Question

这是给出tic tac toe board的获胜者名字（X或Y）的方法。如果没有赢家，则返回空。

玩家X先行：

但在某种特定情况下，如果不起作用： x次：2,0 o播放：1,0 x次：2,1 o戏剧：1,1 x播放2,2 （此时'X'应该获胜但没有任何反应）

o play：1,2 （此时“O”应该获胜但没有任何反应）

x播放任何动作

O被宣布为胜利者。

我无法弄清楚这个漏洞。

public static char checkWinner（char [] [] array）{

// check for row winners
for (int row = 0; row < array.length; row++) {
  // if all three columns are the same then that player wins!
  if (array[row][0] == array[row][1] && array[row][1] == array[row][2]) {
    // return the winner!
    return array[row][0];
  }
}

// check for column winners
for (int col = 0; col < array.length; col++) {
  // if all three rows are the same then that player wins!
  if (array[0][col] == array[1][col] && array[1][col] == array[2][col]) {
    // return our winner!
    return array[0][col];
  }
}

// check for diag winners
if (array[0][0] == array[1][1] && array[1][1] == array[2][2]) {
  return array[0][0];
}

if (array[0][2] == array[1][1] && array[1][1] == array[2][0]) {
  return array[0][2];
}

// otherwise just return an empty character
return ' ';

}

如果有帮助，这是整个代码：

import java.util.Scanner;

public class TicTacToe {

  public static void main(String[] args) {
    // create an empty board
    char[][] board = { { ' ', ' ', ' ' }, { ' ', ' ', ' ' }, { ' ', ' ', ' ' } };

    Scanner input = new Scanner(System.in);

    // start off with player 'X'
    char player = 'X';

    while (true) {
      printBoard(board);

      // ask the player for a location
      System.out.println("Welcome, player " + player);
      System.out.print("Enter a row: ");
      int row = input.nextInt();
      System.out.print("Enter a col: ");
      int col = input.nextInt();

      // check to see if this row is already taken
      if (board[row][col] != ' ') {
        System.out.println("\nThat spot is already taken!");
      } else {
        // otherwise assign the player's token to this spot
        board[row][col] = player;

        // check to see if someone won
        char testWinner = checkWinner(board);

        // do we have a winner?
        if (testWinner != ' ') {
          printBoard(board);
          System.out.println("We have a winner! " + testWinner);
          break;
        } else {
          // switch to the next player
          if (player == 'X') {
            player = 'O';
          } else {
            player = 'X';
          }
        }
      }
    }
  }

  public static char checkWinner(char[][] array) {
    // check for row winners
    for (int row = 0; row < array.length; row++) {
      // if all three columns are the same then that player wins!
      if (array[row][0] == array[row][1] && array[row][1] == array[row][2]) {
        // return the winner!
        return array[row][0];
      }
    }

    // check for column winners
    for (int col = 0; col < array.length; col++) {
      // if all three rows are the same then that player wins!
      if (array[0][col] == array[1][col] && array[1][col] == array[2][col]) {
        // return our winner!
        return array[0][col];
      }
    }

    // check for diag winners
    if (array[0][0] == array[1][1] && array[1][1] == array[2][2]) {
      return array[0][0];
    }

    if (array[0][2] == array[1][1] && array[1][1] == array[2][0]) {
      return array[0][2];
    }

    // otherwise just return an empty character
    return ' ';
  }

  public static void printBoard(char[][] array) {
    System.out.println("\n----------");
    for (int row = 0; row < array.length; row++) {
      System.out.print("|");

      for (int col = 0; col < array.length; col++) {
        System.out.print(array[row][col] + " |");
      }

      System.out.println("\n----------");
    }

    System.out.println();
  }

}

Answer 1

问题在于，当您看到当前组合的所有三个字符都相同时，您不会检查它是否为三个空字符。现在，在您的示例中，它将提供' '，因为x = 0的列填充了' ' s。这种检查所有组合的方式也有点长，我会这样做：

public static char checkWinner(char[][] array) {
    int[][] combinations = {
        {0, 1, 2},
        {3, 4, 5},
        {6, 7, 8},
        {0, 3, 6},
        {1, 4, 7},
        {2, 5, 8},
        {0, 4, 8},
        {2, 4, 6}
    };
    for (int[] co : combinations) {
        char[] cs = new char[] {
            array[co[0] % 3][co[0] / 3],
            array[co[1] % 3][co[1] / 3],
            array[co[2] % 3][co[2] / 3]
        };
        if (cs[0] != ' ' && cs[0] == cs[1] && cs[1] == cs[2]) {
            return cs[0];
        }
    }
    return ' ';
}

Answer 2

checkWinner是否条件忽略空格字符。并且行块在第一次尝试时返回，因为顶行是所有空格。

Tic Tac Toe Board - GetWinner

2 个答案: