在链表C++中的某个位置插入节点

Posted 2023-02-19

【中文标题】在链表C++中的某个位置插入节点

【英文标题】：Insert node at a certain position in a linked list C++

【发布时间】：2014-02-11 15:09:41

【问题描述】：

我正在尝试在某个位置插入一个节点。在我的代码中，位置为 1 的数字仅被插入（基本上在链表的开头），它没有插入位置为 2 的任何数据。 temp2 有问题吗？当我运行程序时，它并没有指向我认为的任何东西。

我知道你们多么讨厌在这里被问到作业问题，但我只是不知道我的程序有什么问题。我只是这方面的初学者，我的老师没有很好地解释链表。

代码如下。

-我得到的输出是 8 7

-我希望它读取 8 6 7 5，其中 6 和 5 插入位置 2

/*
Insert node at a given positon in a linked list.
First element in the linked list is at position 0
*/

#include<stdlib.h>
#include<stdio.h>

struct Node

   int data;
   struct Node* next;
;

struct Node *head;

void Insert(int data, int n)

   Node* temp1 = new Node();
   temp1->data = data;
   temp1->next = NULL;
   if (n == 1)
    temp1->next = head;
    head = temp1;
    return;
   
   Node* temp2 = new Node();
   for (int i = 0; i < n-2; i++)// i feel like it doesn't even go through this loop
    temp2 = temp2->next;
   
   temp1->next = temp2->next;
   temp2->next = temp2;

void print()

    Node* temp = head;
    while(temp != NULL)
        printf("%d ", temp->data);
        temp = temp->next;
    
    printf("\n");

int main()

    head = NULL; //empty linked list
    Insert(7,1); //List: 7     
    Insert(5,2); //List: 7,5   
    Insert(8,1); //List: 8,7,5 
    Insert(6,2); //List: 8,6,7,5      
    print();
system("pause");
 

【问题讨论】：

在Insert函数中，首先你需要到你要插入的节点。，.​​.所以你需要Node * temp2 = head;而不是Node * temp2 = new Node();。而且，此时head可能为NULL..所以需要检查一下。

【参考方案1】：

代码修复

/* 在链表的给定位置插入节点。 链表中的第一个元素位于位置 0 */

#include<stdlib.h>
#include<stdio.h>

struct Node

   int data;
   struct Node* next;
;

struct Node *head;

void Insert(int data, int n)

   Node* temp1 = new Node();
   temp1->data = data;
   temp1->next = NULL;
   if (n == 1)
    temp1->next = head;
    head = temp1;
    return;
   
   Node* temp2 = head ;  //Here Only You need to assign what in head to 
                         //Pointer Variable temp2 and traverse
   for (int i = 0; i < n-2; i++)
    temp2 = temp2->next;
   
   temp1->next = temp2->next;
   temp2->next = temp1;        // linking Whats in Temp1 To Temp2 (next)

void print()

    Node* temp = head;
    while(temp != NULL)
        printf("%d ", temp->data);
        temp = temp->next;
    
    printf("\n");

int main()

    head = NULL; //empty linked list
    Insert(7,1); //List: 7     
    Insert(5,2); //List: 7,5   
    Insert(8,3); //List: 7,5,8 
    Insert(6,1); //List: 6,7,5,8   
    Insert(10,3);//List: 6,7,10,5,8 
    print();
system("pause");
 

【参考方案2】：

在最开始的位置插入一个元素。 列表为空时的 case-1。 case-2 当列表不为空时。

    #include<iostream>

using namespace std;

struct Node
int data;
Node* next; //link == head =stored the address of the next node
;

Node* head;  //pointer to Head node with empty list

void Insert(int y);
void print();

int main()
    head = nullptr; //empty list
    int n,y;
    cout<<"how many number do you want to enter?"<<endl;
    cin>>n;
    for (int i=0;i<n;i++)
        cout<<"Enter the number "<<i+1<<endl;
        cin>>y;
        Insert(y);
        print();
    


void Insert(int y)
    Node* temp = new Node(); //create dynamic memory allocation
    temp->data = y;
    temp->next = head; // temp->next = null; when list is empty
    head = temp;


void print()
    Node* temp = head;
    cout<<"List is: "<<endl;
    while(temp!= nullptr)
        cout<<temp->data<<" ";
        temp = temp->next;
    
    cout<<endl;

【参考方案3】：

Node* insert_node_at_nth_pos(Node *head, int data, int position)
   
    /* current node */
    Node* cur = head;

    /* initialize new node to be inserted at given position */
    Node* nth = new Node;
    nth->data = data;
    nth->next = NULL;

    if(position == 0)
        /* insert new node at head */
        head = nth;
        head->next = cur;
        return head;
    else
        /* traverse list */
        int count = 0;            
        Node* pre = new Node;

        while(count != position)
            if(count == (position - 1))
                pre = cur;
            
            cur = cur->next;            
            count++;
        

        /* insert new node here */
        pre->next = nth;
        nth->next = cur;

        return head;
        

【参考方案4】：

试试这个功能。

节点对象的结构：

class Node

private:
    int data;
    Node *next;

public:
    Node(int);
    ~Node();
    void setData(int);
    int getData();
    void setNext(Node*);
    Node* getNext();
;

函数的实现：

返回状态值始终是一种上帝的做法，此处定义的常量用于调试/记录应用程序的使用情况。

//constants
static int const SUCCESS = 0;
static int const FAILURE = 1;
static int const NULL_OBJ = 2;
static int const POS_EXCEED = 3;

int addAt(int data, int pos)
    Node *tmp = new Node(data);
    if (tmp == NULL)
        //print for debugging only.
        cout << "Object not created. Out of memory maybe" << endl;
        return NULL_OBJ;
    
    if (pos == 0)
        // add at beginning
        tmp->setNext(this->head);
        this->head = tmp;
        return SUCCESS;
    else
        // add element in between or at end
        int counter = 1;
        Node* currentNode = this->head;
        while (counter < pos && currentNode->getNext() != NULL)
            currentNode= currentNode->getNext();
            counter++;
       
       tmp->setNext(currentNode->getNext());
       currentNode->setNext(tmp);
       return SUCCESS;
   
   cout << "Failed due to unknown reason.";
   return FAILURE;

这里的假设是，您将在验证输入（数据和位置）后调用该函数。虽然我们可以验证函数内部的参数，但这不是一个好习惯。

希望这会有所帮助。

【参考方案5】：

 void addToSpecific()
 
 int n;   
 int f=0;   //flag
 Node *temp=H;    //H-Head, T-Tail
 if(NULL!=H)  
 
    cout<<"Enter the Number"<<endl;
    cin>>n;
    while(NULL!=(temp->getNext()))
    
       if(n==(temp->getInfo()))
       
      f=1;
      break;
       
       temp=temp->getNext();
    
 
 if(NULL==H)
 
    Node *nn=new Node();
    nn->setInfo();
    nn->setNext(NULL);
    T=H=nn;
 
 else if(0==f)
 
    Node *nn=new Node();
    nn->setInfo();
    nn->setNext(NULL);
    T->setNext(nn);
    T=nn;
 
 else if(1==f)
 
    Node *nn=new Node();
    nn->setInfo();
    nn->setNext(NULL);
    nn->setNext((temp->getNext()));
    temp->setNext(nn);
 
 

【讨论】：

虽然此代码可能会回答问题，但提供有关它如何和/或为什么解决问题的额外上下文将提高​​答案的长期价值。

【参考方案6】：

只要有这样的东西，你就可以遍历到给定的位置，然后插入：

void addNodeAtPos(int data, int pos)

  Node* prev = new Node();
  Node* curr = new Node();
  Node* newNode = new Node();
  newNode->data = data;

  int tempPos = 0;   // Traverses through the list

  curr = head;      // Initialize current to head;
  if(head != NULL)
  
    while(curr->next != NULL && tempPos != pos)
    
        prev = curr;
        curr = curr->next;
        tempPos++;
    
    if(pos==0)
    
       cout << "Adding at Head! " << endl;
       // Call function to addNode from head;
    
    else if(curr->next == NULL && pos == tempPos+1)
    
      cout << "Adding at Tail! " << endl;
      // Call function to addNode at tail;
    
    else if(pos > tempPos+1)
      cout << " Position is out of bounds " << endl;
     //Position not valid

    else
    
        prev->next = newNode;
        newNode->next = curr;
        cout << "Node added at position: " << pos << endl;
    
 
 else
 
    head = newNode;
    newNode->next=NULL;
    cout << "Added at head as list is empty! " << endl;
 

【参考方案7】：

Node* InsertNth(int data, int position)

  struct Node *n=new struct Node;
  n->data=data;  
  if(position==0)
  // this will also cover insertion at head (if there is no problem with the input)

      n->next=head;
      head=n;
  

  else
  
      struct Node *c=new struct Node;
      int count=1;
      c=head;
      while(count!=position)
      
          c=c->next;
          count++;
      
      n->next=c->next;
      c->next=n;

  
    return ;

【参考方案8】：

我和你一样在插入过程中遇到了一些问题，所以这是我解决问题的代码：

void add_by_position(int data, int pos)
  
        link *node = new link;
        link *linker = head;

        node->data = data;
        for (int i = 0; i < pos; i++)
            linker = linker->next;
        
        node->next = linker;
        linker = head;
        for (int i = 0; i < pos - 1; i++)
            linker = linker->next;
        
        linker->next = node;
        boundaries++;
    

【参考方案9】：

要在特定位置插入k，您需要遍历列表直到位置k-1，然后进行插入。

[您不需要像在代码中那样创建一个新节点来遍历该位置]您应该从头节点开始遍历。