如何刪除添加到C++鏈接列表中的最後一個元素

Question

我第一次使用指針。我有一個程序，它將數字插入鏈表中，打印列表並從列表中刪除特定數字。除了當我嘗試刪除最後插入的號碼時，它工作。

Node.h

#ifndef Node_h 
#define Node_h 

#include <iostream> 
using namespace std; 

class Node 
{ 
public: 
    int data; 
    Node *next; 

public: 
Node(); 
}; 

#endif 

Node.cpp

#include "Node.h" 

Node::Node() 
{ 
} 

LinkedList.h

#ifndef LinkedList_h 
#define LinkedList_h 

#include "Node.h" 

class LinkedList 
{ 
    private: 
    Node *pL; 

public: 
    LinkedList(); 
    void insert(int nr1); 
    void deleteNr(int nr1); 
    void printL(); 
}; 

#endif 

LinkedList.cpp //該程序創建一個 「鏈接列表」的數字

#include "LinkedList.h" 

LinkedList::LinkedList() 
{ 
    pL = NULL; 
} 

void LinkedList::insert(int nr1) 
{ 
    Node *p = new Node; 
    p->data = nr1; 
    p->next = pL; 
    pL = p; 
} 

void LinkedList::deleteNr(int nr1) 
{ 
    Node *p = pL; 
    Node *p2 = pL; 
    while (p != NULL & p->data != nr1) 
    { 
     p2 = p; 
     p = p->next; 
    } 

    if (p != NULL) 
    { 
     p2->next = p->next; 
     delete p; 
    } 
} 

void LinkedList::printL() 
{ 
    Node *p = pL; 

    while (p != NULL) 
    { 
     cout << p->data << "-> "; 
     p = p->next; 
    } 
} 

的main.cpp

#include "LinkedList.h" 

int menu(); 

//////// main ///////// 
int main() 
{ 
    int choice1, nr1; 
    LinkedList lk1; 

    choice1 = menu(); 

    while (choice1 <= 3) 
    { 
     if (choice1 == 1) 
     { 
      cout << "Enter number." << endl; 
      cin >> nr1; 
      lk1.insert(nr1); 
     } 

     else if (choice1 == 2) 
     { 
      cout << "Enter number." << endl; 
      cin >> nr1; 
      lk1.deleteNr(nr1); 
     } 

     else if (choice1 == 3) 
     { 
      lk1.printL(); 
      cout << endl << endl; 
     } 

     else if (choice1 == 4) 
     { 
      cout << "Exit the program." << endl; 
      system("pause"); 
      exit(1); 
     } 

     choice1 = menu(); 
    } // end while loop 
} 

int menu() 
{ 
    int choice1; 

    cout << "1. Insert a number into the linked-list." << endl; 
    cout << "2. Delete a number from the linked-list." << endl; 
    cout << "3. Print the linked-list." << endl; 
    cout << "4. Exit the program." << endl; 
    cout << "Enter choice." << endl; 
    cin >> choice1; 

    return choice1; 
} 

Answer 1

您需要添加代碼來處理給定的輸入對應於列表中的第一項的情況。

void LinkedList::deleteNr(int nr1) 
{ 
    Node *p = pL; 

    if (p != NULL && p->data == nr1) 
    { 
     pL = p->next; 
     delete p; 
     return; 
    } 

    Node *p2 = pL; 
    while (p != NULL && p->data != nr1) 
    { 
     p2 = p; 
     p = p->next; 
    } 

    if (p != NULL) 
    { 
     p2->next = p->next; 
     delete p; 
    } 
} 

Answer 2

你的問題是，通常情況下，P2背後是一覽P一個節點，但如果第一個節點被刪除，那麼第一個while循環中刪除功能有0迭代和p2和p是相同的。頭部被刪除，但pL未更新。它只是指向未分配的內存。這可能會使它看起來像節點未被刪除，或者它可能導致分段錯誤和崩潰。無論哪種方式，這是錯誤的行爲。您需要確保檢查要刪除的節點是第一個節點並更新pL的情況。

嘗試這樣的事情

void LinkedList::deleteNr(int nr1) 
{ 
    Node *p = pL; 
    Node *p2 = pL; 
    if(p != NULL && nr1 == p->data) 
    { 
     pL = p->next; 
     delete p; 
     return; 
    } 

    while (p != NULL && p->data != nr1) 
    { 
     p2 = p; 
     p = p->next; 
    } 

    if (p != NULL) 
    { 
     p2->next = p->next; 
     delete p; 
    } 
} 

如果您希望能夠刪除的nr1所有實例鏈接列表，你需要添加另一個循環：

void LinkedList::deleteNr(int nr1) 
{ 
    Node *p = pL; 
    while(p != NULL && nr1 == p->data) 
    { 
     pL = p->next; 
     delete p; 
     p = pL; 
    } 
    Node *p2 = pL; 

    while (p != NULL) 
    { 
     p2 = p; 
     p = p->next; 
     if(nr1 == p->data) 
     { 
      p2->next = p->next; 
      delete p; 
     } 
    } 
} 

Answer 3

void LinkedList::deleteLast() 
{ 

Node *p = pL; 

if(p == NULL) 
    return; 
else if(p->next == NULL) { 
    p = NULL; 
} 
else { 
    while (p->next->next != NULL) 
    { 
    p = p->next; 
    } 

    p->next = NULL; 
} 
}