C语言链表实现集合的创建，并集，交集，输出

链表实现集合的创建

#include<stdio.h>
#include<stdlib.h>
typedef struct node
{
    int data;
    struct node* next;
}Node;

void add_node(Node** head,int data)
{
    Node* new_node = (Node*)malloc(sizeof(Node)); 
    if(new_node == NULL)
    {
        printf("Error! Memory is not allocated.");
        exit(0);
    }
    new_node->data = data;  
    new_node->next = *head;     
    *head = new_node;        
}

void display(Node* head)
{
    Node* current = head;
    while(current != NULL)
    {
        printf("%d ",current->data);
        current = current->next;
    }
    printf("\n");
}

void set_union(Node* head1,Node* head2)
{
    Node* current1 = head1;
    Node* current2 = head2;
    Node* result = NULL;

    while(current1 != NULL && current2 != NULL)
    {
        if(current1->data < current2->data)
        {
            add_node(&result,current1->data);
            current1 = current1->next;
        }
        else if(current1->data > current2->data)
        {
            add_node(&result,current2->data);
            current2 = current2->next;
        }
        else
        {
            add_node(&result,current1->data);
            current1 = current1->next;
            current2 = current2->next;
        }
    }
    while(current1 != NULL)
    {
        add_node(&result,current1->data);
        current1 = current1->next;
    }
    while(current2 != NULL)
    {
        add_node(&result,current2->data);
        current2 = current2->next;
    }
    display(result);
}

void set_intersection(Node* head1,Node* head2)
{
    Node* current1 = head1;
    Node* current2 = head2;
    Node* result = NULL;

    while(current1 != NULL && current2 != NULL)
    {
        if(current1->data < current2->data)
        {
            current1 = current1->next;
        }
        else if(current1->data > current2->data)
        {
            current2 = current2->next;
        }
        else
        {
            add_node(&result,current1->data);
            current1 = current1->next;
            current2 = current2->next;
        }
    }
    display(result);
}

int main()
{
    Node* set1 = NULL;
    Node* set2 = NULL;

    add_node(&set1,5);
    add_node(&set1,7);
    add_node(&set1,9);
    add_node(&set1,11);
    printf("Set1: ");
    display(set1);

    add_node(&set2,4);
    add_node(&set2,7);
    add_node(&set2,11);
    add_node(&set2,13);
    printf("Set2: ");
    display(set2);

    printf("Union of Set1 and Set2: ");
    set_union(set1,set2);

    printf("Intersection of Set1 and Set2: ");
    set_intersection(set1,set2);

    return 0;
}

输出结果：

Set1: 11 9 7 5 
Set2: 13 11 7 4 
Union of Set1 and Set2: 13 11 9 7 5 4 
Intersection of Set1 and Set2: 11 7