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");
}
答案 0 :(得分：6)
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;
}
}
答案 1 :(得分：1)
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 ;
}
答案 2 :(得分：1)
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;
}
}
答案 3 :(得分：0)
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。
答案 4 :(得分：0)
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++;
}
答案 5 :(得分：0)
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);
}
}
答案 6 :(得分：0)
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
答案 7 :(得分：0)
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。
答案 8 :(得分：0)
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;
}