有两个链表a和b,设结点中包含学号、姓名。从a链表中删去与b链表中有相同 学号的那些结点。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef struct stu{
    char num[6];char name[10];
    struct stu *next;
}List;
int main(int argc, char *argv[]) {
    List a[6]={{"101","wang"},{"103","li"},{"105","chang"},{"106","wei"},{"101","zhang"},{"107","zhao"}};
    List b[3]={{"101","chang"},{"103","mao"},{"105","zhang"}};
    int i,j;
    List *p,*p1,*p2,*pt,*head1,*head2;
    head1=a;head2=b;
    for(p1=head1,i=1;p1<a+6;i++)
    {
        p=p1;
        p1->next=a+i;
        printf("%8s%8s\n",p->num,p->name);
        p1=p1->next;
    }
    p->next=NULL;printf("\n");
    for(p2=head2,i=1;p2<b+3;i++)
    {
        p=p2;
        p2->next=b+i;
        printf("%8s%8s\n",p->num,p->name);
        p2=p2->next;
    }
    p->next=NULL;printf("\n");
    p1=pt=head1;
    while(p1!=NULL)
    {
        p2=head2;
        while(p2!=NULL&&strcmp(p1->num,p2->num)!=0)
            p2=p2->next;
        if(p2!=NULL&& strcmp(p1->num,p2->num)==0)
            if(p1==head1)head1=p1->next;
            else
                pt->next=p1->next;
        else    
            pt=p1;        
        p1=p1->next;
    }
    p1=head1;
    printf("\n\n");
    while(p1!=NULL)
    {
        printf("%8s%8s\n",p1->num,p1->name);
        p1=p1->next;
    }
    return 0;
}

/*
//方法二
    p2=head2;
    while(p2!=NULL) 
    {
        p1=pt=head1;
        while(p1!=NULL)
        {
            if(strcmp(p1->num,p2->num)==0)//如果学号相等，则删除该p1节点 ,p1往后移
            {
                if(p1==head1)
                {
                    head1=p1->next;pt=p1->next;
                }
                else
                    pt->next=p1->next;
            }
            else//否则pt记录其前置节点 ，p1往后移
                pt=p1;
            p1=p1->next;
                        
        }
        p2=p2->next;
    }
*/