Question

我试图编写一个删除链接列表中具有给定值的所有节点的函数，但我想尝试使用递归函数。我认为删除列表中最后一个节点的代码块无意中将整个列表设置为null，但我不知道如何解决这个问题。

编辑：整个列表不应该为空，它应该删除包含给定整数的所有节点。例如，如果给定列表是3-> 4-> 3-> 6并且我们想删除3的所有实例，则列表将是4-> 6

typedef struct nodeStruct {
   int num;
   struct nodeStruct *next;
   struct nodeStruct *prev;
}node;

 node* deleteNode(node* head, int num) {
  node* current = head;
  if (current == NULL) {
      return head;
  } else if (current->num == num) {
      if (current->prev == NULL) {
          head = current->next;//delete first Node
          head->prev = NULL;
          free(current);
          head = deleteNode(head, num);
      } else if (current->next == NULL) {
          free(current);//delete last node
          return head;
      } else {
          current->prev->next = current->next; //delete middle node
          current->next->prev = current->prev;
          free(current);
          head = deleteNode(head, num);
      }
  } else {
      head = deleteNode(current->next, num);
  }
  return head;
}

Answer 1

类似这样的事情

if(head == NULL){
    return NULL;
} else {
    node *rest = deleteNode(head->next, num);
    if(head->num == num){
        if(rest != NULL){
            rest->prev = NULL;
        }
        free(head);
        return rest;
    } else {
        if(rest != NULL){
            rest->prev = head;
        }
        head->next = rest;
        return head;
    }
}

使用递归删除双向链表中的所有节点

1 个答案: