c++ - 链表入队和出队

我在使用c++的链表实现队列的入队和出队时遇到了一些麻烦。我的老师说模板是不可行的，我不能改变他给我们的公共(public)和私有(private)职能。我不断遇到细分错误。我不太了解自己在做什么错。我已经包括了头文件以及入队和出队功能。

header

const int MAX_STRING = 6;

typedef char Element300[MAX_STRING + 1];

class Queue300
{

    public:
        Queue300();
        Queue300(Queue300&);
        ~Queue300();
        void enQueue300(const Element300);
        void deQueue300(Element300);
        void view300();

    private:
        struct Node300;
        typedef Node300 * NodePtr300;
        struct Node300
        {
            Element300 element;
            NodePtr300 next;
        };
        NodePtr300 front, rear;
};

排队

void Queue300::enQueue300(const Element300 input)


{
    NodePtr300 temp = NULL;

    temp = new (std::nothrow) Node300;

    if (temp == NULL)
    {
        cerr << "The queue is full, could not add(enqueue) any more elements." << endl;
    }

    else if (front == NULL && rear == NULL)
    {
        strcpy(temp->element, input);
        rear = temp;
        rear->next = NULL;
        front = rear;
        temp = NULL;
    }

    else
    {
        strcpy(temp->element, input);
        temp = rear->next;
        rear = temp;
        rear->next = NULL;
        temp = NULL;
    }
}

出队

void Queue300::deQueue300(Element300 input)



{
    NodePtr300 temp = NULL;

    if (rear == NULL && front == NULL)
    {
        cerr << "The queue is already empty, could not delete(dequeue) any more elements." << endl;
    }

    else if (front == rear)
    {
        strcpy(temp->element, input);
        temp = front;
        delete temp;
        temp = NULL;
        front = NULL;
        rear = NULL;
    }

    else
    {
        strcpy(temp->element, input);
        temp = front;
        front = front->next;
        temp->next = NULL;
        delete temp;
        temp = NULL;
    }
}

最佳答案

让我们看一下入队:

    if (temp == NULL)
    {
        ...
    }
    else if (front == NULL && rear == NULL)
    {
        ...
    }
    else
    {
        strcpy(temp->element, input); // copy input into temp->element

        // temp points to rear->next (it is always NULL, isn't it?)
        // we lost the Node300 previously pointed by temp.
        temp = rear->next;

        // rear points to the same location as temp (NULL)
        rear = temp;

        // rear->next: dereferencing NULL /!\
        rear->next = NULL;

        temp = NULL;
    }

ASCII艺术的时间。这是入队之前:

  REAR -------------------
                         |
                         v
          [0]--> ... -->[N]-->NULL
           ^
  FRONT ---|

您分配一个[X]引用的节点temp:

temp --> [X]

REAR的下一个必须指向同一节点:

REAR
 |
 v
[N] --> [X]
         ^
         |
        temp

然后，必须更新REAR以引用[X]。
就指针操作而言，您甚至不需要temp指针，但让我们保留它，因为您正在检查节点是否之前已正确分配，这很高兴。

rear->next = temp;
rear = temp;       // or rear = rear->next;
temp->next = NULL; // or rear->next = NULL;

还要注意，您在两个strcpy(temp->element, input);分支中都执行else。您可以在temp == NULL时从函数返回，并在检查front和rear是否均为NULL之前进行复制。