Question

问题1：在列表中的节点中删除＆gt; 3

说明

删除七个列表中的第六个节点，导致仅打印第一个节点和最后一个节点。

可用的节点指针： * next_，* prev_，* data _

删除指定节点的功能在LinkedList.cpp中名称：DeleteNode。

遍历列表以打印节点的函数位于main.cpp中名称：PrintAllNodes

可能的解决方案：

在遍历打印节点时，能够访问main中的Current-＆gt; prev_。

代码：

void LinkedList::DeleteNode( Node* node )
{
    Node *Current = first_; // I want this to be my only Node Ptr Varaible Declaration.
    if ( NULL == first_ )
        std::cout << "Cannot delete from an empty list: \n";
//TRAVERSING WAS/IS A BAD IDEA.............................
    while ( Current != NULL )
    {
        if ( Current->data_ == node->data_ )
        {
            //If Current isn't the head of the list, set prev to next
            if ( Current != first_ )
            {
                Current->prev_        = first_; //statement that follows crashes if this is not assigned.
                Current->prev_->next_ = Current->next_;
            }
            else
            {
                first_ = Current->next_;
                if ( first_ != NULL )
                first_->prev_ = NULL;
            }

            //If Current isn't the tail of the list, set next to prev
            if ( Current->next_ != NULL )
                Current->next_ = Current->prev_;

            else if ( Current->prev_ != NULL )
            Current->prev_->next_ = NULL;

          listLen_--;
          delete Current;
          Current = NULL;
        }
        else
        {
            Current->prev_ = Current;
            Current = Current->next_;
        }
    }
    return;
}

main.cpp中的PrintAllNodes代码：

void PrintAllNodes( LinkedList *LinkedObject, long length = 0 )
{
    const char *Names = NULL; 
    length = LinkedObject->GetListLength();
    Node *GetNode = LinkedObject->GetFirstNode();

    for ( signed short x = 0; x < length; x++ )
    {
        Names = static_cast< NameObject* >( GetNode->data_ )->GetName();
        cout << Names << endl;
        GetNode = GetNode->next_; // traversing
    }
    return;
}

Answer 1

这是你的问题：

Current->prev_  = first_;

您正在做的是在当前连接第一个到最后一个节点之前断开所有节点！（在你的情况下是第七个）

你应该做的只有：

Current->prev_->next_ = Current->next_;
Current->next_->prev_ = Current->prev_; //I think you forgot this
delete Current;
Current = NULL;

如果没有

Current->prev_  = first_;

你遇到了崩溃，那是因为你的Current-＆gt; prev_分配不好。但是将它分配给first_不是解决方案。你应该检查你的其他方法（可能是AddNode），看看你的Current-＆gt; prev_为什么不好。

C ++删除队列中的节点

1 个答案: