题解 | #牛牛的单向链表#

#include<stdio.h>
int main()
{
    int n;
    scanf("%d", &n);
    int arr[1000] = { 0 };
    int i;
    for (i = 0; i < n; i++)
    {
        scanf("%d ", &arr[i]);
    }
    int* p = arr;
    for (i = 0; i < n; i++)
    {
        printf("%d ", *(p + i));
    }
}