链表推回成员函数的实现 [英] Linked List pushback member function implementation

问题描述

我是新手程序员，这是我关于堆栈溢出的第二个问题.

I am a novice programmer and this is my second question on Stack Overflow.

我正在尝试通过使用尾指针为我的链接列表实现推回功能.看起来似乎很简单，但是我有点ing愧，觉得自己忘记了一些东西，或者我的逻辑很棘手.链接列表很难！

I am trying to implement a pushback function for my Linked List by using a tail pointer. It seems straightforward enough, but I have a nagging feeling that I am forgetting something or that my logic is screwy. Linked Lists are hard!

这是我的代码:

template <typename T>
void LinkedList<T>::push_back(const T n)
{
Node *newNode;  // Points to a newly allocated node

// A new node is created and the value that was passed to the function is stored within.
newNode = new Node;
newNode->mData = n;
newNode->mNext = nullptr; 
newNode->mPrev = nullptr;

//If the list is empty, set head to point to the new node.
if (head == nullptr)
{
    head = newNode;
    if (tail == nullptr)
    {
        tail = head;
    }
}
else  // Else set tail to point to the new node.
    tail->mPrev = newNode;
}

感谢您抽出宝贵的时间阅读本文章.

Thank you for taking the time to read this.

推荐答案

您将错误的 mPrev 指向错误的节点.而且，如果前一个 tail 节点不为null即可继续设置列表的前向链，则永远不会设置它的 mNext .

Your pointing the wrong mPrev to the wrong node. And you never set mNext of the prior tail node if it was non-null to continue the forward chain of your list.

template <typename T>
void LinkedList<T>::push_back(const T n)
{
    Node *newNode;  // Points to a newly allocated node

    // A new node is created and the value that was passed to the function is stored within.
    newNode = new Node;
    newNode->mData = n;
    newNode->mNext = nullptr;
    newNode->mPrev = tail; // may be null, but that's ok.

    //If the list is empty, set head to point to the new node.
    if (head == nullptr)
        head = newNode;
    else
        tail->mNext = newNode; // if head is non-null, tail should be too
    tail = newNode;
}

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