Question

我的代码只打印空格而不是每次都提示输入行号和列号，看看空格是否为空。我必须为计算机播放器生成随机数，但不确定我是否正确

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

如果任何电路板值为空，checkTie函数将返回true（严格来说，不是＆＃34; X＆＃34;或＆＃34; O＆＃34;）。 computerPlay函数仅在checkTie的返回值为false时才会继续。其中一个（但不是两个）的逻辑应该被颠倒。我建议更改isTied函数的名称以避免混淆，并反转该函数中的逻辑。这是我的建议（也会将while替换为if，因为它只能运行一次）：

 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);
        }
     }
  }
 }

带有计算机侧随机数发生器的2D TicTacToe

1 个答案: