题解 | #牛牛的单向链表#
牛牛的单向链表
http://www.nowcoder.com/practice/95559da7e19c4241b6fa52d997a008c4
#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));
}
}