需要更好的逻辑TicTacToe

我正在使用Java Swing实现一个TicTacToe GUI应用程序。 获胜的当前逻辑是:

JButton[] button = new JButton[9]; boolean win = false; //logic for player's win if (button[0].getText() == button[1].getText() && button[1].getText() == button[2].getText() && button[0].getText() != "") { Won = true; } else if (button[3].getText() == button[4].getText() && button[4].getText() == button[5].getText() && button[3].getText() != "") { Won = true; } else if (button[6].getText() == button[7].getText() && button[7].getText() == button[8].getText() && button[6].getText() != "") { Won = true; } else if (button[0].getText() == button[3].getText() && button[3].getText() == button[6].getText() && button[0].getText() != "") { Won = true; } else if (button[1].getText() == button[4].getText() && button[4].getText() == button[7].getText() && button[1].getText() != "") { Won = true; } else if (button[2].getText() == button[5].getText() && button[5].getText() == button[8].getText() && button[2].getText() != "") { Won = true; } else if (button[0].getText() == button[4].getText() && button[4].getText() == button[8].getText() && button[0].getText() != "") { Won = true; } else if (button[2].getText() == button[4].getText() && button[4].getText() == button[6].getText() && button[2].getText() != "") { Won = true; } 

这看起来有点笨拙。 玩家的胜利是否有更好的逻辑?

所以,你所拥有的(基本上)是一个2D矩阵(好吧,它是一个虚拟的,但这使得思考更简单)……

您需要的是一种方法,您可以从给定的行和列开始搜索此矩阵的三个单元格。

因此,给定一个起点,您还需要提供有关您要搜索哪个方向的信息(有时候,您想要向后搜索),可能是……

 public boolean matrixWin(int row, int col, int rowDelta, int colDelta) { boolean win = false; String value = button[(row * 3) + col].getText(); if (!value.isEmpty()) { win = true; for (int count = 1; count < 3; count++) { row += rowDelta; col += colDelta; String test = button[(row * 3) + col].getText(); if (test.isEmpty() || !test.equals(value)) { win = false; break; } } } return win; } 

这基本上得到第一个单元格的值,如果它不是空的,它会根据delta值开始在矩阵中移动,并检查每个其他单元格以查看它是否为空或是否与第一个单元格的值匹配。

好吧,这很酷,但我想懒得尝试设置我想检查的每个排列,所以相反,我会做一些辅助方法......

基本上,你要检查三行是否为水平获胜,三列为垂直获胜和左或右对角线......

 public boolean horizontalWin(int row) { return matrixWin(row, 0, 0, 1); } public boolean verticalWin(int col) { return matrixWin(0, col, 1, 0); } public boolean leftDiagonalWin() { return matrixWin(0, 0, 1, 1); } public boolean rightDiagonalWin() { return matrixWin(0, 2, 1, -1); } 

但即便如此,除非我想知道哪一行/对角线/对角线真的赢了,否则我会更容易......

 public boolean horizontalWins() { int row = 0; boolean win = false; do { win = horizontalWin(row); row++; } while (row < 3 && !win); return win; } public boolean verticalWins() { int col = 0; boolean win = false; do { win = verticalWin(col); col++; } while (col < 3 && !win); return win; } 

然后你可以开始......

 public boolean didWin() { return horizontalWins() || verticalWins() || leftDiagonalWin() || rightDiagonalWin(); } 

是的,一个方法调用,但你仍然有权确定“如何”