题解 | #牛牛的双链表求和#

#include <stdio.h>

typedef struct Node {
    int data;
    struct node* next;
} node;
node* cai(int v) {
    node* kk = (node*)malloc(sizeof(node));
    kk->data = v;
    kk->next = NULL;
    return kk;
}
int main() {
    int n;
    scanf("%d", &n);
    node* head1 = NULL;
    node* tail1 = NULL;
    node* head2 = NULL;
    node* tail2 = NULL;
    for (int i = 0; i < n; i++) {
        int o;
        scanf("%d", &o);
        node* kk = cai(o);
        if (head1 == NULL) {
            head1 = kk;
            tail1 = kk;
        } else {
            tail1->next = kk;
            tail1 = kk;
        }
    }
    for (int i = 0; i < n; i++) {
        int o;
        scanf("%d", &o);
        node* kk = cai(o);
        if (head2 == NULL) {
            head2 = kk;
            tail2 = kk;
        } else {
            tail2->next = kk;
            tail2 = kk;
        }
    }
    node* p=head1;
    node* p1=head2;
   
        while(p1!=NULL){
            p1->data=p1->data+p->data;
            p=p->next;
            p1=p1->next;
        }

    p1=head2;
    while(p1!=NULL){
        printf("%d ",p1->data);
        p1=p1->next;
    }


}