题解 | 牛牛的双链表求和

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
struct c
{
    int data;
    struct c *next;
};
int main() 
{
    int n;
    scanf("%d",&n);
    int a[100];
    int b[400];
    int i;
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0;i<n;i++)
    {
        scanf("%d",&b[i]);
    }
    struct c *head1=(struct c *)malloc(sizeof(struct c));
    if(head1==NULL)
    exit(1);
    struct c *head2=(struct c *)malloc(sizeof(struct c));
    if(head2==NULL)
    exit(1);
    head1->next=head2->next=NULL;
    struct c *t=head1,*tt=head2;
    for(i=0;i<n;i++)
    {
        struct c *p=(struct c *)malloc(sizeof(struct c));
        if(p==NULL)
        exit(1);
        struct c *pp=(struct c *)malloc(sizeof(struct c));
        if(pp==NULL)
        exit(1);
        p->data=a[i];pp->data=b[i]+a[i];
        p->next=pp->next=NULL;
        t->next=p;tt->next=pp;
        t=t->next;tt=tt->next;
    }
    struct c *ppp=head2->next;
    while(ppp!=NULL)
    {
        printf("%d ",ppp->data);
        ppp=ppp->next;
    }
    

    return 0;
}