打印双向链接列表

Question

我目前正在学习如何使用链接列表，特别是双向链接列表，并且当我尝试向后打印时遇到了与我的程序有关的问题。

下面是代码的，我需要帮助的部分：

#include <iostream> 

using namespace std; 

struct node 
{ 
    int data; //int to store data in the list 
    node *next; //pointer to next value in list 
    node *prev; //pointer to previous value in list 
}; 

node *appendList(node *current, int newData) //Function to create new nodes in the list 
{ 
    node *newNode; //create a new node 
    newNode = new node; 
    newNode->data = newData; //Assign data to it 
    newNode->next = NULL; //At end of list so it points to NULL 
    newNode->prev = current; //Link new node to the previous value 
    current->next = newNode; //Link current to the new node 
    return newNode; //return the new node 
} 

node *createList(int maxLoop, node *begin, node *current, node *end) //Function to create list 
{ 
    //Allocate the starting node 
    current = new node; 
    current -> data = 1; //First data value is 1 
    current -> next = NULL; //next value is NULL 
    current -> prev = NULL; //previous value is NULL 
    begin = current; //This is the beginning of the list 

    for (int count = 2; count <= maxLoop; count++) //Loop to fill the list 
    { 
     current = appendList(current, count*count); //Create new nodes and fill with square numbers 
    } 
    end = current; //Now we are at the end of the list 
    return begin; //Return begin, this is the problem; I can't return end as well 
} 

void printForward (node *p) //Function to print the list forwards 
{ 
    node *curr = p; //current is the beginning value of the list 
    while (curr != NULL) //Continue while current is not at the end of the list 
    { 
     cout << curr->data << " "; //Print out the data of current 
     curr = curr->next; //Move one value along in the list 
    } 
} 

void printBackward (node *p) //Function to print the list backwards 
{ 
    node *curr = p; //current is the end value of the list 
    while (curr != NULL) //Continue while current is not at the beginning of the list 
    { 
     cout << curr->data << " "; //Print out the data of current 
     curr = curr->prev; //Move one value back in the list 
    } 
} 

int main() 
{ 
    //Initialize current, begin, and end 
    node *current = NULL; 
    node *begin = NULL; 
    node *end = NULL; 
    int maxLoop = 10; //The number of items in the list 

    cout << "The list has now been created." << endl; 
    begin = createList(maxLoop, begin, current, end); //function to create the list 
    cout << "Printed forwards, this list is: "; 
    printForward(begin); //Function to print the list forwards 
    cout << endl; 
    cout << "Printed backwards, this list is: "; 
    printBackward(end); //Function to print the list backwards 
    cout << endl; 
    return 0; 
} 

该计划的目的是创建一个列表，转发打印，向后，插入一个元素，删除一个元素，然后销毁该清单。我已经将它简化为创建，向前打印和向后打印功能。

我遇到的问题是在createList函数中，我正在修改begin和end，但我只能返回一个或另一个。这意味着无论哪个我不返回在主函数中仍然是NULL，因此不指向任何东西。我试着将begin/current/end设置为不等于NULL，但如果我这样做，createList将不起作用。

有没有人有任何想法，我怎么可以修改两者？只需要清楚，在功能中创建的列表HAS TO，在主要初始化它将非常容易。

感谢， 特里斯坦

Answer 1

您的问题是你复制三分球，当你应当参照，即可以将它们传递，使用指针到指针或引用到指针，而不是仅仅复制指针在main最初指向的值。通过你正在做的事情，你无法修改在main中声明的原始指针变量...传递参考将允许你这样做，同时也保持你的函数中的所有列表设置代码。

因此，例如，改变

node* createList(int maxLoop, node *begin, node *current, node *end) 

到

void createList(int maxLoop, node** begin, node** current, node** end) 

并确保采取额外提领顾及你的函数体

最后，你会打电话它像：

createList(maxLoop, &begin, &current, &end); 

然后在createList的函数体内对begin做最后的分配，而不是在main。