从C中的链接列表中删除多个“关键”节点

Question

我在C语言中，需要删除链表中多次出现的“键”字符并返回链表的开头。

仅当“键”不是链接列表的第一个或最后一个节点“字符”时，此功能才能正常工作。 示例…使用键“ a”

fails: a->d->a->m->NULL (throws error) 
fails: t->a->d->a->NULL (throws error) 
passes: d->a->g->n->a->b->NULL (returns d->g->n->b->NULL )

同样，任何带有“键”的立即重复的操作都会失败。 示例…使用键“ a”

fails: d->a->a->a->a->r->n->NULL (returns d->a->a->r->n->NULL)

- - - - - - - - - - - - - - -删除（） - - - - - - - - - ---------------------

node* delete2(char key, node* head)
{
   /*IF NULL*/
   if(!head)
   {
      return head;
   }

   node* prev = NULL;
   node* current = head;

   /*if first node(head) is to be deleted*/
   while (current && current->data == key)
   {
      prev = current;
      current = current->next;
      head = current;
      free(prev);
   }


   /*scan list left to right*/
   while (current)
   {
      if (current->data == key)
      {
         prev->next = current->next;
         free(current);
         current = prev->next;
      }
      prev = current;
      current = current->next;
   }

   return head;
}

Answer 1

应该是这样的：

node * remove_key(char key, node * head)
{
    // remove initial matching elements
    while (head && head->data == key)
    {
        node * tmp = head;
        head = head->next;
        free(tmp);
    }

    // remove non-initial matching elements
    // loop invariant: "current != NULL && current->data != key"
    for (node * current = head; current != NULL; current = current->next)
    {
        while (current->next != nullptr && current->next->data == key)
        {
            node * tmp = current->next;
            current->next = tmp->next;
            free(tmp);
        }
    }

    return head;
}

---

作为一项有趣的心理锻炼，想象一下您有一个“交换”功能（就像C ++一样）：

node * exchange(node ** obj, node * newval)
{ node * tmp = *obj; *obj = newval; return tmp; }

然后，您可以非常简单地编写以下代码：

node * remove_key(char key, node * head)
{
    while (head && head->data == key)
        free(exchange(&head, head->next));

    for (node * current = head; current != NULL; current = current->next)
        while (current->next != nullptr && current->next->data == key)
            free(exchange(&current->next, current->next->next));

    return head;
}

您甚至可以专门研究某种“ exchange_with_next”：

node * exchange_with_next(node ** n) { return exchange(n, (*n)->next); }

Answer 2

第一： prev能在一个不确定的状态：你释放它在第一while和取消引用它在第二while当你做prev->next 。 这就是为什么如果键是第一个字符，则函数失败的原因。

第二：如果您的密钥是最后一个字符，则以下代码：

if (current->data == key)
{
    prev->next = current->next;
    free(current);
    current = prev->next;
}
prev = current;
current = current->next;

将失败，因为您取消引用current但是current为NULL 。

一步步：

if (current->data == key)
{
    prev->next = NULL;// current is the last element so current->next == NULL
    free(current);
    current = prev->next;

}
prev = current;
current = current->next;

if (current->data == key)
{
    prev->next = NULL;
    free(current);
    current = NULL;// because prev->next == NULL
}
prev = current;
current = current->next;

if (current->data == key)
{
    prev->next = NULL;
    free(current);
    current = NULL;
}
prev = NULL;// same again...
current = current->next;

if (current->data == key)
{
    prev->next = NULL;
    free(current);
    current = NULL;
}
prev = NULL;
current = NULL->next;// FAIL!!!