如何删除C中双向链表中的备用节点？

3 回答

• 1

  您从未将“head”分配给列表中的第一个创建元素 . 因此它始终为null . 试试这个：

  if (head == NULL) 
  {
    p->prev = head;
    head = p;      
    q = p;
  }
  

  在你的delete_alt中，你需要这样做：

  while (previous !=NULL && current != NULL) {
      previous->link = current->link; 
      next = current->link;
      free(current);
      if(next) {
        next->prev = previous;
        current = next->link;
      }
      else current = NULL;
      previous = next;
  }
  

  在这里试试：https://repl.it/HK2P/0

  回复于 2024-05-15T13:52:54+08:00
• 0

  你在创建和删除功能中做了一些小错误......

  这是更新的代码试试吧......

  #include <stdio.h>
  #include <stdlib.h>
  
  // using a structure
  typedef struct mynode {
      int data;
      struct mynode *prev;    // to point to previous node
      struct mynode *link;    // to point to next node
  } node;
  node *head = NULL;
  
  // creating the list
  void create() {
      node *p, *q;
      int ch;
      do {
          p = (node *)malloc(sizeof(node));
          printf("enter data\n"); 
          scanf("%d", &p->data);
          p->link = NULL;
          if (head == NULL) 
          {
              p->prev = NULL;
              head = p;
          }
          else
          {
              q = head;
              while (q->link != NULL)
              q = q->link;
              p->prev = q;
              q->link = p;
          }
          printf("create another node?, press 1   ");
          scanf ("%d",&ch);
      } while(ch==1);
  }
  
  //to delete alternate elements
  void delete_Alt() {
      if (head == NULL)
          printf("Empty list...ERROR");
  
      node *previous, *current, *next;    
      previous = head;
      current = head->link;
      while (previous !=NULL && current != NULL) 
      {
          previous->link = current->link; 
          next = current->link;
          free(current);
          if(next) 
          {
              next->prev = previous;
              current = next->link;
          }
          else 
          current = NULL;
          previous = next;
      }
  }
  
  // print the list
  void display() {
      node *temp;
      temp = head;
      while (temp != NULL) {
          printf("%d  ",temp->data);
          temp = temp->link;
      }
      printf("\n");
  }
  
  int main() {
      node *head = NULL;
      create();
      printf("List before deleting is:    ");
      display();
      delete_Alt();
      printf("List after deleting is:     ");
      display();
  return 0;
  }
  

  回复于 2024-05-15T13:52:54+08:00
• 0

  在您的程序中，您只为头部分配了一次值：

  node *head = NULL;
  

  那么它的 Value 不会改变 .

  回复于 2024-05-15T13:52:54+08:00