c++ - 打印链接列表

Question

给定节点的名称，此函数应搜索链接列表。如果在其内部找到，则返回指向该节点的指针，否则返回null。注意:我确定我已经成功编写了此函数。

// getNode

Node *LinkedList::getNode(string name)
{
    Node *temp = head;
    while (temp != NULL)
    {
        if (temp->name.compare(name) == 0)
            return temp;
        temp = temp->next;
    }
    return NULL;
}

给定一个节点，此函数在屏幕上打印:teamName(winScore-loseScore)。例如:UCLA(25-13)或Texas A＆M(31-25)。注意:我确定我已经成功编写了此函数。

// printNode

void LinkedList::printNode(Node *node)
{
    if (node == NULL)
        return;
    else {
        cout << node->name << "(" << node->winScore;
        cout << "-" << node->loseScore << ")";
    }
}

给定一个团队名称，该功能应该以以下格式一张一张地打印其邻接列表中的所有节点(注意:以下仅是一个示例!)这就是我认为错了的地方。

Missouri University beat: New Mexico(52-23), Salisbury (48-31), Virginia (34-9)

void LinkedList::printList(string name)
{
    if (head == NULL)
        cout << "\n Empty list" << endl;
    else {

        Node *temp = head; 
        while (temp != NULL)
        {
            cout << temp->name << " beat: " << temp->name << endl; // Is this right?
            temp = temp->next;
        }
    }    
}

Answer 1

我猜这很接近您想要的:

void LinkedList::printList(string name)
{
    // find the node for the name you supply
    // (or else I don't understand why 'name' is supplied to this function)

    Node *temp = getNode(name);

    if (temp) {                     // node with name found           
        if (temp->next) {           // there's at least one adjacent node
            cout << temp->name << " beat: ";
            while ((temp = temp->next) != nullptr) {
                printNode(temp); 
                if (temp->next) cout << ", ";
            };
            cout << "\n";
        } else {                    // no adjacent nodes
            cout << temp->name << " did not beat anyone\n";
        }
    }
}