Question

就学校实验室而言,我必须编制相关信息清单,然后优先整理这些信息,首先把“高”优先事项从中删除,然后从中删除。我几天来一直试图把这一点推到底,我可以总结一下我的想法。我在试图进行分类时,没有在我的“职业清单”类别中增加除头和面积以外的任何东西,但我似乎都增加了停车场。我只想说这一点,但现在我 st忙。

这里是我迄今为止所做的。希望你们能够理解:

class ListOfMessages
{
    private int m_nSize;
    public Message m_cListStart;
    //public Message m_cNextItem;
    public Message m_cLastItem;

    public ListOfMessages()
    {
        m_nSize = 0;
        m_cListStart = null;
        //m_cNextItem = null;
    }

    public int Count
    {
        get { return m_nSize; }
    }       

    public string Display(Message cMessage)
    {
        return "Message: " + cMessage.m_strMessage + "
Priority: " + cMessage.m_strPriority;
    }

    //list additions
    public int Add(Message newMessage)
    {
        Message nextMessage = new Message();
        //inserts objects at the end of the list
        if (m_nSize == 0)
        {
            m_cListStart = newMessage;
                //m_cLastItem = newMessage;
        }
        else
        {
            Message CurrentMessage = m_cListStart;

            if (newMessage.m_strPriority == "High")
            {

                    if (m_cListStart.m_strPriority != "High")
                    {
                        //Make the object at the start of the list point to itself
                        CurrentMessage.m_cNext = m_cListStart;
                        //Replace the object at the start of the list with the new message
                        m_cListStart = newMessage;

                    }
                else
                {
                    Message LastMessage = null;

                    for (int iii = 0; iii < m_nSize; iii++)//(newmessage.m_strpriority == iii.m_strpriority)
                    //&& (iii.m_cnext == null);)
                    {
                        if (m_cListStart.m_strPriority != "High")
                        {
                            nextMessage = newMessage;
                            nextMessage.m_cNext =
                            CurrentMessage = nextMessage;
                            //LastMessage.m_cNext = CurrentMessage;
                        }
                        LastMessage = CurrentMessage;

                        if (m_cListStart.m_cNext != null)
                            m_cListStart = m_cListStart.m_cNext;
                    }
                }
                //iii = iii.m_cnext;
            }
                    // for (int iii = m_cListStart; ; iii = iii.m_cNext)//(newMessage.m_strPriority == iii.m_strPriority)
                    //    //&& (iii.m_cNext == null);)
                    //{
                    //    //Message lastMessage = iii;
                    //    if (iii.m_strPriority != iii.m_strPriority)
                    //    {
                    //        //iii.m_cNext = newMessage;
                    //        newMessage.m_cNext = iii.m_cNext;
                    //        iii.m_cNext = newMessage;
                    //    }


                    //m_cLastItem.m_cNext = newMessage;
        }
            //m_cLastItem = newMessage;
            return m_nSize++;
    }

    public Message Pop()
    {
        //Message Current = m_cListStart;
        //if the object at the start of the list has another object after it, make that object the start of the list
        //and decrease the size by 1 after popping an object off if there is more than 1 object after the start of the list
        if (m_cListStart.m_cNext != null)
        {
            m_cListStart = m_cListStart.m_cNext;
        }
        if (m_nSize > 0)
            m_nSize--;
        else
            m_cListStart = null;
        return m_cListStart;
        //if (m_cListStart.m_cNext != null)
        //    m_cListStart = m_cListStart.m_cNext;
        //if (m_nSize > 1)
        //    m_nSize--;
        //return m_cListStart;
    }

我为检索这些讯息而pop忙的职能可能需要一些精炼,但现在大多数工作都属于添加功能。我真的在穿过那里黑暗。

是否有任何人知道做我所要求的事情的简单方式?

Answer 1

为什么要写一份与习俗有关的清单如下:

class Node<T> : IComparable<T>
{
   public int Priority {set;get;}
   public T Data {set;get;}
   public Node<T> Next {set;get;}
   public Node<T> Previous {set;get;}

   // you need to implement IComparable here for sorting.
}

这就是你的定义。现在我们需要实施一个联系语言阶级。

你联系的清单类别可以是双重联系的清单,因为你本人对此有怀疑。并且比较容易列入双重链接的清单。

这里的定义是:

class LinkedList<T> : IEnumerable<T> where T: IComparable
{
    public Node<T> Head {set;get;}
    public Node<T> Tail {set;get;}

    // set of constructors
    //.....

    public void Insert(Node<T> node)
    {
       // you can do recursive or iterative impl. very easy.
    }

    // other public methods such as remove, insertAfter, insert before, insert last etc.

    public void Sort()
    {
       // easiest solution is to use insertion sort based on priority.
    }

}

如果你可以创造额外记忆,即:另一个相互关联的清单。插入将处以罚款。为此,你还需要在功能后插入。

我有LinkedList 执行,可加以核对。你只是需要根据优先事项进行分类,你可以使用泡沫、插入或合并。

此外,你还可能想看一下你可以用来执行优先事项的希普。询问是出于这一目的。我有。 Heap 数据结构执行,可以检查。

Answer 2

最为容易的解决办法是将 3<>>>--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

When you add, you add to the end of the correct list. When you remove, you first try to remove from the highest priority list. If that is empty, try the middle one. If even that is empty, use the lowest list.

如果你的优先事项数目不变,这两种情况下的时间复杂性都是O(1)。

友情链接