发现在井字棋赢家（Java中使用2- d阵列）(Finding the winner in Tic-

我无法搞清楚如何显示游戏的赢家在这个井字棋程序。

 import java.util.*;
public class tic
{
    public static void main(String[] args) 
    {
        Scanner in = new Scanner(System.in);

        boolean flag=false;

        char[][] board = 
        {
        {' ', ' ', ' '},
        {' ', ' ', ' '},
        {' ', ' ', ' '}
        };

        boolean done = false;
        int player = 1;
        int row = 0;
        int col = 0;

        while (flag != true) 
        {
          checkForWinner(board);
          System.out.println("Enter the row and column for your next move");
          row = in.nextInt();
          col = in.nextInt();
          if (player == 1) 
          {
            board[row][col] = 'X';
            player = 2;
            checkForWinner(board);
           }
          else 
          { 
            board[row][col] = 'O';
            player = 1;
            checkForWinner(board);
          }
          printBoard(board);
          checkForWinner(board);
        }

        displayWinner(player, flag);  


    }
    public static void printBoard(char[][] board) 
    {
    for (int row = 0; row < 3; row++) 
    {
        for (int col = 0; col < 3; col++) 
        {
            System.out.print("|" + board[row][col] + "|");
        }
        System.out.println();
        System.out.println("-------");
    }
    }
    public static boolean checkForWinner(char[][] board)
   {
       // checkForWinner() method determines if a pattern of data stored
       // in the 2 D char array indicates the a player has won the game.

        boolean flag = false;
        boolean flag1 = false;
        boolean flag2 = false;
        boolean flag3 = false;
        boolean flag4 = false;

        // checks the contents of each row for matching data   
        for (int i = 0; i <= 2; i++)
        {
            if ((board[i][0] == board[i][1] && board[i][1] == board[i][2]) && board[i][2] != ' ') 
                flag1 = true;
        }

         // checks the contents of each column for matching data
        for (int j = 0; j <= 2; j++)
        {
            if ((board[0][j] == board[1][j] && board[1][j] == board[2][j]) && board[2][j] != ' ') 
                flag2 = true;
        }

        // checks the contents of one diagonal for matching data
        if ((board[0][0] == board[1][1] && board[1][1] == board[2][2]) && board[2][2] != ' ') 
                flag3 = true;

        // checks the contents of the other diagonal for matching data
        if ((board[0][2] == board[1][1] && board[1][1] == board[2][0]) && board[2][0] != ' ') 
                flag4 = true;

        // checks if any of the previous conditions evaluated to true        
        if (flag1 == true || flag2 == true || flag3 == true || flag4 == true)
            flag = true;

       // returns true if a winner was found; returns false is no winner     
       return flag;
   } // end of checkForWinner method
   public static void displayWinner(int player, boolean flag)
   {

    if (flag == true)
    {
        int currentplayer;
        currentplayer=player;
        System.out.println("The winner of the game is" +currentplayer);
    }


   }
}

该checkForWinner方法提供给我们的，不能改变，从我已经看到它会检查所有可能的赢状态，而无需考虑，因为这个我有点失去了对如何启动方法的播放器这将显示的赢家。

任何输入，以什么我可以用这种方法做的将是巨大的。

感谢您的关注。

编辑：添加了displayWinner的方法，我试过了，这似乎并没有工作。

呼叫checkForWinner每回合之后。一旦你回到true ，你知道谁是最后一个移动是赢家。

这是因为你已经检查了板前的最后一步，而且已经在当时没有赢家（否则，你会在游戏早期退出）。现在，我们有一个赢家，谁就取得了最后一步一定是谁赢得了球员。

请注意，您当前的通话checkForWinner是没用的，因为你忽略了返回值。目前，该flag仍然未分配。

的通话displayWinner需要采取冠军数量，而不是一个flag 。