Question

我正在尝试在java上编写一个tic-tac-toe游戏。

这是我到目前为止的代码（有效），我真的不明白如何绘制圆圈而不是交叉：

public static void main(String[] args) {
    //Declare variables
    int[][] board = new int[3][3];
    int turns = 0, x, y, mode, score = 0;
    String dec;
    Scanner s = new Scanner(System.in);
    //Constants
    final int EMPTY = 0;
    final int FILL_O = 1;
    final int FILL_X = 2;

    //Ask user for game type
    System.out.println("TicTacToe\n=========");
    System.out.print("(1) or (2) players? ");
    mode = s.nextInt();

    //Draw the board
    StdDraw.setScale(0, 3);
    StdDraw.setPenRadius(0.008);
    for (int c = 0; c <= 3; c++) {
        StdDraw.line(c, 0, c, 3);
    }
    for (int c = 0; c <= 3; c++) {
        StdDraw.line(0, c, 3, c);
    }

    //Two-player Game
    if (mode == 2) {
        //X goes first
        StdDraw.setPenColor(Color.BLUE);
        while (turns < 9) {
            //Draw Os on odd turns
            if (turns % 2 == 1) {
                if (StdDraw.mousePressed()) {
                    x = (int) Math.floor(StdDraw.mouseX());
                    y = (int) Math.floor(StdDraw.mouseY());
                    if (board[x][y] == EMPTY) {
                        StdDraw.circle(x + 0.5, y + 0.5, 0.5);
                        board[x][y] = FILL_O;
                        mode = FILL_O; //Since mode is no longer in use, I 
    reinstate it to track turns
                        StdDraw.show(500);
                        turns++;
                    }
                }
            }

Answer 1

而不是使用StdDraw.text(x, y, board[x][y] + "");测试board[x][y]的值，并调用在坐标x或y处绘制正方形或圆形的函数。

为了说明这一点，请替换

for (int x = 0; x < board.length; x++) {
    for (int y = 0; y < board[x].length; y++) {
        StdDraw.text(x, y, board[x][y] + "");
    }
}

与

for (int x = 0; x < board.length; x++) {
    for (int y = 0; y < board[x].length; y++) {
        if (board[x][y] == 'X') {
            StdDraw.filledSquare(x, y, 0.5 );
        }
        else if (board[x][y] == 'O') {
            StdDraw.filledCircle(x, y, 0.5 );
        }
    }
}

进行一些测试以找到半径的漂亮值，0.5可能不会很好。

Answer 2

该API提供了方法filledCircle和filledRectangle。

public static void  filledCircle(double x, double y, double r) ;

绘制半径为r的实心圆，以（x，y）为中心。

参数：
      - x圆心的x坐标       - y圆心的y坐标
      - r圆的半径

public static void  filledRectangle(double x, double y, double halfWidth, double halfHeight);

绘制一个给定半宽和半高的实心矩形，以（x，y）为中心。

参数：
      - x矩形中心的x坐标       - y矩形中心的y坐标
      - halfWidth是矩形的宽度的一半       - halfHeight是矩形高度的一半

或者看filledSquare ...基本相同的rectamgle概念，但参数值相同..

像

这样的东西

//CENTER
int x = (int) Math.floor(StdDraw.mouseX());
int y = (int) Math.floor(StdDraw.mouseY());
int r = 10; //radius
StdDraw.filledcircle(x, y, r);

并且

//CENTER
int x = (int) Math.floor(StdDraw.mouseX());
int y = (int) Math.floor(StdDraw.mouseY());
int width = 10; 
int height = 10; 
//The method use the center of the shape and half of both "length"
StdDraw.filledRectangle(x, y, height  / 2, height / 2);

Java中的Tic-Tac-Toe游戏

2 个答案: