帶計算機端隨機數發生器的2D TicTacToe

Question

我的代碼只打印空白空格，而不是每次提示輸入行號和列號，並查看空格是否爲空。我要生成電腦玩家的隨機數，但如果我這樣做是不知道正確的

import java.util.*; 
public class TicTacToe 
{ 
public static void main(String[] args) 
{ 
    Scanner in = new Scanner(System.in); 
    int row, column; 
    char player = 'X'; 
    char[][] board = new char[3][3]; 
    for(int i=0;i<3;i++){ 
    for(int j=0;j<3;++j){ 
     board[i][j] = ' '; 
    } 
    } 
    while(!checkWin(board,'O')==true) 
    { 
     computerPlay(board); 
     displayBoard(board); 
     if(checkWin(board,'X')) 
     { 
      System.out.println("Computer Wins"); 
      System.exit(0); 
     } 
     if(checkTie(board)) 
     { 
      System.out.println("Tie game"); 
      System.exit(0); 
     } 
     playerPlay(board); 
     displayBoard(board); 
     if(checkWin(board,'O')) 
     { 
      System.out.println("Player Wins"); 
      System.exit(0); 
     } 
     if(checkTie(board)) 
     { 
      System.out.println("Tie game"); 
      System.exit(0); 
     } 
    } 
    } 

    // Prompt the user for row & column index. Continue asking 
    // until an empty cell is selected. set the cell to 'O' 
    public static void playerPlay(char[][] board) 
    { 
    int row, column; 
    Scanner in = new Scanner(System.in); 
    while(checkTie(board)!= true) 
    { 

    System.out.println("Enter a row and column (0, 1, or 2); for player 
:"); 
    row = in.nextInt(); 
    column = in.nextInt(); 
    } 
} 
// Check by row, column, and diagonals 
public static boolean checkWin(char[][] board,char ch) 
{ 
    if ( board[0][0] == 'O' && board[0][1] == 'O' && board[0][2] == 'O' || // 
    1st row 
     board[1][0] == 'O' && board[1][1] == 'O' && board[1][2] == 'O' || // 
    2nd row 
     board[2][0] == 'O' && board[2][1] == 'O' && board[2][2] == 'O' || // 
    3rd row 
     board[0][0] == 'O' && board[1][0] == 'O' && board[2][0] == 'O' || // 
    1st col. 
     board[0][1] == 'O' && board[1][1] == 'O' && board[2][1] == 'O' || // 
    2nd col. 
     board[0][2] == 'O' && board[1][2] == 'O' && board[2][2] == 'O' || // 
    3rd col. 
     board[0][0] == 'O' && board[1][1] == 'O' && board[2][2] == 'O' || // 
    Diagonal   \ 
     board[2][0] == 'O' && board[1][1] == 'O' && board[0][2] == 'O') // 
    Diagonal /
    { 
    return true; 
    } 
    else { 

    return false; 
    } 
} 

// check for tie. If there no empty cells, then it is a tie 
public static boolean checkTie(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 true; 
     } 
    } 
    } 
    return false; 
} 
// Display the board 
public static void displayBoard(char[][] board) 
{ 
    System.out.println(board[0][0] + " | " + board[0][1] + " | " + board[0][2] 
+ "\n---------"); 
    System.out.println(board[1][0] + " | " + board[1][1] + " | " + board[1][2] 
+ "\n---------"); 
    System.out.println(board[2][0] + " | " + board[2][1] + " | " + board[2][2] 
+ "\n"); 
} 
// Continue generating random values for row and col until an 
// empty cell selected. Set the cell to 'X' 
public static void computerPlay(char[][]board) 
{ 
    while(checkTie(board)!= true) 
    { 
    for (int i=0; i<board.length; i++){ 
     for(int j = 0; j<board.length; j++){ 
      board[i][j] = (char)(Math.random()*10); 

     } 

    } 

    } 

} 


} 

Answer 1

的checkTie函數返回true如有板值爲空（嚴格地說，不是「X」或「O」）。如果checkTie的返回值爲false，則computerPlay函數僅會繼續執行。其中一個（但不是兩個）的邏輯應該顛倒。我建議更改isTied函數的名稱以避免混淆，並反轉該函數中的邏輯。這是我的建議（它也將while替換爲if，因爲它應該只運行一次）：

public static boolean isTied(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 true; 
} 
... 
public static void computerPlay(char[][]board) 
{ 
    if(!isTied(board)) 
    { 
    for (int i=0; i<board.length; i++){ 
     for(int j = 0; j<board.length; j++){ 
      board[i][j] = (char)(Math.random()*10); 
     } 
    } 
    } 
}