使用链表C++的抖音游戏

在我正在读的一本c++书中，我遇到了一个练习，建议我使用链表和数组来做一个井字游戏。我决定先尝试链表，因为数组显然更容易。然而，我一直纠结于如何检查是否有人赢得了比赛。到目前为止，我拥有的是：

struct board
{
    bool square, circle, empty;
    int pos;
    board* next;
};
void startGame(int first, board* fullBoard);
board* getBoard(board* fullBoard);

int main()
{
    int dice, first;
    board* fullBoard = NULL;
    cout << "Welcome to Tic-tac-toe DOS Game. (2 Player version)nn";
    cout << "X is Player 1 and O is Player 2.nI will decide who is starting in the first match...n ";
    srand(time(NULL));
    dice = 1;//rand() % 6 + 1;
    cout << dice;
    if(dice <= 3)
    {
        first = 1;
        cout << "Player 1 is the first!n";
    }
    else
    {
        first = 2;
        cout << "Player 2 is the first!nn";
    }
    system("pause");
    system("cls");
    startGame(first, fullBoard);
}
void startGame(int first, board* fullBoard)
{
    int choice;
    bool isPlaying;
    for(int i = 1; i <= 9; i++)
        fullBoard = getBoard(fullBoard);
    bool isGameOn = true;
    while(isGameOn == true)
    {
        board* current = fullBoard;
        while(current != NULL)
        {
            if(current->empty == true)
                cout << "   " << current->pos;
            else if(current->circle == true)
                cout << "   " << "O";
            else
                cout << "   " << "X";
            if( current->pos == 4 || current->pos == 7)
            {
                cout << "n";
                cout << "-----------------------n";
            }
            else if (current->pos == 1)
                cout << "n";
            else
                cout << "   |";
            current = current->next;
        }

        if(first == 1)
        {
            isPlaying = true;
            while(isPlaying == true)
            {
                cout << "Player 1, please put the number corresponding to the area you want to fill: ";
                cin >> choice;
                while(choice < 1 || choice > 9)
                {
                    cout << "Invalid choice. Please choose a valid option: ";
                    cin >> choice;
                }
                current = fullBoard;
                while(current != NULL && current->pos != choice)
                    current = current->next;
                if(current->empty == true)
                {
                    current->empty = false;
                    current->square = true;
                    isPlaying = false;
                    first = 2;
                }
                else
                    cout << "The field that you chose is already used...n";
            }
        }
        else
        {
            isPlaying = true;
            while(isPlaying == true)
            {
                cout << "Player 2, please put the number corresponding to the area you want to fill: ";
                cin >> choice;
                while(choice < 1 || choice > 9)
                {
                    cout << "Invalid choice. Please choose a valid option: ";
                    cin >> choice;
                }
                current = fullBoard;
                while(current != NULL && current->pos != choice)
                    current = current->next;
                if(current->empty == true)
                {
                    current->empty = false;
                    current->circle = true;
                    isPlaying = false;
                    first = 1;
                }
                else
                    cout << "The field that you chose is already used...n";
            }
        }
        system("cls");
    }

}
board* getBoard(board* fullBoard)
{
    board* newBoard = new board;
    newBoard->empty = true;
    newBoard->circle = false;
    newBoard->square = false;
    newBoard->next = fullBoard;
    if(newBoard->next != NULL)
        newBoard->pos = newBoard->next->pos + 1;
    else
        newBoard->pos = 1;
    return newBoard;

}

正如你所看到的，在我的结构Board上，我有一个名为pos的int，我创建它是为了跟踪整个Board。到目前为止，我能想到的唯一解决方案是检查每个位置。例如：将pos 8与pos 9、7、5和2进行比较，将pos 9与pos 8、7、6、3、5和1进行比较。但我认为这太宽泛了（也许它也很难编码？）。你认为我还有什么选择？

提前感谢，Felipe