Basic operation of C++ linked list
Some basic operations of the linked list are given below:
Constructor of linked list
template <typename T>
List<T>::List()
{
head = new node<T>;
head->next=NULL;
count=0;
}
Destructor of linked list
template <typename T>
List<T>::~List<T>()
{
node<T>* p;
while(head->next != NULL)
{
p = head->next;
head->next = p->next;
delete p;
count--;
}
delete head;
}
Create linked list
template <typename T>
void List<T>::createList(T x)
{
node<T>* rear = head;
while (rear->next!=NULL)
rear = rear->next;
node<T>* p = new node<T>;
p->data = x;
p->next = NULL;
rear->next = p;
rear = p;//这一步似乎可要可不要
count++;
}
Linked list insertion
template <typename T>
void List<T>::insert(T x,int num)
{
if (num>count+1)
{
cout << "No location to inserte!" << endl;
return;
}
int n=0;
node<T>* tem=head;
while (num-1!=n)
{
tem=tem->next;
n++;
}
node<T>* p=new node<T>;
p->next=tem->next;
tem->next=p;
p->data=x;
count++;
}
Delete of linked list
template <typename T>
void List<T>::del(int n)
{
int num=count-n,k=0;
node<T>* p=head;
while(num!=k)
{
p=p->next;
k++;
}
p->next=p->next->next;
count--;
}
In-place inversion of linked list
Regarding the two inversions, I want to talk about it a little bit. In-place inversion is to change the direction of next and finally realize the flip. A simple diagram is roughly as follows:
template <typename T>
void List<T>::reverse1()
{
node<T>* q=NULL;
node<T>* temp;
head=head->next;//这一步操作是避免原来head->next->data随机取值导致逆置的链表多出一个随机数。
while(head!=NULL)
{
temp=head;
head=head->next;
temp->next=q;
q=temp;
}
node<T>* new_head=new node<T>;//构建一个头结点,否则逆置之后的head->next指向第二个数，观察print()函数可知，打印的第一项是head->next。
new_head->next=temp;
head = new_head;
}
The head of the linked list is reversed
This one also draws a rough diagram, I hope you can compare with the above picture, and be inspired:
template <typename T>
void List<T>::reverse2()
{
node<T>* p = head->next;
node<T>* q;
head->next=NULL;
while(p)
{
q=p;
p=p->next;
q->next=head->next;
head->next=q;
}
}
The complete code is as follows
#include <iostream>
using namespace std;
template <typename T>
struct node
{
T data;
node* next;
};
template <typename T>
class List
{
public:
List();//构造一个链表
~List();//析构一个链表
void createList(T x);//创建一个链表
void insert(T x,int num);//这特定的位置插入结点
void del(int n);//删除链表倒数第 n 个结点
int length();//求链表的长度
void reverse1();//就地转置
void reverse2();//头插法逆置
void print();//打印结果
private:
int count;
node<T>* head;
};
template <typename T>
List<T>::List()
{
head = new node<T>;
head->next=NULL;
count=0;
}
//这个链表的析构是不是正确的呀？首先头结点好像没有析构，其次最后一步能不能直接delete head，head好像不是new出来的。
template <typename T>
List<T>::~List<T>()
{
node<T>* p;
while(head->next != NULL)
{
p = head->next;
head->next = p->next;
delete p;
count--;
}
delete head;
}
template <typename T>
void List<T>::createList(T x)
{
node<T>* rear = head;
while (rear->next!=NULL)
rear = rear->next;
node<T>* p = new node<T>;
p->data = x;
p->next = NULL;
rear->next = p;
rear = p;//这一步似乎可要可不要
count++;
}
template <typename T>
int List<T>::length()
{
return count;
//如果没有设置count这个变量，也可以用下面这种方法：
// int count=0;
// while(head->next!=NULL)
// {
// head=head->next;
// count++;
// }
// return count;
}
template <typename T>
void List<T>::insert(T x,int num)
{
if (num>count+1)
{
cout << "No location to inserte!" << endl;
return;
}
int n=0;
node<T>* tem=head;
while (num-1!=n)
{
tem=tem->next;
n++;
}
node<T>* p=new node<T>;
p->next=tem->next;
tem->next=p;
p->data=x;
count++;
}
template <typename T>
void List<T>::del(int n)
{
int num=count-n,k=0;
node<T>* p=head;
while(num!=k)
{
p=p->next;
k++;
}
p->next=p->next->next;
count--;
}
// void reverse();
// void connect();//两个有序的链表合并
// T answer();//求中间结点n
template <typename T>
void List<T>::print()//打印结果
{
node<T>* p = head->next;
while(p->next!=NULL)
{
cout << p->data << " ";
p=p->next;
}
cout << p->data << endl;
}
template <typename T>
void List<T>::reverse1()
{
node<T>* q=NULL;
node<T>* temp;
head=head->next;//这一步操作是避免原来head->next->data随机取值导致逆置的链表多出一个随机数。
while(head!=NULL)
{
temp=head;
head=head->next;
temp->next=q;
q=temp;
}
node<T>* new_head=new node<T>;//构建一个头结点,否则逆置之后的head->next指向第二个数，观察print()函数可知，打印的第一项是head->next。
new_head->next=temp;
head = new_head;
}
template <typename T>
void List<T>::reverse2()
{
node<T>* p = head->next;
node<T>* q;
head->next=NULL;
while(p)
{
q=p;
p=p->next;
q->next=head->next;
head->next=q;
}
}
int main()
{
List<int> list;
for(int i=0;i<5;i++)
list.createList(i);
cout << "The length of this list is " << list.length() << endl << "The list is as follows: ";
list.print();
cout << "-------------" << endl;
list.insert(400,6);
cout << "The length of this list is " << list.length() << endl << "The list is as follows: ";
list.print();
cout << "-------------" << endl;
list.del(1);
cout << "The length of this list is " << list.length() << endl << "The list is as follows: ";
list.print();
cout << "-------------" << endl;
list.reverse1();
cout << "The length of this list is " << list.length() << endl << "The list is as follows: ";
list.print();
cout << "-------------" << endl;
list.reverse2();
cout << "The length of this list is " << list.length() << endl << "The list is as follows: ";
list.print();
return 0;
}
The results of the operation are attached below
For related knowledge of linked lists, you can refer to my other blog, and attach its link
below: link .
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