你删除链表最好用双链表，单链表在删除时还麻烦，要用两个指针一前一后的，否则在删除时无法对要删除的前一个节点进行操纵，本人也是在学数据结构，经验不多，你的那个p指针赋值为NULL了，后来的那个p->next=p1->next又是怎么回事呢？

#include <stdio.h>
#include <string.h>
#define LA 4
#define LB 5
#define NULL 0
struct student
{
    int num;
    char name[8];
    struct student *next;
}a[LA],b[LB];


int main(void)
{
    struct student a[LA]={{101,"wang"},{102,"li"},{105,"zhang"},{106,"wei"}};
    struct student b[LB]={{103,"zhang"},{104,"ma"},{105,"chen"},{107,"guo"},{108,"lui"}};
    int i;
    struct student *p,*p1,*p2,*head1,*head2;
    /*初始化*/
    head1=a;
    head2=b;
    printf("\nlist  a:\n");
    for(p1=head1,i=1;p1<a+LA;i++)
    {
        p=p1;
        p1->next=a+i;
        printf("%8d%8s\n",p1->num,p1->name);
        p1=p1->next;
    }
    p->next=NULL;
    printf("\nlist  b:\n");
    for(p2=head2,i=1;p2<b+LB;i++)
    {
        p=p2;
        p2->next=b+i;
        printf("%8d%8s\n",p2->num,p2->name);
        p2=p2->next;
    }
    p->next=NULL;
    printf("\n");
    
    /*删除*/
    p1=head1;
    while(p1 != NULL)
    {
        
        p2=head2;
        while(p2 != NULL && p1->num != p2->num)
            p2=p2->next;
        if(p2!=NULL && p1->num == p2->num) // 如果p2指向NULL，会访问出错，添加条件p2!=NULL
        {
            if(p1==head1)
                head1=p1->next;
            else
                p->next=p1->next;
        }
        p=p1;
        p1=p1->next;
    }
    
    
    p1=head1;
    printf("\nresult:\n");
    while(p1!=NULL)
    {
        printf("%7d %7s\n",p1->num,p1->name);
        p1=p1->next;
    }
    
    return 0;
}

[ 本帖最后由 xxwpk007 于 2010-8-8 23:34 编辑 ]

[ 本帖最后由 xxwpk007 于 2010-8-9 00:18 编辑 ]