代码自己真正敲才发现自己掌握的一点都不好。
实现了单链表的基本操作
#include "stdio.h" struct student { long num; float score; struct student *next; }; //结构体
1，创建单链表
struct student * create_link() { struct student *p=NULL,*head=NULL,*p1=NULL; int n,i=1; printf("please input the numble of nodes you want to create:"); scanf("%d",&n); //n是输入的链表的个数 while(i<=n) { p=(struct student*)malloc(sizeof(struct student)); //分配空间 scanf("%ld %f",&p->num,&p->score); if(head==NULL) { head=p; p1=p; i++; continue; //退出本次循环 } p1->next=p; //关键，是所有结点连接的关键 p1=p1->next; i++; } p1->next=NULL; //最后让最后的节点指向空 return head;
}
2，插入结点（在第i个结点后面插入）
struct student *insert_alink(struct student * head,int i,long n,float s) { struct student *p,*p1,*pre; p=(struct student*)malloc(sizeof(struct student)); p->num=n; p->score=s; int j=1; if(i<1) //判断i<0的情况 { printf("the potition is error"); return head; } if(head==NULL) //i不小于零，就将插入的结点作为首结点，而不管i的大小（其实也可以直接退出，找不到该结点） { head=p; return head; } pre=head; p1=pre->next; while(j<i) {
pre=p1; if(pre==NULL)
break; p1=p1->next; j++; } if(pre!=NULL) { p->next=p1; pre->next=p; } else printf("the potition is error"); return head; } 3，两个链表合并（其实也是插入链表的问题，将head1插入到head的第i个位置后面）
struct student * two_links(struct student * head,struct student *head1,int i) { int j=1; struct student *p=head,*p1=head1,*pre,*pre1; if(head==NULL || head1==NULL) { printf("there are at least one linked list is null"); return head; } while(j<=i) //使pre指向带插入结点，p是pre的后继结点，链表插在pre和p之间 { pre=p; if(pre==NULL) //若是pre为空，则说明i的值太大超出了head链表的结点个数 { printf("the potision is errir\n"); return head; } p=p->next; j++; } while(p1!=NULL) //pre是倒数是最后一个不为空的结点，p1为空 { pre1=p1; p1=p1->next;
}
if(i==0)
{
pre1->next=head;
return head1;
} //如果i==0,则插在表头
//否则
pre1->next=p;
pre->next=head1;
return head;
}
4,计算链表结点个数
int size_stu(struct student * head)
{ int i;
struct student * p=head;
if(head==NULL)
return 0;
while(p!=NULL)
{
p=p->next;
i++;
}
return i;
}
5，删除
struct student * delete_alink(struct student * head,int i)
{ struct student * p1=head,*pre;
if(head==NULL || i<1)
{
printf("the head is invalid or the position is error\n");
return head;
}
int j=1;
if(i==1)
{
head=head->next;
return head;
}
while(j<i)
{ pre=p1;
p1=p1->next;
j++;
}
if(p1!=NULL)
{
pre->next=p1->next;
free(p1);
}
else
printf("the potition is error\n");
return head;
}
6,逆序
struct student * nixun(struct student * head)
{
struct student *p1=NULL,*p2=NULL,*p3=NULL;
p1=head;
p2=head->next;
p3=p2->next;
while(p2!=NULL)
{
p1->next=p3;
p2->next=head;
head=p2;
p2=p3;
if(p2==NULL)
break;
p3=p3->next;
}
return head;
}
