SDUT ACM OJ 实验二链表
作者：互联网
A ：顺序建立链表
#include<stdio.h>
#include<stdlib.h>
struct node
{
int data;
struct node *next;
}*head,*tail,*p,*q;
int main ()
{
head=(struct node *)malloc(sizeof(struct node));
head->next=NULL;
tail=head;
int n,i;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
p=(struct node *)malloc(sizeof(struct node));
scanf("%d",&p->data);
p->next=NULL;
tail->next=p;
tail=p;
}
head=head->next;
while(head->next!=NULL)
{
printf("%d ",head->data);
head=head->next;
}
printf("%d\n",head->data);
}
B ：逆序建立链表
#include <stdio.h>
#include <stdlib.h>
struct node
{
int data;
struct node *next;
}*head,*tail,*q,*p;
int main()
{
int n,i;
scanf("%d",&n);
head=(struct node *)malloc(sizeof(struct node));
head->next=NULL;
tail=head;
for(i=0;i<n;i++)
{
p=(struct node *)malloc(sizeof(struct node));
scanf("%d",&p->data);
p->next=tail->next;
tail->next=p;
}
head=head->next;
while(head->next!=NULL)
{
printf("%d ",head->data);
head=head->next;
}
printf("%d\n",head->data);
}
C ： 链表的结点插入
#include<stdio.h>
#include<stdlib.h>
struct node
{
int data;
struct node *next;
}*head,*tail,*p,*q;
int main()
{
int n,i,a;
while(scanf("%d",&n)!=EOF)
{
head=(struct node *)malloc(sizeof(struct node));
head->next=NULL;
for(i=0;i<n;i++)
{
p=(struct node *)malloc(sizeof(struct node));
scanf("%d %d",&a,&p->data);
q=head;
while(a--&&q->next!=NULL)
{
q=q->next;
}
p->next=q->next;
q->next=p;
}
head=head->next;
while(head->next!=NULL)
{
printf("%d ",head->data);
head=head->next;
}
printf("%d\n",head->data);
}
}
D ：单链表中重复元素的删除
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct node
{
int data;
struct node *next;
};
struct node*creat()
{
struct node *tail,*p,*head;
head = (struct node*)malloc(sizeof(struct node));
head->next=NULL;
tail = head;
while(1)
{
p = (struct node*)malloc(sizeof(struct node));
scanf("%d",&p->data);
if(p->data==-1)break;
p ->next = NULL;
tail->next = p;
tail = p;
}
return (head);
}
struct node*ncreat(int n)
{
struct node *head,*p;
head = (struct node*)malloc(sizeof(struct node));
head->next = NULL;
for(int i=1;i<=n;i++)
{
p = (struct node*)malloc(sizeof(struct node));
scanf("%d",&p->data);
p->next = head->next;
head->next = p;
}
return (head);
}
void show(struct node*head)
{
struct node *p;
p =head->next;
while(p)
{
if(p->next!=NULL)
{
printf("%d ",p->data);
}
else
{
printf("%d\n",p->data);
}
p = p->next;
}
}
int del(struct node *head,int n)
{
struct node *p,*q,*t;
p = head->next;
while(p)
{
q = p;
t = q->next;
while(t)
{
if(p->data==t->data)
{
q->next = t->next;
free(t);
t = q->next;
n--;
}
else
{
q = t;
t = q->next;
}
}
p = p->next;
}
return n;
}
void nizhi(struct node*head)
{
struct node *p,*q;
p = head->next;
head ->next = NULL;
q = p->next;
while(p)
{
p->next = head->next;
head->next = p;
p = q;
if(q)
{
q = q->next;
}
}
}
int main()
{
struct node *head;
head = (struct node*)malloc(sizeof(struct node));
int n;
scanf("%d",&n);
head=ncreat(n);
printf("%d\n",n);
show(head);
int m = del(head,n);
printf("%d\n",m);
show(head);
return 0;
}
E ：链表的逆置
#include <stdio.h>
#include <stdlib.h>
struct node
{
int data;
struct node *next;
}*head,*tail,*q,*p;
int main()
{
int x;
head=(struct node *)malloc(sizeof(struct node));
head->next=NULL;
tail=head;
while(~scanf("%d",&x)&&x!=-1)
{
p=(struct node *)malloc(sizeof(struct node));
p->data=x;
p->next=tail->next;
tail->next=p;
}
head=head->next;
while(head->next!=NULL)
{
printf("%d ",head->data);
head=head->next;
}
printf("%d\n",head->data);
}
F ：有序链表的归并
#include<stdio.h>
#include<stdlib.h>
struct node
{
int data;
struct node *next;
}*head1,*head,*tail,*p,*q;
int main()
{
int m,n;
scanf("%d %d",&m,&n);
head=(struct node *)malloc(sizeof(struct node));
head1=(struct node *)malloc(sizeof(struct node));
head->next=head1->next=NULL;
tail=head;
while(m--)
{
p=(struct node *)malloc(sizeof(struct node));
scanf("%d",&p->data);
p->next=tail->next;
tail->next=p;
tail=p;
}
tail=head1;
while(n--)
{
p=(struct node *)malloc(sizeof(struct node));
scanf("%d",&p->data);
p->next=tail->next;
tail->next=p;
tail=p;
}
p=head->next;
tail=head;
head->next=NULL;
free(head1);
while(p&&q)
{
if(p->data>q->data)
{
tail->next=q;
tail=q;
q=q->next;
}
else
{
tail->next=p;
tail=p;
p=p->next;
}
}
if(p)
{
tail->next=p;
}
else
{
tail->next=q;
}
p=head->next;
printf("%d",p->data);
p=p->next;
while(p)
{
printf(" %d",p->data);
p=p->next;
}
printf("\n");
}
G ：单链表的拆分
#include <stdio.h>
#include <stdlib.h>
int main()
{
int b[100002],c[200002],i,x=0,y=0,n,a;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a);
if(a%2==0)
{
b[x]=a;
x++;
}
else
{
c[y]=a;
y++;
}
}
printf("%d %d\n",x,y);
for(i=0;i<x-1;i++)
{
printf("%d ",b[i]);
}
printf("%d\n",b[x-1]);
for(i=0;i<y-1;i++)
{
printf("%d ",c[i]);
}
printf("%d\n",c[y-1]);
}
H ：双向链表
#include<stdio.h>
#include<stdlib.h>
struct node
{
int data;
struct node *next,*pre;
}*head,*r,*p,*q;
int main ()
{
head=(struct node *)malloc(sizeof(struct node));
head->next=NULL;
head->pre=NULL;
r=head;
int m,n,x,i;
scanf("%d %d",&n,&m);
for(i=0;i<n;i++)
{
p=(struct node *)malloc(sizeof(struct node));
scanf("%d",&p->data);
p->next=r->next;
p->pre=r;
r->next=p;
r=p;
}
for(i=0;i<m;i++)
{
scanf("%d",&x);
p=head->next;
while(p->data!=x){p=p->next;}
if(p->pre==head){printf("%d\n",p->next->data);}
else if(p->next==NULL){printf("%d\n",p->pre->data);}
else printf("%d %d\n",p->pre->data,p->next->data);
}
}
I ： 约瑟夫问题
//这道题我没有用链表写，因为链表太冗长繁杂
//放心考试一样过
#include<stdio.h>
int main()
{
int n,m,i,p=0;
scanf("%d %d",&n,&m);
for(int i=2;i<=n;i++)
{p=(p+m)%i;}
printf("%d",p+1);
}
J ： 不敢死队问题
//这道题我也没有用链表写，你懂的
#include<stdio.h>
#include<stdlib.h>
int main ()
{
int n;
while(~scanf("%d",&n))
{
int m=5,i=0,p,q,l=0;
while(++i<=n)
{
p=i*m;
while(p>n)
p=p-n+(p-n-1)/(m-1);
l++;
if(p==1)
printf("%d\n",l);
q=p;
}
}
}
如果学习了c++，就可以轻松的完成这些题目，虽然考试不一定能用，但是提前学习有助于减轻大二数据结构课业负担。
比如说这几道题就可以用c++简单优美的书写：
1.建立顺序链表
#include<bits/stdc++.h>
using namespace std;
int main()
{
list<int> l;//建立一个链表
int n,x;
cin>>n;
while(n--)
{
cin>>x;
l.emplace_back(x);//在链表的头节点后面依次添加数据
}
list<int>::iterator it=l.begin();//定义一个迭代器来遍历链表
for(;it!=l.end();it++)
{
if(it!=l.begin()) cout<<" ";
cout<<*it;
}
}
2.链表的逆序（建立逆序链表，这俩一样）
#include<bits/stdc++.h>
using namespace std;
int main()
{
list<int> l;
int n,x;
cin>>n;
while(n--)
{
cin>>x;
l.emplace_back(x);
}
l.reverse(); //可以看到除了这里其他的地方都跟正序没什么区别
list<int>::iterator it=l.begin();
for(;it!=l.end();it++)
{
if(it!=l.begin()) cout<<" ";
cout<<*it;
}
}
3.有序链表的归并
#include<bits/stdc++.h>
using namespace std;
int main()
{
list<int> l;
int n,m,x;
list<int>::iterator it;
cin>>n>>m;
n+=m;
while(n--)
{
cin>>x;
l.push_front(x);
}
l.sort(); //排序就这样简单，一行完成
for(it=l.begin();it!=l.end();it++)
{
if(it!=l.begin()) cout<<" ";
cout<<*it;
}
}
其它题目想研究的同学可以自己学习学习，研究研究。
不懂得可以私信我，随时解答。
程序有错误请私信我，以及时改正。感谢！
标签：node,head,struct,int,ACM,next,链表,SDUT,data 来源： https://blog.csdn.net/weixin_52604835/article/details/122813809