php小编草莓为您带来最新的java问答专栏,本期将探讨java中处理tic tac toe(井字棋)游戏的相关问题。无论您是初学者还是有经验的开发者,都可以在这里找到有关java中处理井字棋游戏的实用技巧和解决方案。让我们一起深入了解这个有趣的话题,提升自己在java编程领域的技能!
我目前正在处理中开发一个简单的 tic-tac-toe 游戏,但我在两个方面遇到了问题:
展示位置问题: x 和 o 符号未正确放置在网格上。位置似乎是随机的,我不确定为什么。每当我点击一个框时,它似乎没有放置在那个协调的框中
鼠标点击问题: 我已经设置了左键单击放置 x,右键单击放置 o,但它似乎没有按预期工作。
这是我当前的代码:
// Declare a 3x3 grid of TicTacToeBox objects TicTacToeBox[][] grid = new TicTacToeBox[3][3]; // gameStatus: // 0 - Display Home screen // 1 - Display Tic Tac Toe grid // 2 - Display Game over screen int gameStatus = 0; // Determine which player's turn it is int currentPlayer = 1; void setup() { size(600, 600); displayHomeScreen(); } void draw() { // Draw the appropriate screen based on gameStatus if (gameStatus == 1) { background(255); displayGrid(); } else if (gameStatus == 2) { background(0); displayGameOver(); } } void mousePressed() { // Check the gameStatus and respond to mouse clicks accordingly if (gameStatus == 1) { float boxSize = width / 3.0; int col = floor(mouseX / boxSize); int row = floor(mouseY / boxSize); // Check if the box is valid and empty if (isValidBox(row, col) && grid[row][col].symbol == ' ') { // Place X or O based on the mouse button and currentPlayer if (mouseButton == LEFT && currentPlayer == 1) { grid[row][col].symbol = 'X'; currentPlayer = 2; } else if (mouseButton == RIGHT && currentPlayer == 2) { grid[row][col].symbol = 'O'; currentPlayer = 1; } } } else if (gameStatus == 0 && mouseX > 250 && mouseX < 350 && mouseY > 250 && mouseY < 300) { // Transition to the game screen when PLAY is clicked gameStatus = 1; } } void displayGrid() { float boxSize = width / 3.0; // Loop through the grid and draw each TicTacToeBox for (int row = 0; row < 3; row++) { for (int col = 0; col < 3; col++) { if (grid[row][col] == null) { grid[row][col] = new TicTacToeBox(row, col, boxSize, ' '); } grid[row][col].draw(); } } } // Check if the row and column are clear to place boolean isValidBox(int row, int col) { return row >= 0 && row < 3 && col >= 0 && col < 3; } void displayHomeScreen() { // Display the home screen with instructions background(255); fill(0); textAlign(CENTER, TOP); textSize(50); text("Tic-Tac-Toe", width/2, 100); textSize(25); fill(0); text("Click PLAY to start", width/2, 200); noFill(); rect(250, 250, 100, 50); textSize(20); fill(0); text("PLAY", width/2, 265); } void displayGameOver() { // Display the game over screen with a prompt to play again fill(255, 0, 0); textAlign(CENTER, TOP); textSize(50); text("GAME OVER!", width/2, 100); textSize(25); fill(0, 0, 255); text("CLICK TO PLAY AGAIN", width/2, 200); } class TicTacToeBox { float x; float y; float boxSize; char symbol = ' '; TicTacToeBox(float x, float y, float boxSize, char symbol) { this.x = x; this.y = y; this.boxSize = boxSize; this.symbol = symbol; } void draw() { stroke(0); noFill(); rect(x * boxSize, y * boxSize, boxSize, boxSize); textAlign(CENTER, CENTER); textSize(32); fill(0); float symbolX = x * boxSize + boxSize/2; float symbolY = y * boxSize + boxSize/2; text(symbol, symbolX, symbolY); } }
以下源代码演示了使用 rect 类数组创建 3x3 网格来跟踪鼠标按下情况。也许类似的技术可以用在您的游戏中。
rect[] r; final int _numcols = 3; final int _numrows = 3; final int _wndw = 400; final int _wndh = 400; class rect { int x, y, w, h; boolean leftpressed = false; boolean rightpressed = false; // constructor rect(int xpos, int ypos, int wide, int ht) { x = xpos; y = ypos; w = wide; h = ht; } void display(int id) { fill(255); // background color rect(x, y, w, h); fill(0); // text color textsize(18); text(str(id), x + w - 18, y + 18); } } void rectgrid(int left, int top, int w, int h, int vg, int hg) { int id = 0; // build by row for (int k = 0; k < _numrows; k++) { for (int j = 0; j < _numcols; j++) { int x = left + j*(w+vg); int y = top + k*(h+hg); r[id] = new rect(x, y, w, h); id++; } } } void setup() { size(_wndw, _wndh); background(0, 0, 245); r = new rect[_numrows*_numcols]; rectgrid(0, 0, _wndw/_numcols, _wndh/_numrows, 0, 0); } void draw() { for (int i = 0; i < r.length; i++) { r[i].display(i); // display each object if (r[i].leftpressed == true) { text("x", r[i].x + r[i].w/2, r[i].y + r[i].h/2); } if (r[i].rightpressed == true) { text("o", r[i].x + r[i].w/2, r[i].y + r[i].h/2); } } } void mousepressed() { for (int i = 0; i < r.length; i++) { if ((mousex >= r[i].x) && (mousex <= r[i].x +r[i].w) && (mousey >= r[i].y) && (mousey <= r[i].y + r[i].h)) { println("id =", i); if (mousebutton == left) { r[i].leftpressed = true; } if (mousebutton == right) { r[i].rightpressed = true; } } } }
建议: 您的网格不需要二维数组;一维数组工作得很好并且不太复杂。我将向网格类添加两个布尔值(leftpressed 和 rightpressed)并删除符号参数。网格类的“draw()”方法可能应该重命名为“display()”之类的方法,因为“draw”是处理中的关键字,可以避免混淆。可以使用下面所示的技术安全地删除方法 displaygrid() 和 isvalidbox()。主要代码更改应该在 mousepressed() 中,因为它无法正常工作。循环遍历网格中的每个框将正确捕获鼠标按钮单击,此时您可以检查单击的是鼠标右键还是左鼠标按钮,并且可以将相应的布尔值设置为“true”。然后,主“draw()”将使用此信息绘制“x”或“o”。我知道这听起来很多,但这些更改是解决您的问题的一种方法。您修改后的源代码如下所示:
// Declare a 3x3 grid of TicTacToeBox objects Grid[] g = new Grid[9]; // gameStatus: // 0 - Display Home screen // 1 - Display Tic Tac Toe grid // 2 - Display Game over screen int gameStatus = 0; // Determine which player's turn it is int currentPlayer = 1; class Grid { float x; float y; float boxSize; boolean leftPressed = false; boolean rightPressed = false; Grid(float x, float y, float boxSize) { this.x = x; this.y = y; this.boxSize = boxSize; } void display() { stroke(0); noFill(); rect(x, y, boxSize, boxSize); } } void setup() { size(600, 600); displayHomeScreen(); // initialize array float boxSize = width / 3.0; int id = 0; for (int k = 0; k < 3; k++) { for (int j = 0; j < 3; j++) { float x = j*boxSize; float y = k*boxSize; g[id] = new Grid(x, y, boxSize); id++; } } } void draw() { // Draw the appropriate screen based on gameStatus if (gameStatus == 1) { background(255); for (int i = 0; i < g.length; i++) { g[i].display(); // Display each object if (g[i].leftPressed == true) { text("X", g[i].x + g[i].boxSize/2, g[i].y + g[i].boxSize/2); } if (g[i].rightPressed == true) { text("O", g[i].x + g[i].boxSize/2, g[i].y + g[i].boxSize/2); } } } else if (gameStatus == 2) { background(0); displayGameOver(); } } void mousePressed() { // Check the gameStatus and respond to mouse clicks accordingly if (gameStatus == 1) { for (int i = 0; i < g.length; i++) { if ((mouseX >= g[i].x) && (mouseX <= g[i].x +g[i].boxSize) && (mouseY >= g[i].y) && (mouseY <= g[i].y + g[i].boxSize)) { println("id =", i); if (mouseButton == LEFT) { g[i].leftPressed = true; } if (mouseButton == RIGHT) { g[i].rightPressed = true; } } } } else if (gameStatus == 0 && mouseX > 250 && mouseX < 350 && mouseY > 250 && mouseY < 300) { // Transition to the game screen when PLAY is clicked gameStatus = 1; } } void displayHomeScreen() { // Display the home screen with instructions background(255); fill(0); textAlign(CENTER, TOP); textSize(50); text("Tic-Tac-Toe", width/2, 100); textSize(25); fill(0); text("Click PLAY to start", width/2, 200); noFill(); rect(250, 250, 100, 50); textSize(20); fill(0); text("PLAY", width/2, 265); } void displayGameOver() { // Display the game over screen with a prompt to play again fill(255, 0, 0); textAlign(CENTER, TOP); textSize(50); text("GAME OVER!", width/2, 100); textSize(25); fill(0, 0, 255); text("CLICK TO PLAY AGAIN", width/2, 200); }
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