我无法删除链表中的某个节点

Question

我有下一个链表：

#include <iostream>
#include <string>
#include <cstddef>
using namespace std;
//Node class
class Node
{
public:
string name;
string age;
Node *next;
Node *prev;
Node(string name, string age)
{
this->name = name;
this->age = age;
this->next = NULL;
this->prev = NULL;
}
};
//Doubly circular Linked list class
class List
{
private:
Node *first;
Node *last;
public:
List()
{
this->first = NULL;
this->last = NULL;
}
void deleteNode(string name)
{
Node *start = this->first;
if (start == NULL)
{
return;
}
Node *curr = start, *prev_1 = NULL;
while (curr->name != name)
{
if (curr->next == start)
{
printf("nList doesn't have node with value = %d", name);
return;
}
prev_1 = curr;
curr = curr->next;
}
if (curr->next == start && prev_1 == NULL)
{
(start) = NULL;
free(curr);
return;
}
if (curr == start)
{
prev_1 = (start)->prev;
start = (start)->next;
prev_1->next = start;
(start)->prev = prev_1;
free(curr);
}
else if (curr->next == start)
{
prev_1->next = start;
(start)->prev = prev_1;
free(curr);
}
else
{
Node *temp = curr->next;
prev_1->next = temp;
temp->prev = prev_1;
free(curr);
}
}
void add(string name, string age)
{
Node *node = new Node(name, age);
if (this->first == NULL)
{
this->first = node;
this->first->next = node;
this->first->prev = node;
this->last = node;
}
else
{
this->last->next = node;
node->next = this->first;
node->prev = this->last;
this->last = node;
this->first->prev = this->last;
}
}
void print()
{
Node *aux;
aux = this->first;
while (aux != NULL)
{
cout << aux->name << ", " << aux->age<<endl;
aux = aux->next;
if (aux == this->first)
{
break;
}
}
}
};
int main(int argc, char const *argv[])
{
List list;

list.add("David", "25");
list.add("James", "45");
list.add("Charles", "78");
list.add("Katty", "96");
cout<<"List without delete any node"<<endl;
list.print();
//trying to delete a certain node
list.deleteNode("David");
cout<<"printing list without David because I deleted him above"<<endl;
list.print();
return 0;
}

在我的函数deleteNode中，我试图发送一个名称将其从链表中删除，但当运行我的代码时，它不起作用，它只显示下一个：

List without delete any node
David, 25
James, 45
Charles, 78
Katty, 96

我的预期输出是：

List without delete any node
David, 25
James, 45
Charles, 78
Katty, 96
printing list without David because I deleted him above
James, 45
Charles, 78
Katty, 96

我不知道还能做什么，我希望你能帮助我，谢谢。

Answer 1

对于初学者来说，free是释放用new分配的指针的错误方法。请改用delete。此外，通常最好将字符串作为常量引用传递。

deleteNode函数看起来非常复杂。让我为你简化一下。告诉我你对此的看法：

void deleteNode(const string& name)
{
Node *start = this->first;
while ((start != nullptr) && (start->name != name))
{
start = start->next;
}
if (start == nullptr)
{
return;
}
Node* prev = start->prev;
Node* next = start->next;
if (prev != nullptr)
{
prev->next = next;
}
else
{
first = next;
}
if (next != nullptr)
{
next->prev = prev;
}
else
{
last = prev;
}
delete start;
}

同样，add可以简化为：

void add(const string& name, const string& age)
{
Node *node = new Node(name, age);

if (this->last == NULL)   // first insert
{
first = node;
last = node;
}
else
{
last->next = node;
node->prev = last;
last = node;
}
}