双指针如何用于链表实现？

Question

以下代码工作正常：

#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);

}

但是当我尝试以下操作时，什么都没有发生，并且 head 与所有操作一起保持为 nullptr。 我所做的只是将单指针传递给 function 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其工作？ 第二个程序是否仅向函数发送 head 的副本，仅此而已？

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);
    }
}

您不能通过main function 中的push_back更改指针head的值。 我们可以通过传递它的指针或对另一个 function 的引用来更改 object，但不能通过传递自身！ 所以每次head = new Node(data); （尝试更换head ）实际上没有在调用push back()并导致 memory 溢出的 function 中更换head