Question

我已经完成了一个井字游戏，但我唯一的问题是我无法确定游戏中的赢家/领带。我有一个专门用于寻找胜利者的方法，但它不起作用。我究竟做错了什么？这是代码：

import java.util.Scanner;

public class TicTacToe {

    public static final int X = 1;
    public static final int Y = 2;

    public static void main(String[] args) {

        final int X = 1;
        final int Y = 2;

        int[][] board = {{0, 0, 0},
        {0, 0, 0},
        {0, 0, 0}
        };

        int count = 0;

        while (!checkWinner(board)) {
            printBoard(board);

            if (count % 2 == 0) {
                System.out.println("X's turn");
                move(X, board);
            } else {
                System.out.println("Y's turn");
                move(Y, board);
            }

            System.out.println();
            count++;
        }

    }

    public static boolean checkWinner(int[][] board) {
        if (board[0][0] == board[1][0] && board[1][0] == board[2][0] && (board[0][0] == "X" || board[0][0] == "Y")) {
            return true;
        } else if (board[0][1] == board[1][1] && board[1][1] == board[2][1] && (board[0][1] == "X" || board[0][1] == "Y")) {
            return true;
        } else if (board[0][2] == board[1][2] && board[1][2] == board[2][2] && (board[0][2] == "X" || board[0][2] == "Y")) {
            return true;
        } else if (board[0][0] == board[0][1] && board[0][1] == board[0][2] && (board[0][0] == "X" || board[0][0] == "Y")) {
            return true;
        } else if (board[1][0] == board[1][1] && board[1][1] == board[1][2] && (board[1][0] == "X" || board[1][0] == "Y")) {
            return true;
        } else if (board[2][0] == board[2][1] && board[2][1] == board[2][2] && (board[2][0] == "X" || board[2][0] == "Y")) {
            return true;
        } else if (board[0][0] == board[1][1] && board[1][1] == board[2][2] && (board[0][0] == "X" || board[0][0] == "Y")) {
            return true;
        } else if (board[2][0] == board[1][1] && board[1][1] == board[0][2] && (board[2][0] == "X" || board[2][0] == "Y")) {
            return true;
        } else {
            return false;
        }
    }

    public static void printBoard(int[][] board) {
        for (int i = 0; i < 3; i++) {
            System.out.print("| ");
            for (int j = 0; j < 3; j++) {
                if (board[i][j] == TicTacToe.X) {
                    System.out.print("X");
                } else if (board[i][j] == TicTacToe.Y) {
                    System.out.print("Y");
                } else {
                    System.out.print("-");
                }
                System.out.print(" | ");
            }
            System.out.println();
        }
    }

    public static void move(int player, int[][] board) {
        Scanner myScan = new Scanner(System.in);

        System.out.print("Row: ");
        int x = myScan.nextInt();
        System.out.print("Column: ");
        int y = myScan.nextInt();

        board[x][y] = player;

    }

}

我不需要为此程序添加其他方法，因为这就是我所需要的。

Answer 1

如果没有获胜者 AND 主席未满，您应该将null循环中的条件替换为

let foo:number = null; // Error foo is not nullable

我建议您重新访问while方法，虽然该格式有效但效率很低，您可以找到一种更简单，更有趣的方法来实现它。例如，您可以利用以下内容：

while(!checkWinner(board) && /*board is not full*/) {
//...
}

if(checkWinner(board)) //→ there's a winner;
else //it's a draw;

Answer 2

您的代码是否编译没有任何错误？

我看不出这是如何起作用的？ board [1] [0] ==“X”

board是一个整数数组！您应该看到“不兼容的操作数类型int和String”！

Answer 3

您的数组是整数类型，但您可以在函数中将其与＆＃34; X＆＃34;等字符串进行比较。和＆＃34; Y＆＃34;。你必须替换... ==＆＃34; X＆＃34;与... == TicTacToe.X和... ==＆＃34; Y＆＃34;在checkWinner（）函数的所有if语句中使用... == TicTacToe.Y。

Tic-Tac-Toe检查获胜者

3 个答案: