线性表的有序合并之【直接在head1的后面接上head2】（与我的第一篇文章雷同）

#include <stdio.h>
#include <stdlib.h>
struct linklist
{
    int data;
    struct linklist *next;
};
struct linklist* create_head(int len)//创建链表
{
    struct linklist *head;
    struct linklist *p, *q;
    p = (struct linklist*)malloc(sizeof(struct linklist));
    int i;
    head = p;
    for(i = 0; i < len; i++)
    {
        q = (struct linklist*)malloc(sizeof(struct linklist));
        p->next = q;
        scanf("%d", &q->data);//还是选择保留头节点为空白，防止对head操作不当
        p = q;
    }
    p->next = NULL;//即最后的q = NULL
    return head;
}
void inte(struct linklist *head1, struct linklist *head2)
{
    struct linklist *p;
    p = head1;
    while(p->next)
    {
        p = p->next;
    }
    p->next = head2->next;//链表的值存储是从head->next开始的
}
void sort_linklist(struct linklist *head)
{
    struct linklist *p, *q;
    p = head;
    int temp;
    for(p = head->next; p != NULL; p = p->next)
    {
        for(q = p; q != NULL; q = q->next)
        {
            if(p->data > q->data)//升序排列
            {
                temp = p->data;
                p->data = q->data;
                q->data = temp;
            }
        }
    }
}
void print_linklist(struct linklist *head)
{
    struct linklist *p;
    p = head;
    while(p)
    {
        printf("%d ", p->data);
        p = p->next;
    }
}
int main()
{
    int N, M;
    while(~scanf("%d", &N))//易错点1：要记得是多次输入哦，不然你就会答案错误呜呜呜呜呜
    {
         //输入两个链表的长度
        struct linklist  *head1, *head2;
        head1 = create_head(N);
        scanf("%d", &M);
        head2 = create_head(M); //分别定义两个链表
        inte(head1, head2); //合并两个链表,直接在head1的后面加上head2
        sort_linklist(head1);//排序
        print_linklist(head1->next);//打印排序好的链表
        printf("\n");//易错点2：每次输出后要换行，不然你就会格式错误呜呜呜
    }
    return 0;
}