
}LinkNode;
//创建链表
LinkNode *creat_link(int n){
    int i;
    LinkNode *head = (LinkNode *)malloc(sizeof(LinkNode));
    LinkNode *tail;
    head->next = NULL;
    scanf("%d", &head->num);
    tail = head;
    for (i = 0; i < n-1; ++i)
    {
        LinkNode *p = (LinkNode *)malloc(sizeof(LinkNode));
        scanf("%d", &p->num);
        p->next = NULL;
        tail->next = p;
        tail = tail->next;
    }
    return head;
}
//链表对应数据相加操作
void add_node(int n, const LinkNode *head_1, LinkNode *head_2){
    int i;
    LinkNode *tail_1 = head_1;
    LinkNode *tail_2 = head_2;
    for(i = 0; i < n; ++i){
        tail_2->num = tail_2->num + tail_1->num;
        tail_1 = tail_1->next;
        tail_2 = tail_2->next;
    }
}

int main(int argc, char const *argv[])
{
    int n,i;
    scanf("%d", &n);
    LinkNode *head_1 = creat_link(n);//创建链表a
    LinkNode *tail;
    LinkNode *head_2 = creat_link(n);//创建链表b
    add_node(n, head_1, head_2);
    while(tail != NULL){
        printf("%d ", tail->num);
        tail = tail->next;
    }
    printf("\n");
    return 0;
}