链表中的搜索函数 - C++

Search Function in Linked List - C++

提问by Ryan

My aim is to produce a function that searches for a number already in the list and print that it has been found.
My initial idea was to follow my remove function which searches through the list until it finds a number (to then delete).
This seemed the logical way to code the search function. If this isn't correct how would I modify it to search through my list and display that a number has been found?
I have nodes *head, *current and *temp as well as node pointer nextand numberas the data type in a class on a .h file.
Thank you.
NOTE- I used my remove() function under the search() function.

我的目标是生成一个函数来搜索列表中已有的数字并打印它已找到。
我最初的想法是按照我的删除功能搜索列表，直到找到一个数字（然后删除）。
这似乎是对搜索功能进行编码的合乎逻辑的方式。如果这不正确，我将如何修改它以搜索我的列表并显示已找到一个数字？
我有节点 *head、*current 和 *temp 以及节点指针next和number作为 .h 文件上类中的数据类型。
谢谢你。
注意- 我在 search() 函数下使用了我的 remove() 函数。

#include <iostream>                                                 
#include <string>                                                   
#include <fstream>                                                  
#include "LinkedList.h"

using namespace SDI;

int main()
{
    LinkedList menu;

    menu.insert(5);                     
    menu.insert(4);
    menu.insert(2);
    menu.insert(3);
    menu.insert(8);
    menu.remove(4);
    menu.reverse();
    menu.display();
    menu.search(2);
    system("pause");

};


LinkedList::LinkedList()            
{
    head = NULL;
    current = NULL;
    temp = NULL;
};


LinkedList::~LinkedList()           
{

};


void LinkedList::insert(int add)                                    //insert function, data is stored in add from function body
{
    Node* newnode = new Node;                                       //definition of add node, make new node and make node* point to it
    newnode->next = NULL;                                           //point and set up to last node in the list (nothing)
    newnode->number = add;                                          //adds data to list

    if (head != NULL)                                               //if head is pointing to object then we have list
    {
        current = head;                                             //make current pointer point to head
        while (current->next != NULL)                               //check to see if end at list, is it the last node?
        {
            current = current->next;                                //advances current pointer to end of list
        }
        current->next = newnode;                                    //adds new node next to value already stored
    }
    else
    {
        head = newnode;                                             //if we don't have element in list
    }
};


void LinkedList::remove(int remove)                                 //remove function, data is stored in remove from function body
{
    Node* remove1 = NULL;                                           //searches through for same value in remove and deletes
    temp = head;
    current = head;
    while (current != NULL && current->number != remove)            //check if current node is one we want to delete...if not advance current pointer to next one
    {
        temp = current;                                             //keep temp pointer one step behind
        current = current->next;                                    //advance to next node, traverse list till at the end
    }
    if (current == NULL)                                            //pass through whole list and value not found
    {
        std::cout << "N/A\n";
        delete remove1;                                             //removes spare number floating around in memory
    }
    else
    {
        remove1 = current;                                          //pointing to value we want to delete
        current = current->next;                                    //advances current pointer to next node
        temp->next = current;                                       //stops hole that occurs in list, patches this up
        if (remove1 == head)                                        //if pointer is pointing to front of list
        {
            head = head->next;                                      //advance the head to next
            temp = NULL;
        }

        delete remove1;
    }
};


void LinkedList::search(int searchNum)
{
    Node* searchnumber = nullptr;
    temp = head;
    current = head;

    while (current != NULL && current->number != searchNum)
    {
        temp = current;
        current = current->next;
    }
    if (current != NULL)
    {
        searchnumber = current;
        current = current->next;
        std::cout << "-" << searchnumber << " Found";
    }
    else
    {
        std::cout << "N/A";
    }
};


void LinkedList::display()
{
    current = head;                                                 //point to start of list

    while (current != NULL)                                         //while it points to something in list
    {
        std::cout << current->number;                               //display list starting from start
        current = current->next;                                    //advance to next pointer
    }
};


void LinkedList::reverse()
{
    Node *new_head = nullptr;                                       //create new head as we want it to start from last element

    for (current = head; current;)                                  //same as display, ask it to go through list from head then outside loop assign to new head and switch sides
    {
        temp = current;                                             //keep temp pointer one step behind
        current = current->next;                                    //goes through each element in the list
        temp->next = new_head;                                      //scrolls through backwards from new head
        new_head = temp;                                            
    }

    head = new_head;                                                //assign head to new head
};

回答by Mohammad Ali Akber

Your search algorithm seems to be wrong. Change it to :

您的搜索算法似乎是错误的。将其更改为：

if (current != NULL) // (current == NULL) is wrong because it means the value wasn't found
{
    searchnumber = current;
    current = current->next;
    std::cout << "-" << searchnumber->number << " Found"; // here searchnumber is the node's address. You need to print its value, so use searchnumber->number
}

And you don't need to remove nodes till you find the desired value.
You can just use your search algorithm to find if a number already in the list. If that's what you want.

在找到所需值之前，您无需删除节点。
您可以使用搜索算法来查找列表中是否已有数字。如果那是你想要的。

回答by Andrii

While a list is unordered a comparison of search algorithms doesn't have any sense. Simply iterate over all nodes one by one and apply match criteria.

虽然列表是无序的，但比较搜索算法没有任何意义。只需一一迭代所有节点并应用匹配标准。