单链表插入，删除，查询，输出的方式

野性酷女 2022-05-26 22:41 169阅读 0赞

下面这个例子包含单链表的头或尾还有中间增加节点，删除节点，析构new出的空间，及其他代码。

\#include<iostream>

\#include<string>  
using namespace std;  
class Student            \\\\student类相当于节点Node中储存数据的date  
\{  
public:  
int id;  
string name;  
Student() \{\}  
Student(int id, string name) :id(id), name(name) \{\}  
void display()  
\{  
cout << id << ' ' << name << endl;  
\}  
\};  
class Node  
\{  
public:  
Student data;  
Node \*next;     \\\\指向下一节点的指针  
Node()            \\\\不含参的构造函数  
\{  
next = nullptr;     
\}  
Node(Student s)  
\{  
data = s;  
next = nullptr;  
\}  
Node(Node \*p)  
\{  
this->data = p->data;  
this->next = nullptr;  
\}  
void display()  
\{  
data.display();  
\}  
\};  
class Link  
\{  
public:  
Node \* head;  \\\\定义头结点  
Link()  
\{  
head = new Node();     
\}  
void addNodeTail(Node \*p)  
\{  
Node \*q = head;  
while (q->next != nullptr)  
\{  
q = q->next;  
\}  
//q->next = p;  
Node \*temp = new Node(p);    \\\\ new出新空间后，\*temp=\*p  
q->next = temp;  
\}  
void addNodeHead(Node \*p)   \\\\在头部插入节点  
\{  
//p->next = head->next;      
//head->next = p;                \\\\这些方法也行  
Node \*temp = new Node(p);  
temp->next = head->next;  
head->next = temp;  
\}  
void addNodeByOrder(Node \*p)  \\\\在中间插入节点  
\{  
Node \*q1 = head;  
Node \*q2 = head->next;  
while (q2 != nullptr && q2->data.id < p->data.id)  
\{  
q1 = q1->next;  
q2 = q2->next;  
\}  
Node \*temp = new Node(p);  
temp->next = q1->next;  
q1->next = temp;  
\}  
void deleteByID(int id)     \\\\通过查询id删除  
\{  
Node \*q1 = head;  
Node \*q2 = head->next;  
while (q2 != nullptr && q2->data.id != id)  
\{  
q1 = q1->next;  
q2 = q2->next;  
\}  
if (q2 != nullptr)  
\{  
q1->next = q2->next;  
delete q2;  
\}  
  
  
\}  
void display()    \\\\输出  
\{  
Node \*q = head->next;//q指向第二个节点  
while (q != nullptr)  
\{  
q->display();  
q = q->next;  
\}  
\}  
~Link()    \\\\ 析构  
\{  
Node \*q = head;  
while (q != nullptr)  
\{  
Node \*temp = q->next;  
delete q;  
q = temp;  
\}  
\}  
\};  
int main()  
\{  
Node n1(Student(1, "AAA"));  
Node n2(Student(2, "BBB"));  
Node n3(Student(3, "CCC"));  
Link link;  
link.addNodeByOrder(&n3);  
link.addNodeByOrder(&n1);  
link.addNodeByOrder(&n2);  
  
  
link.display();  
cout << endl;  
return 0;

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