在链接列表C ++中的某个位置插入节点

Question

我正在尝试在某个位置插入一个节点。在我的代码中，位置1的数字仅被插入（基本上在链表的开头）并且它没有插入位置2的任何数据。是否有一些错误的temp2？当我运行程序时，它并没有指向我想的任何东西。

我知道你们有多讨厌这里被问到的家庭作业问题，但我不知道我的课程有什么问题。我只是一个初学者而且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");
} 

Answer 1

只需要这样的东西，你遍历到给定的位置，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;
 }
}

Answer 2

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 ;
}

Answer 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;
    }    
}

Answer 4

要插入特定位置k，您需要遍历列表直到位置k-1，然后执行插入操作。

[您不需要像在代码中那样创建一个新节点来遍历该位置]您应该从头节点遍历。

Answer 5

我在插入过程中遇到了一些问题，所以这里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++;
    }

Answer 6

 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);
 }
 }

Answer 7

以下链接可以帮助您在不同类型的链接列表中的特定位置插入节点：

单一链接列表：Click Here

循环链接列表：Click Here

双重链接列表：Click Here

循环双重链接列表：Click here

Answer 8

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();
};

功能的实现：

返回状态值始终是上帝的做法，此处定义的常量用于调试/记录应用程序的使用。

//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;
}

此处的假设是，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。

希望这有帮助。

Answer 9

在最开始的位置插入元素。 列表为空时的情况1。 case-2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;
}