打印雙向鏈接列表

Question

我目前正在學習如何使用鏈接列表，特別是雙向鏈接列表，並且當我嘗試向後打印時遇到了與我的程序有關的問題。

下面是代碼的，我需要幫助的部分：

#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; 
} 

該計劃的目的是創建一個列表，轉發打印，向後，插入一個元素，刪除一個元素，然後銷燬該清單。我已經將它簡化爲創建，向前打印和向後打印功能。

我遇到的問題是在createList函數中，我正在修改begin和end，但我只能返回一個或另一個。這意味着無論哪個我不返回在主函數中仍然是NULL，因此不指向任何東西。我試着將begin/current/end設置爲不等於NULL，但如果我這樣做，createList將不起作用。

有沒有人有任何想法，我怎麼可以修改兩者？只需要清楚，在功能中創建的列表HAS TO，在主要初始化它將非常容易。

感謝， 特里斯坦

Answer 1

您的問題是你複製三分球，當你應當參照，即可以將它們傳遞，使用指針到指針或引用到指針，而不是僅僅複製指針在main最初指向的值。通過你正在做的事情，你無法修改在main中聲明的原始指針變量...傳遞參考將允許你這樣做，同時也保持你的函數中的所有列表設置代碼。

因此，例如，改變

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

到

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

並確保採取額外提領顧及你的函數體

最後，你會打電話它像：

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

然後在createList的函數體內對begin做最後的分配，而不是在main。