链表

智商抓鸡,链表看得我纠结了一天,继续note

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只换名不换内容问题

struct student *sort(struct student *head)
{
    struct student *p1,*p2,*temp;
    p1=head;
    p2=head->next;
    printf("before change:p1:%ld,%d;p2:%ld,%d\n",p1->num,p1->score,p2->num,p2->score);
    temp=p2;
    p2=p1;
    p1=temp;
    printf("after change:p1:%ld,%d;p2:%ld,%d\n",p1->num,p1->score,p2->num,p2->score);
    return head;
}

运行结果

before change:p1:1,11;p2:2,22
fter change:p1:2,22;p2:1,11

 

表明可以通过

temp=p2;
p2=p1;
p1=temp;
的方法来交换名称而不交换内容

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交换名称与内容问题

链表头交换

struct student *sort(struct student *head)
{
    struct student *p1,*p2,*p3,*temp,*temp_name;
    p1=head;
    p2=head->next;
    p1->next=p2->next;
    p2->next=p1;
    head=p2;
    return head;
}

运行结果:

换前

1 11
2 22
3 33
4 44

换后
2 22
1 11
3 33
4 44

链表中及链表尾

struct student *sort(struct student *head)
{
    struct student *p1,*p2,*p3,*temp;
    p3=head;
    p1=p3->next;
    p2=p1->next;

    temp=(struct student *)malloc(LEN);
    temp->next=p3->next;
    p3->next=p1->next;
    p1->next=p2->next;
    p2->next=temp->next;
    return head;
}

 运行结果1(链表中)

换前

1 11
2 22
3 33
4 44

换后
1 11
3 33
2 22
4 44

运行结果2(链表尾)

换前

1 11
2 22
3 33
4 44

换后
1 11
3 33
2 22
4 44

此法可行

必须先给temp分配一个空间,否则会造成不可预测情况

此处带出另外一个问题

定义一个自定义struct的指针后,使用前要执行以下其中一步

1.分配空间

2.指向某个已经分配空间的struct

3.指向NULL

做了前两个操作的才能访问或者修改结构内的变量,如p->num等

---

 结合以上两种方法可以做到只换内容不换名字

struct student *sort(struct student *head)
{
    struct student *p1,*p2,*p3,*temp,*temp_name;
    p3=head;
    p1=p3->next;
    p2=p1->next;

    printf("before change num:\np1:%ld,%d\np2:%ld,%d\n",p1->num,p1->score,p2->num,p2->score);
    print(head);
    temp=(struct student *)malloc(LEN);
    temp->next=p3->next;
    p3->next=p1->next;
    p1->next=p2->next;
    p2->next=temp->next;
    printf("after change num:\np1:%ld,%d\np2:%ld,%d\n",p1->num,p1->score,p2->num,p2->score);
    print(head);

    printf("before change num and name:\np1:%ld,%d\np2:%ld,%d\n",p1->num,p1->score,p2->num,p2->score);
    print(head);
    temp_name=p2;
    p2=p1;
    p1=temp_name;
    printf("after change num and name:\np1:%ld,%d\np2:%ld,%d\n",p1->num,p1->score,p2->num,p2->score);
    print(head);
    return head;
}

运行结果:

before change num:
p1:2,22
p2:3,33
1 11
2 22
3 33
4 44
after change num:
p1:2,22
p2:3,33
1 11
3 33
2 22
4 44
before change num and name:
p1:2,22
p2:3,33
1 11
3 33
2 22
4 44
after change num and name:
p1:3,33
p2:2,22
1 11
3 33
2 22
4 44

 

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最后我的链表冒泡排序法

struct student *sort(struct student *head)//from small to large
{
    struct student *temp,*tempname,*tempcounter,*p1,*p2,*p3;
    int counter=0,i,j;
    temp=(struct student *)malloc(LEN);
    //Çó½áµãÊý
    tempcounter=head;
    while(tempcounter->next!=NULL)
    {
        tempcounter=tempcounter->next;
        counter++;
    }
    //sort
    for(i=0;i<counter;i++)
    {
        p1=head;
        p2=head->next;
        if(p1->num > p2->num)
        {
            p1->next=p2->next;
            p2->next=p1;
            head=p2;
        }
        p1=head->next;
        p2=p1->next;
        p3=head;
        for(j=0;j<counter-i-1;j++)
        {
            if(p1->num > p2->num)
            {
                //exchange all
                temp->next=p3->next;
                p3->next=p1->next;
                p1->next=p2->next;
                p2->next=temp->next;
                //exchange name
                tempname=p2;
                p2=p1;
                p1=tempname;
            }
            p1=p1->next;
            p2=p2->next;
            p3=p3->next;
        }
    }
    free(temp);
    return head;
}

 

还是VM厉害,VM的方法更灵活简单

struct student *sort(struct student *head)
{
    struct student *h2,*t,*p;
    h2=NULL;p=head;
    while(p!=NULL){
        t=p;p=p->next;t->next=NULL;
        h2=insert(h2,t);
    }
    return h2;
}

老师给的insert函数是按顺序插入的，此法是用新链表，按顺序读取原链表，然后insert进新链表中，自然是按照顺序排了

 

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再补，这题弄得太复杂了。。根本不需要换结点那么复杂，只需要对num和score直接修改就可以了。。今天好煞笔。。