题解 | #牛牛的双链表求和#

#include <stdio.h>
#include <stdlib.h>

//链表结点的定义
struct Node {
    int num;
    struct Node* Next;
};

//创建头结点
struct Node* creatList() {
    struct Node* headnode;
    headnode = (struct Node*)malloc(sizeof(struct Node));
    headnode->num = 0;
    headnode->Next = NULL;
    return headnode;
}

//创建结点
struct Node* creatnode(int num) {
    struct Node* newnode;
    newnode = (struct Node*)malloc(sizeof(struct Node));
    newnode->num = num;
    newnode->Next = NULL;
    return newnode;
}

int main() {
    int n, num;
    scanf("%d\n", &n);

    struct Node* a = creatList();
    struct Node* q = a;
    struct Node* p;

    for (int i = 0; i < n; i++) {
        scanf("%d ", &num);
        p = creatnode(num);

        //尾插法
        q->Next = p;
        q = p;
    }

    struct Node* b = creatList();
    q = b;
    for (int i = 0; i < n; i++) {
        scanf("%d ", &num);
        p = creatnode(num);

        //尾插法
        q->Next = p;
        q = p;
    }

    for ( p = a->Next, q = b->Next; p != NULL ; p = p->Next, q = q->Next) {
        printf("%d ", p->num + q->num);
    }
    return 0;
}