有序列表的归并

/*
发现写长代码的时候写分成小部分去实现很有必要，容易差错
*/
#include<stdio.h>  
#include<stdlib.h>  
struct node  
{  
    int num;  
    struct node* next;  
 
};  
struct node * input(int n)  
{  
    int i;  
    struct node *head,*tail,*p;  
    head = tail = (struct node *)malloc(sizeof(struct node));  
    head->next = NULL;  
 
    for(i = 0; i < n; i++)  
    {  
        p = (struct node *)malloc(sizeof(struct node));  
        p->next = NULL;  
        scanf("%d", &p->num);  
        tail->next = p;  
        tail = tail->next;  
    }  
 
    return head;  
}  
 
int mix(struct node *head1,struct node *head2)  
{  
    struct node *tail,*p1,*p2;  
    p1 = head1->next;  
    p2 = head2->next;  
    free(head2);  
    tail = head1;  
    while(p1&&p2)  
    {  
        if(p1->num < p2->num)  
        {  
            tail->next = p1;  //tail是游动指针
            tail = p1;  //tail徐游动，就像建立新的链表一样
            p1 = p1->next;  
        }  
        else 
        {  
            tail->next = p2;  
            tail = p2;  
            p2 = p2->next;  
        }  
    }  
    if(p1)  
    {  
        tail->next = p1;  
    }  
    else 
    {  
        tail->next = p2;  
    }  
    p1 = head1->next;  
    while(p1->next != NULL)  
    {  
        printf("%d ",p1->num);  
        p1 = p1->next;  
    }  
    printf("%d",p1->num);  
    return 0;  
}  
 
int main()  
{  
    struct node *head1,*tail,*head2,*p;  
    int n1,n2;  
 
    scanf("%d%d",&n1,&n2);  
 
    head1 = input(n1);  
    head2 = input(n2);  
    p = head1->next;  
 
    mix(head1,head2);  
}