删除双向循环链表中节点的正确方法

Question

The purpose of this code is to manage insertion and deletion and visualisation. I just want to know if I'm doing everything correctly, let me know if there are more possible ways to do this. This is my first attempt, i didn't follow any tutorial.

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



typedef struct Node {
  int n;
  struct Node *next;
  struct Node *prev;

}TNode;
typedef TNode* Node;



void NewNode(Node *pp, int n)
{
  Node temp, last;

  temp = (Node)malloc(sizeof(struct Node));

  temp->n = n;
  temp->next = temp;
  temp->prev = temp;

  if(*pp != NULL)
    {
      last = (*pp)->prev;
      temp->next = (*pp);
      temp->prev = last;
      last->next = (*pp)->prev = temp;
    }

  *pp = temp;

}

void ViewList(Node head)
{
  if(head == NULL)
  {
    return;
  }
  Node node = head->prev;
 do
  {
    printf("Curr: %d\n", node->n);
    node = node->prev;
  }while(node != head->prev);
}

void ReadData(Node * head, int * n)
{
  printf("\nInsert a number:");
  scanf("%d", n);
  NewNode(head, *n);
}

Node SearchNode(Node head)
{
  int d;
  printf("\nElement to Delete:");
  scanf("%d", &d);

  while(head != NULL)
    {
      if(head->n == d)
        {
          return head;
        }
      head = head->next;
    }
  printf("\nNo Element [%d] Found", d);
  return NULL;
}

void Delete(Node * head)
{
  Node del = SearchNode(*head);

       if(*head == NULL || del == NULL)
        {
          return;
        }
      if(*head == del && del->next == *head)
      {
        *head = NULL;
        free(del);
        return;
      }
      if(*head == del)
      {
        *head = del->next;
        del->prev->next = *head;
        (*head)->prev = del->prev;
        free(del);
        return;
      }
      if((*head)->prev == del)
        {
          (*head)->prev = del->prev;
          del->prev->next = *head;
          free(del);
          return;
        }
        del->next->prev = del->prev;
        del->prev->next = del->next;
        free(del);
}

int Menu()
{
  int c;

  printf("\n*** M E N U ***\n"
     "1 - New Node\n"
     "2 - View List\n"
     "3 - Delete\n"
     "0 - Exit\n"
     "\n>> ");
  scanf(" %d", &c);

  return c;
}

int main()
{
  int c,n;
  Node head = NULL;

  do {
    c = Menu();

    switch (c)
    {
      case 1: ReadData(&head, &n); break;
      case 2: ViewList(head); break;
      case 3: Delete(&head); break;
      default: c = 0;
    }

  } while (c != 0);

  return 0;
}

How can i test if this is a real circular doubly linked list and not a simple doubly linked list?

Answer 1

Your program works fine, the only real bug I detected is in SearchNode : if the element is not present in the list, you enter an infinite loop. Your list is obviously a circular list and therefore you need to check if you have got back to the head in the function, which means that the element you're searching for is not in the list:

Corrected version:

Node SearchNode(Node head)
{
  int d;
  printf("\nElement to Delete:");
  scanf("%d", &d);

  Node start = head;     // remember where we started

  while (head != NULL)
  {
    if (head->n == d)
    {
      return head;
    }
    head = head->next;

    if (head == start)   // if we are back to where we started
      break;             // the element hasn't been found and we stop the loop
  }

  printf("\nNo Element [%d] Found", d);
  return NULL;
}

There are also some design flaws, that don't prevent the program from working correctly:

One of them is this: The n variable is not used outside ReadData , so it's pointless to declare it in main and pass it's pointer to ReadData .

Corrected version:

void ReadData(Node * head)
{
  int n;  // declare n locally
  printf("\nInsert a number:");
  scanf("%d", &n);
  NewNode(head, n);
}

Call it like this: ReadData(&head) and remove int n; from main .