打印双向链表

Question

我目前正在学习如何使用链表，特别是双向链表，当我尝试向后打印时，我的程序遇到了问题。

以下是我需要帮助的代码部分：

#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;
}

该程序的目的是创建一个列表，向前，向后打印，插入元素，擦除元素，然后销毁列表。我已将其削减为仅创建，向前打印和向后打印功能。

我遇到的问题是，在 createList 函数中，我正在修改开始和结束，但我只能返回一个或另一个。这意味着无论我不返回哪个，在 main 函数中仍然是 NULL，因此不会指向任何内容。我尝试将开始/当前/结束设置为不等于 NULL，但如果我这样做，createList 将无法正常工作。

有没有人对我如何修改两者有任何想法？需要明确的是，列表必须在函数中创建，只需在主要中初始化它就非常容易了。

谢谢特里斯坦

Answer 1

你的问题是你正在复制指针，而你应该通过引用传递它们，即使用指针到指针或引用到指针，而不仅仅是复制指针最初指向的值main。 通过您正在执行的操作，您无法修改在 main 中声明的原始指针变量......通过引用传递将允许您执行此操作，同时将所有列表设置代码保留在函数中。

例如，改变

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

自

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

然后确保在函数主体中考虑额外的取消引用

最后，您可以这样称呼它：

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

并做最后的分配，begin createList的功能内部，而不是main。