Question

对于我的作业，我必须制作一个方法来检查TicTacToeArray变量并确定是否有人赢了。特别是，如果有任何列，行或专业，则有赢家游戏板的对角线完全由Xs或Os填充。当检测到胜利者时 scoreTTT（）应将获胜者变量设置为“X”或“O”，具体取决于谁赢了。如果X或O都没有获胜，则获胜者变量应保留“*”。

到目前为止，我有这个：

公共课TicTacToe {

public static void main(String[] args){



}

//state variables
static char[][] TicTacToeArray;  //the game board
static int step = 0;             //the current step number
static char winner = '*';        //who has won (X/O/*)        *=nobody
static char player = 'X';        //whose turn it is (X/O)   *=nobody

//Creates a game board of size n x n and resets state variables to
//their initial conditions for a new game.
public static void startTTT(int n){
    TicTacToeArray = new char [n][n];
    for(int i = 0; i < n; i++){
        for(int j=0; j < n; j++){
            TicTacToeArray [i][j] = '*';
    }
}

step = 0;
winner = '*';
player = 'X';

}

public static void displayTTT(){
    String row;

    int n = TicTacToeArray.length;

    //now I'm priting row0
    row = "       Column";
    System.out.println(row);

    //row 1
    row = "       ";
    for (int i=0; i<n; i++){
        row = row + " "+ i;

    }
    row = row + "  TicTacTow";
    System.out.println(row);

    //row 2
    row = "      +";
    for (int i=0; i<n; i++){
    row = row + "--";

    }

    System.out.println(row +"  Step = " + step);

    //row 3
    row = "    0 |" ; 
    for (int i=0; i<n; i++){
    row = row + " " + TicTacToeArray [0][i];
    }
    System.out.println(row + "  Player = " + player);

    //row 4
    row = "Row 1 |";
    for (int i=0; i<n; i++){
    row = row + " " + TicTacToeArray[1][i];
    }
    System.out.println(row);


    //row 5
    row = "";
    for( int i=2;i<n;i++){
    row = "    " + i + " |" ;
        for( int j=0; j < n; j++){
    row += " " + TicTacToeArray[i][j];
    if (j == n)
    System.out.println(row);

    }
    if(i == n-1)
    row += "  Winner = " + winner;
    System.out.println(row);
    }



}

//Updates a position on the game board, increments the step counter,
//and toggles the player from X to O (or vica versa). This method should
//test for invalid input (see assignment document) before changing
//the game state. If no error is encountered, it performs the update
//and returns true. Otherwise it returns false.
public static boolean updateTTT(char sym, int row, int col){
    if (sym != 'X' && sym != 'O'){
            return false;
    }
    if(row < 0 || col < 0 || row >= TicTacToeArray.length || col >= TicTacToeArray.length){
            return false;
    }   



    if (TicTacToeArray[row][col] == '*')
    TicTacToeArray [row][col] = sym;
    else
    return false;

    // toggle player
    for(;;){
    if (player =='X'){
    player = 'O';
    break;
    }
    if(player == 'O'){
    player = 'X';
    break;
    }
    }
    //inc step count
    step +=1;


    return true;
}

//（这是注释掉的/我到目前为止对于这个方法，其余的代码应该工作） // public static void scoreTTT（）{

//for(int i=0; i < TicTacToeArray.length; i++){
    //for(int j =0; j < TicTacToeArray.length; j++)
    //if (TicTacToeArray[i][j] ==  TicTacToeArray[i][j+1])




}

我认为我需要制作3个不同的嵌套循环来检查对角线然后检查行/列，它们还必须检查任何数组大小的完整行/列，例如4 x 4.我只是我不确定如何让循环遍历整个行/列/对角线。谢谢你的帮助。

Answer 1

你是对的，你有三种不同的情况。

首先检查所有列，然后检查所有行，然后检查对角线。

第一个应该是一个循环（对于每一列，检查每一行）。第二个应该像第一个（只是反向列和行）。第三种情况很简单，你只有两个选项，左上角到右下角，右上角到左下角。只需确保所有相应的框都相同。

我会提供更具体的细节，但这是作业。

Answer 2

char[][] board = new char[][] {
        {'x', 'o', 'x'},
        {'x', 'o', 'o'},
        {'o', 'o', 'x'}
};

//check rows/cols
for (int i = 0; i < board.length; i++) {
    String row = "";
    String col = "";
    for (int j = 0; j < board.length; j++) {
        row += board[i][j];
        col += board[j][i];
    }
    System.out.println("Row: " + row);
    System.out.println("Col: " + col);
}

//check diagonals -- the logic in this loop could be folded into the previous one
String diag1 = ""; //top-left to bottom-right
String diag2 = ""; //bottom-left to top-right
for (int i = 0; i < board.length; i++) {
    diag1 += board[i][i];
    diag2 += board[board.length-i-1][board.length-i-1];
}
System.out.println("Diag1: " + diag1);
System.out.println("Diag2: " + diag2);

你必须弄清楚如何检查获胜者。

如何确定TicTacToe（Java）的获胜者

2 个答案: