在列表末尾插入一个整数并在第n个位置删除

Question

因此，在我的链表列表程序中，我想要它做的是要求用户输入多少个数字，然后输入数字，然后添加这些数字数字在列表的末尾。然后，它将打印列表。之后，用户将选择列表中元素的位置来删除，列表将再次被打印。

#include <iostream> 
using namespace std; 

struct Node{ 

int data; 
    Node* link; 
}; 

Node* head; 

void Insert(int data){ //insert an integer at the end of the list 

    Node* temp = new Node(); 
    Node* temp2 = new Node(); 

    temp->data = data; 
    temp->link = NULL; 
    if(head = NULL){ 
    head = temp; 
    return; 
    } 
    temp2 = head; 
    while(temp2->link != NULL){ 
    temp2 = temp2->link; 
    } 

    temp2->link = temp; 

} 

void Delete(int n){ //delete an integer at nth position 

    Node* temp1 = new Node(); 
    temp1 = head; 
    if(n == 1){ //if the first node is to be deleted 
    head = temp1->link; //now head points to second node 
    delete temp1; //delete first node 
    return; 
    } 

    for(int i = 0; i < n-2; i++){ 
    temp1 = temp1->link; //temp1 points to (n-1)th node 
    } 
    Node* temp2 = temp1->link; //temp2 points to nth node 
    temp1->link = temp2->link; // pointing to (n+1)th node 
    delete temp2; //deleting nth node 

} 

void Print(){ //print out the list 
    Node* printNode = head; 
    cout << "List: "; 

    while(printNode != NULL){ 
    cout << printNode->data; 
    cout << " "; 
    printNode = printNode->link; 
    } 

    cout << "\n"; 
} 

int main(){ 

    int x, count, n; 

    head = NULL; //start with an empty list 

    cout << "How many numbers? " << endl; 
    cin >> count; 

    for(int i = 0; i < count; i++){ 

    cout << "Enter number: "; 
    cin >> x; 
    Insert(x); 
    } 

    Print(); 

    cout << "Enter position to delete: "; 
    cin >> n; 

    Delete(n); 
    Print(); 

    return 0; 

} 

接受第一个数字后，程序停止工作。我可以知道我在哪里做错了代码，我能做些什么来使代码更有效率？提前致谢。

Answer 1

您可能需要重新考虑您的插入功能。代码崩溃的部分是在while循环插入期间。如果你想temp2保存数据，那么你需要动态地为它分配空间。但是，您只是将其用作位置指示器（以遍历列表） - 那么为什么您需要分配空间以指向头部或列表中的任何其他节点位置？

这里是我怎么会插入到表（在课程后面）：

void Insert(int data){ //insert an integer at the end of the list 

    Node* temp = new Node(); 

    // This is to ensure that temp was created -> Also called defensive programming. 
    if (!temp) 
    { 
     cout << "We did not have enough space alloted to dynamically allocate a node!" << endl; 
     exit(1); 
    } 

    temp->data = data; // Bad nominclature for program; Don't use the same name twice. 
    temp->link = NULL; 

    if (head == NULL) 
    { 
     head = temp; 
    } 
    else 
    { 
     // This is to help traverse the linked list without having to 
     // manipulate the position of what head points to. 
     Node *Pos_Indicator = head; 

     while (Pos_Indicator->link != NULL) 
     { 
      Pos_Indicator = Pos_Indicator->link; 
     } 

     // We are at the end of the list, it is now safe to add. 
     Pos_Indicator->link = temp; 

     // Should probably have a check here for whether it was successful or not. 
    } 

} 

我能够编译并没有其他问题，运行您的代码完成。让我知道如果这有帮助！

编辑：。或者你知道(head = NULL)到(head == NULL)也工作:(

Answer 2

对我而言

大捂脸，只是一个小错误代码已得到纠正

if(head == NULL){ 
    head = temp; 
    return; 
}