Question

我的列表结构如下：

struct Node {
    Node *next;
    Node *prev;
    T datum;
};
Node *first;   // points to first Node in list, or 0 if list is empty
Node *last;    // points to last Node in list, or 0 if list is empty

我试图做以下事情：

void pop_front()
{
     //copied from lecture slides
     assert(!empty());
     Node *victim = first;
     first = first->next;
     if(first != 0)
     {
         first->prev = 0;
     }
     delete victim;
     victim=0;
}

问题是，当我执行删除受害者行时，它会给我一个内存泄漏。我不知道出了什么问题。

编辑：这就是我添加节点的方式：

 //MODIFIES: this
    //EFFECTS:  inserts i into the front of the list
    void pushit_tofront(const T &datum)
    {
        //if list is empty, update both the first and last element
        //copied from lecture slides
        Node *p = new Node;
        p->datum = datum;
        p->next = first;
        if(empty())
        {
            first = last = p;
        }
        else
        {
            first = p;
        }

    }

Answer 1

尝试将此代码用于节点

 class Node {
   public:
        int get() { return object; }; // returns the value of the element
        void set(int object) { this->object = object; }; // set the value of the element

        Node* getNext() { return nextNode; }; // get the address of the next node
        void setNext(Node* nextNode) // set the address of the next node
         { this->nextNode = nextNode; };

        Node* getPrev() { return prevNode; }; // get the address of the prev node
        void setPrev(Node* prevNode) // set the address of the prev node
         { this->prevNode = prevNode; };

 private:
        int object; // it stores the actual value of the element
        Node* nextNode; // this points to the next node
        Node* prevNode; // this points to the previous node
     };

双重链接列表看起来像是未完成的。但你只需在其中添加删除功能。

class DoublyList
{
public:
        List();
        void add (int addObject);
        int get();
        bool next();
        void start();
        void remove();


private:
        int size;
        Node * headNode;
        Node * currentNode;
        Node * lastCurrentNode;
};

双重链接列表删除它将从任何位置删除。如果条件对你有所帮助，你已经询问了最后两个节点的第一个节点。

DoublyList::remove(){
    Node *removeNode= currentNode;

    if((currentNode->getNext()!=null)&&(currentNode->getprev())!=headNode)){ // it will remove the node in middle
    lastCurrentNode->setNext(currentNode->getNext());
    (currentNode->getNext())->setPrev(currentNode->getPrev);
    currentNode= currentNode->getNext();
    }

    if((currentNode->getprev())!=headNode)&&(currentNode->getNext()==null)){   // it will remove the node if it is last node
    lastCurrentNode->setNext(null);
    currentNode= lastCurrentNode;
    lastCurretnNode= currentNode->getPrev();
    }

    if((currentNode->getprev())==headNode)&&(currentNode->getNext()==null)){ //if it is at the start and next node is null
        headNode->setNext(null);
    lastCurrentNode= headNode;
    currentNode= headNode;
    }

        if((currentNode->getprev())==headNode)&&(currentNode->getNext()!=null)){  //if it is at the start and next node is not null
        headNode->setNext(currentNode->getNext());
        (currentNode->getNext())->setPrev(headNode);
    lastCurrentNode= headNode;
    currentNode= currentNode->getNext();
    }



    delete removeNode;
    size--;
}

我认为您正在尝试使用双向链接列表进行堆栈。但你的问题并不清楚。所以我已经在所有条件下编写了删除方法。我没有测试过这段代码。你必须自己做。如果你有任何错误可以联系我。

Answer 2

关于学术诚实问题，我不打算发布解决方案，因为这是一项受欢迎的作业。

但是，我会给你一些指示：）

任何时候你做一个双向链表的变异你（在你的情况下）最多有六个指针需要担心。

curr->next = ?
curr->prev = ?
next->prev = ?
next->next = ?
head = ?
tail = ?

如果您明确确保所有指针都得到妥善管理，您可能会解决当前的问题。

如何删除双向链表中的第一个节点？

2 个答案: