如何创建可以创建多个链表的单个函数

Question

如代码所示，我必须使用2个类似的函数来创建2个链表。难道没有一种方法可以用一个函数创建任意多的列表吗？我尝试使用struct Node **p和struct Node *p作为函数的参数，但没有成功有人能帮我用这个功能创建多个链表吗我想创建一个append函数，而不是一个insert函数，它也要求位置。

#include <iostream>
using namespace std;
struct  Node
{
int data = 10 ;
struct Node *next;
} *first , *second , *third;
void Display(struct Node *p)
{
while (p)
{
cout<<p->data<<" ";
p = p->next ;
}
cout<<"n";
}
void Append_1(int elem)
{
Node* t , *last;
t = new Node;
t->data = elem;
t->next = NULL;
if(first == 0)
first = last = t;
else 
{
last->next = t;
last = t;
}
}
void Append_2(int elem)
{
Node* t , *last;
t = new Node;
t->data = elem;
t->next = NULL;
if(second == 0)
second = last = t;
else 
{
last->next = t;
last = t;
}
}
//void SortMerge(struct Node *p , struct Node *q);
int main()
{
Append_1(3);
Append_1(7);
Display(first);
Append_2(10);
Append_2(14);
Append_2(21);
Display(second);
//SortMerge(first , second);
Display(third);

return 0;
}

Answer 1

您可以创建这样的类：

struct Node{
int data;
Node* next;
Node* previous;
};
class Graph{
public:
Graph(int = 0);
~Graph();
void display_left_right();
void display_right_left();
void append(int);
void append_at_pos(int,int);
void prepend(int);
int get_num_elt();
int get_data_at_pos(int);
private:
Node* head;
Node* tail;
int num_elt=0;
};
Graph::Graph(int first_data){
head = new Node;
head->next = NULL;
head->previous = NULL;
head->data = first_data;
tail = head;
num_elt++;
}
Graph::~Graph(){
Node* main_traverser = head;
while(main_traverser){
main_traverser = head->next;
delete head;
head = main_traverser;
}
std::cout <<"Graph deleted!" << std::endl;
}
void Graph::display_left_right(){
Node* traverser = head;
while(traverser != NULL){
std::cout << traverser->data << " ";
traverser = traverser->next;
}
std::cout << std::endl;
}
void Graph::display_right_left(){
Node* traverser = tail;
while(traverser != NULL){
std::cout << traverser->data << " ";
traverser = traverser->previous;
}
std::cout << std::endl;
}
void Graph::append(int new_data){
Node* add = new Node;
add->data = new_data;
add->next = NULL;
add->previous = tail;
tail->next = add;
tail = add;
num_elt++;
}
void Graph::append_at_pos(int pos, int new_data){
if(pos > num_elt+1 || pos<=0){std::cout << "Wrong position!" << std::endl; return;}
if(pos==1){
prepend(new_data);
return;
}
if(pos==num_elt+1){
append(new_data);
return;
}
Node* add = new Node;
Node* traverser = head;
add->data = new_data;
for(int i=0; i<pos-2; i++){
traverser = traverser->next;
}
add->next = traverser->next;
add->previous = traverser;
traverser->next->previous = add;
traverser->next = add;
}
void Graph::prepend(int new_data){
Node* add = new Node;
add->next = head;
add->previous = NULL;
add->data = new_data;
head->previous = add;
head = add;
num_elt++;
}
int Graph::get_num_elt(){
return num_elt;
}
int Graph::get_data_at_pos(int pos){
Node* traverser = head;
if(pos <=0 || pos> num_elt){std::cout << "Wrong position!" << std::endl; return 0;}
for(int i=0; i<pos-1; i++){
traverser = traverser->next;
}
return traverser->data;
}
main(){
Graph a(2);
a.append(3);
a.append(4);
a.prepend(1);
a.display_left_right();
a.append_at_pos(1,6);
a.display_left_right();
std::cout << "data at 1: " << a.get_data_at_pos(1) << std::endl;
}

Answer 2

当你说"创建多个链表；我认为您的意思是为链表创建节点，您有2个附加函数。我认为你有这两个函数的原因是你不知道从哪里开始遍历你的链表。出于这个原因，我认为在您的主函数中，您应该声明链表的头，一个单独的节点是开始。将它的data和next设置为null，然后将head值传递给一个函数，这样它就可以从head开始遍历。这里有一个通用的附加函数，它在末尾添加一个节点，其中参数是对头节点的引用，以及新节点的值：

void append(Node ** head, int new_data)
{
Node * select_node = * head;
// select node is set to the head node, and will traverse until it is at the end
while (select_node -> next != NULL)
{
// select node is set to the next node until it is NULL (end of linked list)
select_node = select_node -> next;
}
// now that select node is the last node, we need to make it's next value a node
// and that node should be a new node (allocated in heap) with the value of the input value
//and the next value be NULL (because it's the end of the linked list)
Node * next_node = new Node();
next_node -> data = new_data;
next_node -> next = NULL;
select_node -> next = next_node;
}