假设一个单循环链表，其结点含有三个域pre、data、link。其中data为数据域；pre为指针域，它的值为空指针（NIL）；link为指针域，它指向后继结点。请设计算法，将此表改成双向循环链表。 使用C语言写出完整代码

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

typedef struct Node {
    int data;
    struct Node* pre;
    struct Node* link;
} Node;

void AddNode(Node** head, int data) {
    Node* new_node = (Node*)malloc(sizeof(Node));
    new_node->data = data;
    new_node->pre = NULL;
    new_node->link = NULL;
    if (*head == NULL) {
        *head = new_node;
        new_node->link = *head;
        new_node->pre = *head;
    } else {
        Node* temp = *head;
        while (temp->link != *head) {
            temp = temp->link;
        }
        temp->link = new_node;
        new_node->pre = temp;
        new_node->link = *head;
        (*head)->pre = new_node;
    }
}

void ConvertToDoubleLinkedList(Node** head) {
    Node* temp = *head;
    Node* prev = NULL;
    while (temp != NULL) {
        temp->pre = prev;
        prev = temp;
        temp = temp->link;
    }
    temp = *head;
    while (temp->link != *head) {
        temp = temp->link;
    }
    temp->link = *head;
    (*head)->pre = temp;
}

void Display(Node* head) {
    if (head == NULL) {
        printf("List is empty!\n");
        return;
    }
    Node* temp = head;
    do {
        printf("%d ", temp->data);
        temp = temp->link;
    } while (temp != head);
    printf("\n");
}

int main() {
    Node* head = NULL;
    AddNode(&head, 1);
    AddNode(&head, 2);
    AddNode(&head, 3);
    AddNode(&head, 4);
    AddNode(&head, 5);
    Display(head);
    ConvertToDoubleLinkedList(&head);
    Display(head);
    return 0;
}