Question

这是我到目前为止所做的，但是它没有在测试文件中工作。基本上会跳到其他地方if(cnode == preposition)

void LinkedList::Delete(Node *PrePosition) {

    Node *cnode = head;
    Node *pnode = NULL;
    while (cnode != NULL) {
        if (cnode->value != NULL) {
            if (pnode == NULL) {
                //if there is not previous node
                head = cnode->next;
            }
            else if(cnode == PrePosition){
                //if there is previous node
                cout << endl << "Deleting: " << cnode << endl;
                pnode->next = cnode->next;
            }
        }
        else {
            //dont delete
            pnode = cnode;
        }
        cnode = cnode->next;

    }

}

Answer 1

1：从上一个节点中取出指针，并在要删除的节点之后将其指向下一个节点

2：删除从前一个节点到当前节点的指针

3：删除从下一个节点到当前节点的指针（如果它是双向链表）

Answer 2

单个链表中的三个删除案例：

删除第一个节点

void delete_first()
{
    node *temp=new node;
    temp=head;
    head=head->next;
    delete temp;
}

删除最后一个节点

void delete_last()
{
    node *current = new node;
    node *previous = new node;
    current=head;
    while(current->next != NULL)
    {
      previous = current;
      current = current->next;  
    }
    tail = previous; // if you have a Node* tail member in your LinkedList
    previous->next = NULL;
    delete current;
}

删除特定位置（您的情况）

void LinkedList::delete_position(int pos)
{
    node *current=new node;
    node *previous=new node;
    current=head;
    for(int i=1; i < pos; i++) //or i = 0; i < pos-1
    {
       previous=current;
       current=current->next;
    }
    previous->next=current->next;
    delete current;
}

^^来自codementor ^^

但是如果您的函数签名打算delete_node(Node* nodeToDelete) [PrePosition在这种情况下不是一个好名字]并且您希望删除传递给函数的节点而不知道它在列表中的位置，我们可以像这样修改delete_position（）：

void LinkedList::delete_node(Node* nodeToDelete)
{
    node *current= head;
    node *previous= nullptr;

    if (head == nodeToDelete){
        head = nodeToDelete->next;
        delete nodeToDelete;
        return
    }//else
    while(current != nodeToDelete)
    {
        previous = current;
        current = current->next
    }
    previous->next = current->next;
    delete nodeToDelete;
}

同样在您的原始代码中，如果它正在跳过您提到的行，当cnode中包含非空值时，pnode始终为null。

Answer 3

这是完整的代码

    class SportShoe  {
    private:
        struct nodeSport {
            int ShoeID;
            char BrandShoe[SIZE]; 
            char TypeShoe[SIZE];
            char ColourShoe[SIZE];
            int SizeShoe;
            float PriceShoe; 
            nodeSport *last;
            };
            nodeSport *first = NULL; 

    public:
        int MenuSportShoe();
        void AddSportShoe();
        void DisplaySportShoe();
        void DeleteSportShoe();
        static void ExitSportShoe();
    };

   int SportShoe::MenuSportShoe() {
     int OptionSportShoe = 0;

    cout << endl;
    cout << "Please select from the menu:" << endl;
    cout << ":: 1 :: Add item to shoe list" << endl;
    cout << ":: 2 :: Display shoes list" << endl;
    cout << ":: 3 :: Delete item from the list" << endl;
    cout << ":: 4 :: Back" << endl;
    cout << "=>> ";
    cin >> OptionSportShoe;

    while (OptionSportShoe == 1){
        AddSportShoe();
    }

    while (OptionSportShoe == 2){
        DisplaySportShoe();
      }

    while (OptionSportShoe == 3){
        DeleteSportShoe();
    }

    while (OptionSportShoe == 4){
        ExitSportShoe();
    }

    return 0;
  }

void SportShoe::AddSportShoe() {    
    nodeSport *tempShoe1, *tempShoe2; 

    tempShoe1 = new nodeSport;
    cout << "Please enter the Shoe ID : (eg. 43210) " << endl;
    cout << "=>> ";
    cin >> tempShoe1->ShoeID;

    cout << "Please enter the Shoe Brand: (eg. Adidas) " << endl;
    cout << "=>> ";
    cin.sync();
    cin.getline(tempShoe1->BrandShoe,SIZE);

    cout << "Please enter the Shoe Type : (eg. Running) " << endl;
    cout << "=>> ";
    cin.sync();
    cin.getline(tempShoe1->TypeShoe,SIZE);

    cout << "What is the Shoe Colour : (eg. Grey) " << endl;
    cout << "=>> ";
    cin.sync();
    cin.getline(tempShoe1->ColourShoe,SIZE);

    cout << "Please enter Shoe Size : (eg. 9) " << endl;
    cout << "=>> ";
    cin >> tempShoe1->SizeShoe; 

    cout << "Please enter the price of the Shoe : (eg. RM123.45) " << endl;
    cout << "=>> RM ";
    cin >> tempShoe1->PriceShoe;


    tempShoe1->last = NULL;  


    if (first == NULL)  
        first = tempShoe1;
    else  
    {
        tempShoe2 = first; 
        while (tempShoe2->last != NULL) 
            tempShoe2 = tempShoe2->last;

        tempShoe2->last = tempShoe1;
    }

    system("PAUSE");
    MenuSportShoe();
  }

void SportShoe::DisplaySportShoe() {
    nodeSport *tempShoe1;
    tempShoe1 = first;

    while(tempShoe1){
        cout << "ID : " << tempShoe1->ShoeID << endl;
        cout << "Brand : " << tempShoe1->BrandShoe << endl;
        cout << "Type : " << tempShoe1->TypeShoe << endl;
        cout << "Colour : " << tempShoe1->ColourShoe << endl;
        cout << "Size : " << tempShoe1->SizeShoe << endl;
        cout << "Price : " << tempShoe1->PriceShoe << endl;
        cout << endl;
        tempShoe1 = tempShoe1->last;
    }

    system("PAUSE");
    MenuSportShoe();
  }

 void SportShoe::DeleteSportShoe(){
    nodeSport *tempShoe1, *tempShoe2; 
    int DataShoe;
    tempShoe2 = tempShoe1 = first;

    if(tempShoe1 == NULL)
    {
        cout << "\nList is empty!" << endl;
        system("PAUSE");
        MenuSportShoe();
    }

    while(tempShoe1 != NULL)
    {
        cout << "\nEnter the Shoes ID to be deleted: (eg. 123) ";
        cin >> DataShoe;

        tempShoe2 = tempShoe1;
        tempShoe1 = tempShoe1->last;

        if(DataShoe == tempShoe1-> ShoeID){ 
            if(tempShoe1 == first)  {
                first = first->last;
                cout << "\nData deleted ";
            }

            else{
                tempShoe2->last = tempShoe1->last;
                if(tempShoe1->last == NULL){
                    tempShoe2 = tempShoe2;
                }
                cout << "\nData deleted ";
            }

            delete(tempShoe1);

            system("PAUSE");
            MenuSportShoe();
        }

        else{
            cout << "\nRecord not Found!!!" << endl;
            system("PAUSE");
            MenuSportShoe();
        }
    }
  }

  void SportShoe::ExitSportShoe(){
    int sepatu;

    cout << endl;
    cout << "Please choose the option below."<<endl;
    cout << ":: 1 :: Sport Shoe." << endl;
    cout << ":: 2 :: Ladies High Heel." << endl;
    cout << ":: 3 :: Exit" << endl;
    cout << "=>> ";
    cin >> sepatu;

    while(sepatu == 1){
        SportShoe listShoe;
        listShoe.MenuSportShoe();
    }

    while(sepatu == 2){
        HighHeel listShoe;
        listShoe.MenuHighHeel();
    }

    while(sepatu == 3){
        cout << "Thank you. Till we meet again."<< endl;
        exit(1);
    }

  }

  main() {

    cout << "Hello! Welcome to MySepatu Online Shop administrator."<< endl;
    cout << endl;

    SportShoe::ExitSportShoe();
    HighHeel::ExitHighHeel();

    return 0;
  }

Answer 4

public class linkedList {
    int count = 0;
    
    class Node {
        int element;
        Node next;
        Node(int element) {
            this.element = element;
        }
    } 
    
    Node head = null;
    Node tail = null;
    
    public void addNode(int Object) {
        Node newNode = new Node(Object);
        if (head == null) {
            head = tail = newNode;
        } else {
            tail.next = newNode;
            tail = newNode;
        }
    }
    
    public void Display() {
        Node current = head;
        while (current!=null) {
            System.out.println(current.element);
            count ++;
            current = current.next;
        }
    }
    
    public void Length() {
        System.out.println(count);   
    }
    
    public void Remove(int node) {
        Node curr = head;
        while (curr!=null) { // looping the nodes
            if (curr.element == node ) {
                curr.element = curr.next.element;
                curr = curr.next;
                // To fix the Duplicates
                while (curr!= tail) {
                    curr.element = curr.next.element;
                    curr = curr.next;
                }
                RemoveEnd();
                break;
            }
            curr = curr.next;
        }
    }
    
    public void RemoveEnd() {
        Node current3 = head;
        while (current3.next != tail) {
            current3 = current3.next;
        }
        tail = current3;
        tail.next = null;
    }
}

如何删除链表中的节点？

4 个答案: