在C ++中以相反的顺序打印我的链表 [英] Printing my linked list in reverse order in C++

问题描述

所以我是C ++的新手，今天我决定坐下来了解链表如何工作。我有很多乐趣，迄今为止，但我遇到了一个问题，当尝试以相反的顺序打印我的链接列表（不颠倒链表的顺序！）

So I'm fairly new to C++ and today I decided to sit down and understand how linked lists work. I'm having a lot of fun doing it so far, but I've encountered a problem when trying to print my linked list in reverse order (not reverse the order of the linked list!)

此外，我想要这样做而没有双重链接列表：

Also, I wanted to do this without having a double linked list:

#include <iostream>
#include <string>

using namespace std;

class LinkedList
{
    public:
        LinkedList()
        {
            head = NULL;
        }

        void addItem(string x)
        {
            if(head == NULL)
            {
                head = new node();
                head->next = NULL;
                head->data = x;
            } else {
                node* temp = head;
                while(temp->next != NULL)
                    temp = temp->next;

                node* newNode = new node();
                newNode->data = x;
                newNode->next = NULL;
                temp->next = newNode;
            }
        }
        void printList()
        {
            node *temp = head;
            while(temp->next != NULL)
            {
                cout << temp->data << endl;
                temp = temp->next;
            }
            cout << temp->data << endl;
        }

        void addToHead(string x)
        {
            node *temp = head;
            head = new node;
            head->next = temp;
            head->data = x;
        }

        int countItems()
        {
            int count = 1;
            for(node* temp = head; temp->next != NULL; temp = temp->next)
                ++count;
            return count;
        }

        void printReverse()
        {
            node* temp2;
            node* temp = head;
            while(temp->next != NULL)
                temp = temp->next;

            //Print last node before we enter loop
            cout << temp->data << endl;

            for(double count = countItems() / 2; count != 0; --count)
            {
                //Set temp2 before temp
                temp2 = head;
                while(temp2->next != temp)
                    temp2 = temp2->next;
                cout << temp2->data << endl;

                //Set temp before temp2
                temp = head;
                while(temp->next != temp2)
                    temp = temp->next;
                cout << temp->data << endl;
            }
            cout << "EXIT LOOP" << endl;
        }

    private:
        struct node
        {
            string data;
            node *next;
        }

    *head;
};

int main()
{
    LinkedList names;

    names.addItem("This");
    names.addItem("is");
    names.addItem("a");
    names.addItem("test");
    names.addItem("sentence");
    names.addItem("for");
    names.addItem("the");
    names.addItem("linked");
    names.addItem("list");

    names.printList();

    cout << endl;

    names.addToHead("insert");

    names.printList();

    cout << endl;

    cout << names.countItems() << endl;

    cout << "Print reverse: " << endl;
    names.printReverse();
    cout << endl;

    return 0;
}

现在我不知道为什么我的代码崩溃了，任何帮助是赞赏！

Now I'm not sure exactly why my code crashes, any help is appreciated!

谢谢！

推荐答案

printList ，你还必须检查 head == NULL ，否则你正在接受指向 NULL的指针的成员。以下应该工作。

Within printList, you have to also check for head == NULL, otherwise you are acessing members of a pointer pointing to NULL. The following should work.

    void printList()
    {
        node *temp = head;
        while(temp != NULL) // don't access ->next
        {
            cout << temp->data << endl;
            temp = temp->next;
        }
    }

在 printReverse / code>我真的不明白为什么你花一半的元素打印和打印两个元素在每次迭代。但是，你真的不需要一个for循环。你可以简单地停止， temp == head 在你的循环后，从那时你只是打印了头。

In printReverse() I really can't understand why you take half of the counts of the elements to print and print two elements in every iteration. However, you really don't need a for-loop here. You can simply stop as soon as temp == head after your loop, since then you just printed the head. And only print one element, the one whose next pointer points to the previously printed element.

另一个递归尝试解决问题的方法如下：

Another, recursive, attempt to solve the problem looks like this:

    void printReverse()
    {
        printReverseRecursive(head);
    }
    void printReverseRecursive(node *n)
    {
        if(n) {
            printReverseRecursive(n->next);
            cout << n->data << endl;
        }
    }

这篇关于在C ++中以相反的顺序打印我的链表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！