java - (Java) 使用二维数组的 Tic-Tac-Toe 游戏

Question

在类里面，我们的作业是创建一个二维数组并围绕它创建一个井字游戏。除了当整个棋盘已满并且游戏是平局时显示之外，我已经完成了所有操作。我尝试了一些方法，但没有找到解决方案，我需要一些帮助...这是我的代码:

import java.util.Scanner;

public class TicTacToe {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int row, column;
        char player = 'X';

        //create 2 dimensional array for tic tac toe board
        char[][] board = new char[3][3];
        char ch = '1';
        for (int i = 0; i < 3; i++){
            for (int j = 0; j < 3; j++) {
                board[i][j] = ch++;
            }
        }
        displayBoard(board);
        while(!winner(board) == true){

            //get input for row/column
            System.out.println("Enter a row and column (0, 1, or 2); for player " + player + ":");
            row = in.nextInt();
            column = in.nextInt();

            //occupied
            while (board[row][column] == 'X' || board[row][column] == 'O') {
                System.out.println("This spot is occupied. Please try again");
            }
            //place the X
            board[row][column] = player;
            displayBoard(board);

            if (winner(board)){
                System.out.println("Player " + player + " is the winner!");
            }

            //time to swap players after each go.
            if (player == 'O') {
                player = 'X';

            }
            else {
                player = 'O';
            }
            if (winner(board) == false) {
            System.out.println("The game is a draw. Please try again.");

        }

    }

    private static void displayBoard(char[][] board) {
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[i].length; j++) {
                if (j == board[i].length - 1) System.out.print(board[i][j]);
                else System.out.print( board[i][j] + " | ");
            }
            System.out.println();
        }


    }
    //method to determine whether there is an x or an o in the spot
    public static Boolean winner(char[][] board){
        for (int i = 0; i< board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == 'O' || board[i][j] == 'X') {
                    return false;
                }
            }
        }

        return (board[0][0] == board [0][1] && board[0][0] == board [0][2]) ||
            (board[0][0] == board [1][1] && board[0][0] == board [2][2]) ||
            (board[0][0] == board [1][0] && board[0][0] == board [2][0]) ||
            (board[2][0] == board [2][1] && board[2][0] == board [2][2]) ||
            (board[2][0] == board [1][1] && board[0][0] == board [0][2]) ||
            (board[0][2] == board [1][2] && board[0][2] == board [2][2]) ||
            (board[0][1] == board [1][1] && board[0][1] == board [2][1]) ||
            (board[1][0] == board [1][1] && board[1][0] == board [1][2]);
    }
}

我想要输出说板已满，但我什么也没得到。这是我输出的最后一行，正如您所看到的，我当前的策略不起作用，因为它继续要求输入。 -->

输入行和列(0、1或2)；对于玩家 X:2 0X |哦| X哦|哦| XX | X |氧输入行和列(0、1 或 2)；对于玩家 O:

Answer 1

首先:

 while (board[row][column] == 'X' || board[row][column] == 'O') {
            System.out.println("This spot is occupied. Please try again");
        }

这将创建一个无限循环，因为 row 和 column 不应更改，您应该要求新的输入!

还有

public static Boolean winner(char[][] board){
    for (int i = 0; i< board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            if (board[i][j] == 'O' || board[i][j] == 'X') {
                return false;
            }
        }
    }

一旦您点击“O”或“X”，您将以错误退出方法(没有获胜者)

您可能想要检查的是是否每个位置都被占用

public static Boolean winner(char[][] board){
   //Boolean which is true until there is a empty spot
   boolean occupied = true;
   //loop and check if there is empty space or if its a draw
    for (int i = 0; i< board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            //Check if spot is not 'O' or not 'X' => empty 
            if (board[i][j] != 'O' || board[i][j] != 'X') {
                occupied = false;
            }
        }
    }
    if(occupied)
        return false;
   //Check if someone won
    return (board[0][0] == board [0][1] && board[0][0] == board [0][2]) ||
        (board[0][0] == board [1][1] && board[0][0] == board [2][2]) ||
        (board[0][0] == board [1][0] && board[0][0] == board [2][0]) ||
        (board[2][0] == board [2][1] && board[2][0] == board [2][2]) ||
        (board[2][0] == board [1][1] && board[0][0] == board [0][2]) ||
        (board[0][2] == board [1][2] && board[0][2] == board [2][2]) ||
        (board[0][1] == board [1][1] && board[0][1] == board [2][1]) ||
        (board[1][0] == board [1][1] && board[1][0] == board [1][2]);
}

现在将检查是否有获胜者或平局

Occupied == true == tie == return false

Winner == return true

但是你有三种状态:

• 获胜
• 领带
• 未完成

使用更改后的方法，在获胜之前您将无法完成游戏。

原因:

 while(!winner(board) == true)

这使得游戏在没有获胜者的情况下继续运行(winner() 将为 false，因为所有内容都已被占用或没有获胜者)

while(!false==true) => while(true) 

您可以编写类似于 Winner 的方法，但它只检查棋盘上是否有空位:

public static Boolean hasEmptySpot(char[][] board){
   //loop and check if there is empty space 
    for (int i = 0; i< board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            if (board[i][j] != 'O' && board[i][j] != 'X') {
                return true;
            }
        }
    }
    return false;
}

//New code 
while(hasEmptySpot(board) || !winner(board)){
          //Your code for the game here
     ....
    }

当没有空位时，这将结束游戏当你完成游戏后，你可以调用winner(board)，如果你平局或获胜，它就会返回!

通过创建hasEmptySpot()，您可以将获胜者方法更改为

public static Boolean winner(char[][] board){
    return (board[0][0] == board [0][1] && board[0][0] == board [0][2]) ||
        (board[0][0] == board [1][1] && board[0][0] == board [2][2]) ||
        (board[0][0] == board [1][0] && board[0][0] == board [2][0]) ||
        (board[2][0] == board [2][1] && board[2][0] == board [2][2]) ||
        (board[2][0] == board [1][1] && board[0][0] == board [0][2]) ||
        (board[0][2] == board [1][2] && board[0][2] == board [2][2]) ||
        (board[0][1] == board [1][1] && board[0][1] == board [2][1]) ||
        (board[1][0] == board [1][1] && board[1][0] == board [1][2]);
}

为什么？因为您完成了游戏并且知道只有两种可能的结果:获胜或平局。

希望这对你有一点帮助。

编辑我自己也有逻辑错误!

第一个错误:您仍然需要在游戏运行时检查是否有获胜者忘记了这一点!

while(hasEmptySpot(board) || !winner(board)){
}

现在，当有获胜者或没有空位时，这将退出游戏循环

第二个错误:在 hasEmptySpot() 中

 if (board[i][j] != 'O' && board[i][j] != 'X') {
                return true;

不是

 if (board[i][j] != 'O' || board[i][j] != 'X') {
                return true;

在上面的示例中修复了它。

对于给您带来的不便，我们深表歉意!