题解 | #牛牛的单向链表#

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

    typedef struct node{
        int data;
        struct node *next;
    }node;

    node *createlist()//创建一个头节点，头节点的指针域置空
    {
        node *head = malloc(sizeof(node));
        if(head == NULL)
        {
            printf("内存分配失败\n");
            exit(1);
        }
        head->next = head;
        return head;
    }
    void insertlist(node *head,int data)
    {
        node *new = malloc(sizeof(node));
        if(new == NULL)
        {
            printf("out of memory");
            exit(1);
        }
        new->data = data;
        new->next = NULL;

        node *p = NULL;
        for(p = head; p->next != head; p = p->next);
        new->next = p->next;
        p->next = new;
    }

    void output(node *head)
    {
        if (head->next == head) {
            return;
        }
        node *p = head->next;
        while (p != head) {
            printf("%d ", p->data);
            p = p->next;
        }
    }

    int main(void)
    {
        node *head = createlist();//创建一个头节点
        int n,i,j;
        scanf("%d", &n);
        int arr[n];

        for (i = 0; i < n; i++) {
            scanf("%d", &arr[i]);
        }

        //将数据插入链表
        for(j = 0; j < n; j++)
        
            insertlist(head,arr[j]);
        output(head);
        
        return 0;
    }