c++ - 链表反向函数 OOP C++

Question

我正在为我的链表编写反向函数，但我不知道如何将 head 指向列表中新的第一个数字。在 main 中声明 head 为 ListNode* 并通过引用传递它是行不通的，因为我还有很多其他函数不想修改。还有其他方法吗？这是我的头文件:

// Specification file for the NumberList class
#ifndef NUMBERLIST_H
#define NUMBERLIST_H

class NumberList
{
private:
   // Declare a structure for the list
   struct ListNode
   {
      double value;           // The value in this node
      struct ListNode *next;  // To point to the next node
   };

   ListNode *head;            // List head pointer

public:
   // Constructor
   NumberList()
      { head = nullptr; }

   // Destructor
   ~NumberList();

   // Linked list operations
   void appendNode(double);
   void insertNode(double);
   void deleteNode(double);
   int searchList(double);
   void reverseList() const;
   void displayList() const;
};
#endif

这里是反向函数。放置 head = previousNode 将不起作用:

void NumberList::reverseList() const
{
    ListNode *currentNode;              // points to current node
    ListNode *previousNode = nullptr;   // stores address of previous node
    ListNode *nextNode = nullptr;       // stores the address of next node
    ListNode *tempNode = nullptr;       // 

    currentNode = head;

    while (currentNode != nullptr)
    {
        cout << "current value: " << currentNode->value << endl;
        nextNode = currentNode->next;
        currentNode->next = previousNode;
        previousNode = currentNode;
        currentNode = nextNode;
    }
    //head = previousNode;
}

*编辑:在 main() 中添加

#include <iostream>
#include <fstream>
using namespace std;
#include "NumberList.h"

int main()
{
    NumberList nList;
    ifstream inFile;
    double number;
    double searchedNum;

    inFile.open("input.txt");

    if (!inFile)
    {
        cout << "File could not be opened!" << endl;
    }
    else
    {
        while (inFile >> number)
        {
            nList.appendNode(number);
        }

        cout << "The list after file reading:" << endl;
        nList.displayList();

        cout << "What number would you like to search for? ";
        cin >> searchedNum;
        cout << "Position: " << nList.searchList(searchedNum) << endl;

        cout << "The list after reversing:" << endl;
        nList.reverseList();
        nList.displayList();
    }



    return 0;
}

Answer 1

这是一个迭代版本(我用 //Added * 标记了更改):

void NumberList::reverseList()
{
    ListNode *currentNode;              // points to current node
    ListNode *previousNode = nullptr;   // stores address of previous node
    ListNode *nextNode = nullptr;       // stores the address of next node

    currentNode = head;
    ListNode* temp; // Added*
    while (currentNode != nullptr)
    {
        temp = currentNode->next; // Added *
        nextNode = currentNode->next;
        currentNode->next = previousNode;
        previousNode = currentNode;
        currentNode = temp; // Added *
    }
}