C++双链表反向打印

Question

我正在尝试编写一个反向打印函数作为双向链表的一部分。以下是我编写的相关函数:

void PLAYER::AddNode(int addID, std::string addName){

nodePtr n = new node; //creates a new node pointer
n->next = NULL;       //Make next null
n->prev = NULL;      // this will set this to be the ending node
n->ID = addID;      //These two lines pass the information into the node
n->name = addName;  // ID# and Name Information

if(head != NULL){  // This checks to see if a list is set up.
    curr = head;    // Make this point to the head.
    while(curr->next != NULL){ // Loops through until the NULL is found
        curr = curr->next;
    }
    curr->next = n; //Make the currnet node point to N
    n->prev = curr; //Make the previous node connect to curr
    n->next = tail; // connect new node to the tail.

}
else{
    head = n;   //If there is no list, this makes N the first node.

}

这是对要使用的函数进行原型(prototype)设计的类。

class PLAYER
{
public:  // Functions go inside PUBLIC
    PLAYER();
    void AddNode(int addID, std::string addName);
    void DeleteNode(int delPlayer);
    void SortNode();
    void PrintList();
    void InsertHead(int AddID, std::string addName);
    void PrintReverse();

private:  //Variables go into here
   typedef struct node{
            // ...
            std::string name;
            int ID;
            node* next;
            node* prev;
    }* nodePtr;


    nodePtr head, curr, temp, prev, test, tail;


};

最后我尝试创建一个反向遍历函数来向后打印。

void PLAYER::PrintReverse()
{
    curr = head; 
    while(curr->next != NULL) //Get to the end of the list
    {
        curr = curr->next;
    }
    while(curr->prev != NULL)    //Work backward and print out the contents
    {
        std::cout << curr->ID << " " << curr->name << endl;
        curr = curr->prev;
    }
}

我想做的是在 PrintReverse() 函数中让它通过尾指针初始化，但是我无法弄清楚要添加到 PrintReverse() 和 AddNode() 的函数以获得新节点尾部指向。

这是我在这里发布的第一个问题，我希望我涵盖了我的所有基础。感谢您提供我能找到的任何帮助。

编辑:

感谢您的所有意见。我正在重新学习数据结构，是的，这是我自己做的一些功课，目的是让逻辑重新开始流动。

我今晚回家后会进行更改。

Answer 1

需要考虑以下更改。

PrintReverse 函数不需要正向传递来获取tail。

void PLAYER::PrintReverse()
{
    curr = tail; 
    while(curr != NULL)    //Work backward and print out the contents
    {
        std::cout << curr->ID << " " << curr->name << endl;
        curr = curr->prev;
    }
}

AddNode 函数中尾部的处理方式存在问题。查看注释包含 [CHANGED] 和 [ADDED] 的行:

if(head != NULL){  // This checks to see if a list is set up.
    curr = head;    // Make this point to the head.
    while(curr->next != NULL){ // Loops through until the NULL is found
        curr = curr->next;
    }
    curr->next = n; //Make the currnet node point to N
    n->prev = curr; //Make the previous node connect to curr
    n->next = NULL; // [CHANGED]: we want the last node not to have a successor.
}
else{
    head = n;   //If there is no list, this makes N the first node.
}

tail = n;       // [ADDED]: the last node added is the new tail. 

然而，更简单的解决方案是再次避免前向传递，并从尾部开始。

if(tail != NULL){  // This checks to see if a list is set up.
    tail->next = n; //Make the old tail node point to N
    n->prev = tail; 
    n->next = NULL; 
}
else{
    head = n;   //If there is no list, this makes N the first node.
}

tail = n;       // The last node added is the new tail.