题解 | #牛牛的单向链表#

#include<stdio.h>
#include<stdlib.h>
typedef struct node
{
	int date;
	struct node* next;
}node;
node* creat(int n)
{
	node* head, * p;
	head = (node*)malloc(sizeof(node));
	head->date = 0;
	head->next = NULL;
	p = head;
	for (int i =0; i <n; i++)
	{
		node* newnode = (node*)malloc(sizeof(node));
		scanf("%d", &newnode->date);
		newnode->next = NULL;
		p->next = newnode;
		p = p->next;
	}
	return head;
}
void printnode(node* head)
{
	node* p;
	p = head->next;
	while (p != NULL)
	{
		printf("%d ", p->date);
		p = p->next;
	}
}
int main()
{
	int n=1, i;
	node* head;
	scanf("%d", &n);
	head = creat(n);
	printnode(head);
	return 0;
}