题解 | #牛牛的单向链表#

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

typedef struct link { int data; struct link *next; }Link;

int main() { Link p= (Link)malloc(sizeof(Link)); Link temp=p;// = (Link)malloc(sizeof(Link)); int len;

scanf("%d",&len);

for (int i = 0; i < len; i++)
{
	Link *a = (Link*)malloc(sizeof(Link));
	scanf("%d", &(a->data));
	//a->data = arr[i];
	a->next = NULL;
	temp->next = a;
	temp = temp->next;
}
temp = p;
while (temp->next)
{
	temp = temp->next;
	printf("%d ", temp->data);
	
}

return 0;

}