双指针如何用于链表实现

Question

以下代码运行良好：

#include <iostream>
using namespace std;
struct Node{
int data;
struct Node *next;
Node(int data, Node *next = nullptr){
this->data = data;
this->next = next;
}
};
void push_back(Node **head, int data){
if(*head == nullptr){
*head = new Node(data);
}
else{
Node *current = *head;
while(current->next != nullptr){
current = current->next;
}
current->next = new Node(data);
}
}

void Print(Node **head){
Node *current = *head;
while(current != nullptr){
cout << current->data << " ";
current = current->next;
}
cout << endl;
}
int main(){
Node *head = nullptr;
push_back(&head, 5);
push_back(&head, 2);
push_back(&head, 1);
push_back(&head, -7);
Print(&head);
}

但当我尝试咆哮时，什么也没发生，头和所有的操作都保持为null。我所做的只是向函数传递单指针而不是双指针：

#include <iostream>
using namespace std;

struct Node{
int data;
struct Node *next;
Node(int data, Node *next = nullptr){
this->data = data;
this->next = next;
}
};
void push_back(Node *head, int data){
if(head == nullptr){
head = new Node(data);
}
else{
Node *current = head;
while(current->next != nullptr){
current = current->next;
}
current->next = new Node(data);
}
}

void Print(Node *head){
Node *current = head;
while(current != nullptr){
cout << current->data << " ";
current = current->next;
}
cout << endl;
}
int main(){
Node *head = nullptr;
push_back(head, 5);
push_back(head, 2);
push_back(head, 1);
push_back(head, -7);
Print(head);
}

我不明白为什么我需要双指针才能使它工作？第二个程序是否只向函数发送了一份头的副本，而没有其他内容？

Answer 1

void push_back(Node *head, int data){
if(head == nullptr){
head = new Node(data);
}
else{
Node *current = head;
while(current->next != nullptr){
current = current->next;
}
current->next = new Node(data);
}
}

在main函数中，不能通过push_back来更改指针head的值。我们可以通过将对象的指针或引用传递给另一个函数来更改对象，但不能通过传递它自己！因此，每次head = new Node(data);(尝试更改head(实际上都不会更改调用push back()并导致内存溢出的函数中的head