Question

我是一名新手程序员，这是我在Stack Overflow上的第二个问题。

我正在尝试使用尾指针为我的链表实现回送功能。这似乎很简单，但我有一种唠叨的感觉，我忘记了某些东西，或者说我的逻辑很棘手。链接列表很难！

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;
}

感谢您抽出宝贵时间阅读本文。

Answer 1

您将错误的mPrev指向错误的节点。并且如果先前的mNext节点为非空，则永远不会设置tail以继续列表的前向链。

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;
}

Linked List pushback成员函数实现

1 个答案: