湖南大学CG题库（数据结构与算法）题解

前言：湖南大学2022数据结构与算法课程答案

实验一：预测线性表的实现

1、基于链表的线性表实现（题解）

题目描述：

题解：

#include<bits/stdc++.h>
using  namespace  std;

//断言工具函数：如果val为假则输出s然后中断程序
void  Assert(bool  val,string  s){
   
        if(!val){
   
                cout<<"断言失败："<<s<<endl;
        }
}

template  <typename  E>  class  List  {
     //  List  ADT
private:
        void  operator  =(const  List&)  {
   }            //  Protect  assignment
        List(const  List&)  {
   }                      //  Protect  copy  constructor
public:
        List()  {
   }                    //  Default  constructor
        virtual  ~List()  {
   }  //  Base  destructor

        //  Clear  contents  from  the  list,  to  make  it  empty.
        virtual  void  clear()  =  0;

        //  Insert  an  element  at  the  current  location.
        //  item:  The  element  to  be  inserted
        virtual  void  insert(const  E&  item)  =  0;

        //  Append  an  element  at  the  end  of  the  list.
        //  item:  The  element  to  be  appended.
        virtual  void  append(const  E&  item)  =  0;

        //  Remove  and  return  the  current  element.
        //  Return:  the  element  that  was  removed.
        virtual  E  remove()  =  0;

        //  Set  the  current  position  to  the  start  of  the  list
        virtual  void  moveToStart()  =  0;

        //  Set  the  current  position  to  the  end  of  the  list
        virtual  void  moveToEnd()  =  0;

        //  Move  the  current  position  one  step  left.  No  change
        //  if  already  at  beginning.
        virtual  void  prev()  =  0;

        //  Move  the  current  position  one  step  right.  No  change
        //  if  already  at  end.
        virtual  void  next()  =  0;

        //  Return:  The  number  of  elements  in  the  list.
        virtual  int  length()  const  =  0;

        //  Return:  The  position  of  the  current  element.
        virtual  int  currPos()  const  =  0;

        //  Set  current  position.
        //  pos:  The  position  to  make  current.
        virtual  void  moveToPos(int  pos)  =  0;

        //  Return:  The  current  element.
        virtual  const  E&  getValue()  const  =  0;
};

//  Singly  linked  list  node
template  <typename  E>  class  Link  {
   
public:
        E  element;            //  Value  for  this  node
        Link  *next;                //  Pointer  to  next  node  in  list
        //  Constructors
        Link(const  E&  elemval,  Link*  nextval  =NULL)  {
   

element=elemval;
next=nextval;

        }
        Link(Link*  nextval  =NULL)  {
   

next=nextval;

        }
};

template  <typename  E>  class  LList:  public  List<E>  {
   
private:
        Link<E>*  head;              //  Pointer  to  list  header
        Link<E>*  tail;              //  Pointer  to  last  element
        Link<E>*  curr;              //  Access  to  current  element
        int  cnt;                              //  Size  of  list

        void  init()  {
                   //  Intialization  helper  method
        curr=tail=head=new Link<E>;
        cnt=0;
        }

        void  removeall()  {
         //  Return  link  nodes  to  free  store

while(head!=NULL){
   
curr=head;
head=head->next;
delete curr;
}

        }

public:
        LList(int  size=100)  {
   
                init();        //  Constructor
        }
        ~LList()  {
   
                removeall();        //  Destructor
        }

        //  使用空格分隔输出线性表中的所有数据，并最终换行
        //无数据时直接输出空行
void  print(){
   
        Link<E>* tmp=head->next;
        while(tmp!=NULL){
   
        cout<<tmp->element<<' ';
        tmp=tmp->next;
        }
        cout<<endl;
        }

        void  clear()  {
   
                removeall();        //  Clear  list
                init();
        }

        //  Insert  "it"  at  current  position
void  insert(const  E&  it)  {
   
        curr->next=new Link<E>(it,curr->next);
        if(tail==curr)tail=curr->next;

        cnt++