TicTacToe.java checkWinner 方法存在问题

Question

所以我创建了一个 checkWinner 方法，使用 'row' 和 'col' 私有变量，这样我就可以找到二维数组中的 'curPlayer' 位置。

import java.util.Scanner;

public class TicTacBoard
{
  private char[][] board;  // 2-D array of characters
  private char curPlayer; // the player whose turn it is (X or O)
  // added so I can locate the current player location in the board
  private int row;
  private int col;

  // Constructor: board will be size x size
  public TicTacBoard(int size)
  {
    board = new char[size][size];

    // initialize the board with all spaces:
    for(row=0; row < board.length; row++)
      for(col=0; col < board[row].length; col++)
        board[row][col] = ' ';

    curPlayer = 'X';  // X gets the first move
  }

  public void playGame()
  {
    display();
    do
    {
      takeTurn();
      display();
    }while(!checkWinner(row, col));
  }

  ///////  display  ////////
  // Display the current status of the board on the
  // screen, using hyphens (-) for horizontal lines
  // and pipes (|) for vertical lines.
  public void display()
  {
    System.out.println();
    dispRow(0);
    System.out.println("-----");
    dispRow(1);
    System.out.println("-----");
    dispRow(2);
    System.out.println();
  }

  // Display the current status of row r of the board
  // on the screen, using hyphens (-) for horizontal
  // lines and pipes (|) for vertical lines.
  private void dispRow(int r)
  {
    System.out.println(board[r][0] + "|" + board[r][1]
                       + "|" + board[r][2]);
  }

  ///////  takeTurn  ////////
  // Allow the curPlayer to take a turn.
  // Send output to screen saying whose turn
  // it is and specifying the format for input.
  // Read user's input and verify that it is a
  // valid move.  If it's invalid, make them
  // re-enter it.  When a valid move is entered,
  // put it on the board.
  public void takeTurn()
  {
    Scanner scan = new Scanner(System.in);
    int row, col;
    boolean invalid;

    do{
      invalid = false; // assume correct entry
      System.out.println("It is now " + curPlayer + "'s turn.");
      System.out.println("Please enter your move in the form row column.");
      System.out.println("So 0 0 would be the top left, and 0 2 would be the top right.");
      row = scan.nextInt();
      col = scan.nextInt();

      if(row < 0 || col < 0 || row > 2 || col > 2)
      {
        System.out.println("Invalid entry: row and column must both be between 0 and 2 (inclusive).");
        System.out.println("Please try again.");
        invalid = true;
      }
      else if(board[row][col] != ' ')
      {
        System.out.println("Invalid entry: Row " + row + " at Column " + col
                           + " already contains: " + board[row][col]);
        System.out.println("Please try again.");
        invalid = true;
      }
    }while(invalid);
    // Now that input validation loop is finished, put the move on the board:
    board[row][col] = curPlayer;

    // Switch to the other player (take turns):
    if(curPlayer == 'X')
      curPlayer = 'O';
    else
      curPlayer = 'X';
  }

  // If the game is over, print who won (if anyone),
  // and return true.  If the game is not over, return false.
  public boolean checkWinner(int row, int col)
  {
    // YOUR CODE GOES HERE
    int x = row;
    int y = col;
    // board length is always 3 here
    // check winner on column
    for (int i = 0; i < board.length; i++) {
      if (board[x][i] != curPlayer)
        break;
      if (i == board.length - 1)
        System.out.println("Player " + curPlayer + " wins!");
        return true;
    }
    
    //check winner on row
    for (int i = 0; i < board.length; i++) {
      if (board[i][y] != curPlayer)
        break;
      if (i == board.length - 1)
        System.out.println("Player " + curPlayer + " wins!");
        return true;
    }

    // checks winner on diagonal up 
    if (x == y) {
      for (int i = 0; i < board.length; i++) {
        if (board[i][i] != curPlayer)
          break;
        if (i == board.length - 1)
          System.out.println("Player " + curPlayer + " wins!");
          return true; 
      }
    }

    // check winner on diagonal down
    if (x + y == board.length - 1){
      for (int i = 0; i < board.length; i++) {
        if (board[i][(board.length-1)-i] != curPlayer)
          break;
        if (i == board.length - 1)
          System.out.println("Player " + curPlayer + " wins!");
        return true;
      }
    }

    // checks if board is full
    for (int i = 0; i < board.length; i++) {
      for (int j = 0; j < board.length; j++) {
        if (board[i][j] == '-')
          System.out.println("Nobody won, game ends in a draw!");
          return true;
      }
    }
    return false;
  }
}

代码有效，但我在检查时得到了这个：

 | | 
-----
 | | 
-----
 | | 

It is now X's turn.
Please enter your move in the form row column.
So 0 0 would be the top left, and 0 2 would be the top right.
2 0

 | | 
-----
 | | 
-----
X| | 

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index 3 out of bounds for length 3
    at TicTacBoard.checkWinner(TicTacBoard.java:126)
    at TicTacBoard.playGame(TicTacBoard.java:43)
    at Main.main(Main.java:14)

我以为板子的长度总是3，位置在0到3之间。这个错误有什么解决办法吗？有没有更有效的方法来做到这一点？请告诉我！

Answer 1

您有一个“隐藏”问题 - 也就是说，您在 takeTurn 方法中使用局部变量隐藏实例字段 row 和 col。

在当前状态下...

// Constructor: board will be size x size
public TicTacBoard(int size) {
    board = new char[size][size];

    // initialize the board with all spaces:
    for (row = 0; row < board.length; row++) {
        for (col = 0; col < board[row].length; col++) {
            board[row][col] = ' ';
        }
    }

    curPlayer = 'X';  // X gets the first move
}

在构造函数有运行后，row和col将是3，但在takeTurn中，你定义row和col 作为局部变量...

public void takeTurn() {
    Scanner scan = new Scanner(System.in);
    int row, col;
    boolean invalid;

这意味着，当您在 playGame 方法中调用 checkWinner 时...

public void playGame() {
    display();
    do {
        takeTurn();
        display();
    } while (!checkWinner(row, col));
}

您正在传递实例字段值（3/3）并且一切都中断了。

因此，“快速”解决方案是从 takeTurn

中删除 row/col 的本地声明

public void takeTurn() {
    Scanner scan = new Scanner(System.in);
    //int row, col;
    boolean invalid;

您也可以在构造函数中修复此问题，但要制作 row/col 局部变量

for (int row = 0; row < board.length; row++) {
    for (int col = 0; col < board[row].length; col++) {
        board[row][col] = ' ';
    }
}

但在某些时候，您需要更新播放器的 row/col 值，但我可能会考虑从 takeTurn 传回此信息而不是尝试使用实例字段。

Answer 2

您的 if 语句中还有一个微妙但常见的错误。如果没有方括号、{和}，当上面的条件语句为真时，只会执行紧跟在if语句之后的行。但是，您的缩进表示您期望不同的行为。

例如，您的第一个 for 循环是：

for (int i = 0; i < board.length; i++) {
  if (board[x][i] != curPlayer)
    break;
  if (i == board.length - 1)
    System.out.println("Player " + curPlayer + " wins!");
    return true;
}

这里，当if语句为真时，只执行System.out.println()行。 return true; 语句的缩进表示您希望它仅在 println() 中运行，仅当条件为真时。

return true; 行不依赖于前面的 if 语句，因为它不在括号内并且 if 语句仅运行行紧随其后。这意味着 for 循环只会运行一次迭代，因为 return 行是 STAND-ALONE 并且每次都执行，而不管那些 if语句求值。

您应该始终，始终，始终在您的 if 语句中添加方括号，即使它们是“one-liners”。考虑到这一点，我希望它看起来更像：

for (int i = 0; i < board.length; i++) {
  if (board[x][i] != curPlayer) {
    break;
  }
  if (i == board.length - 1) {
    System.out.println("Player " + curPlayer + " wins!");
    return true;
  }
}

现在 return 行仅在前面的 if 语句为真时执行。

TicTacToe.java checkWinner 方法存在问题

TicTacToe.java issues with checkWinner method

java

multidimensional-array