c++ - C ++删除队列中的节点

Question

问题1：删除列表中的节点> 3

描述：

删除 7 个列表中的第 6 个节点会导致仅打印第一个和最后一个节点。

可用节点指针：*next_、*prev_、*data_

删除指定节点的函数在LinkedList.cpp 名称：DeleteNode。

遍历列表打印节点的函数在 main.cpp 名称：PrintAllNodes

可能的解决方案：

遍历打印节点时能够在 main 中访问 Current->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 PrintAllNodes 的代码：

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节点之前断开所有节点并将第一个节点连接到最后一个节点！（你的情况是第七个）

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_;

你会崩溃，那是因为你的 Current->prev_ 没有分配好。但是将其分配给 first_ 不是解决方案。你应该检查你的其他方法（可能是 AddNode），看看为什么你的 Current->prev_ 不好。