题解 | #牛牛的单向链表#

#include <stdio.h>
#include <stdlib.h>
struct List {
    int data;
    struct List* node;
};
int main() {
    int a, n;
    scanf("%d", &n);
    int arr[n];
    for (a = 0; a < n; a++) {
        scanf("%d", &arr[a]);
    }
    struct List* p = NULL;
    struct List* head = NULL;
    head = malloc(sizeof(struct List));
    struct List* end = head;
    for (a= 0; a < n; a++) {
        p = malloc(sizeof(struct List));
        p->data = arr[a];
        end->node = p;
        end = p;
    }
    end->node = NULL;

    struct List* x = head->node;
    for (a = 0; a < n; a++) {
        printf("%d ", x->data);
        x = x->node;
    }

    return 0;
}