Question

我在C中，需要删除链接列表中多次出现的'key'字符，并返回链表的头部。

只有当'key'不是第一个或最后一个节点'char'时，此功能才能正常工作，链表。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 )

此外，任何带有“钥匙”的东西都会立即重复失败。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

它应该是这样的：

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练习，想象一下你有一个＆＃34;交换＆＃34;功能（就像C ++一样）：

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

然后你可以非常简单地编写这段代码：

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;
}

你甚至可以专注于某种＆＃34; exchange_with_next＆＃34;：

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

Answer 2

首先： prev可能处于未确定状态：您在第一个while中释放它并在第二个while中取消引用{ {1}}。如果密钥是第一个字符，这就是函数失败的原因。

第二：如果您的密钥是以下代码的最后一个字符：

prev->next

将失败，因为您取消引用if (current->data == key) { prev->next = current->next; free(current); current = prev->next; } prev = current; current = current->next;，但current为current。

一步一步：

NULL

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;

删除多个＆＃39;键＆＃39;来自C中链表的节点

2 个答案: