Question

嗨大家我真的需要帮助完成我的任务。我试过这样做，但我无法弄清楚。

启动板：

结果欲望：

但每次执行时，只要有胜利者，我总会得到以下信息：

最后，如何计算赢得最多的玩家的胜率？

限制：

您必须在井字游戏板上使用2D阵列。（我知道我没有使用2D数组，因为我无法弄清楚如何使编译器接受来自2D数组的1-9输入。）
您无法使用类来解决此问题。

如果您愿意，可以使用静态方法。

import java.util.Scanner;
public class TicTacToe {    public static String answer = "yes";
public static Scanner input = new Scanner(System.in);
public static String getName(int noPlayer) {
    System.out.print("Enter name of Player " + noPlayer + ": ");
    return input.next();
}

public static int getMove(String board[], String player) {
    printBoard(board);
    System.out.print("Enter move for " + player + ": ");
    int move = input.nextInt() - 1;
    while (moveTaken(board, move)) {
        System.out.println("Move taken.");
        System.out.print("Enter move for " + player + ": ");
        move = input.nextInt() - 1;
    }
    return move;
}
public static String gameResult(String board[]) {
    final int checkWin[][] = {{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[] i: checkWin) {
        if (board[i[0]].equals(board[i[1]]) && board[i[0]].equals(board[i[2]]) && board[i[1]].equals(board[i[2]])) {
            if (board[i[0]].equals("O")) {
                return "X wins";
            }
            else {
                return "O wins";
            }
        }
    }
    if (!board[0].equals("1") && !board[1].equals("2") && !board[2].equals("3") &&
        !board[3].equals("4") && !board[4].equals("5") && !board[5].equals("6") &&
        !board[6].equals("7") && !board[7].equals("8") && !board[8].equals("9")) {
        return "draw";
    }
    return "not completed";
}
public static boolean moveTaken(String board[], int move) {
    if (board[move].equals("O") || board[move].equals("X")) {
        return true;
    }
    return false;
}
public static void printBoard(String board[]) {
    System.out.println(" " + board[0] + " " + board[1] + " " + board[2] +
                       "\n" +
                       " " + board[3] + " " + board[4] + " " + board[5] +
                       "\n" +
                       " " + board[6] + " " + board[7] + " " + board[8]);
}
public static void conclusion(String result, String pO, String pX) {
    if (result.equals("O wins")) {
        System.out.println(pO + " wins!");

    }
    else if (result.equals("X wins")) {
        System.out.println(pX + " wins!");
    }
    else {
        System.out.println("Draw.");
    }
}
public static void main(String args[]) {

      String X = getName(1);
      String O = getName(2);

  int count = 1;
  do{   

    String nextPlayer = X;
    String board[] = {"1", "2", "3", "4", "5", "6", "7", "8", "9"};
    int move;
        while (gameResult(board).equals("not completed")) {
            move = getMove(board, nextPlayer);
            if (nextPlayer == X) {
                board[move] = "X";
                nextPlayer = O;
            }
            else {
                board[move] = "O";
                nextPlayer = X;
            }
        }
    System.out.println("game up to date: " + count);

   conclusion(gameResult(board), X, O);

    System.out.println("Do you want to play again? yes or no");
     answer = input.next();
     count++; 


}
    while(answer.equalsIgnoreCase("yes"));




  }
}

Answer 1

您只能在执行移动之前在getMove中打印电路板。因此，永远不会显示最后一步。

您应该在游戏结束时添加对printBoard的通话。

这里有一个可以添加它的地方：

...
System.out.println("game up to date: " + count);
printBoard(board);
conclusion(gameResult(board), X, O);
...

无法在我的java tictactoe游戏中获得正确的输出

1 个答案: