Java:连接4中对角线
我最近做了一个Connect4游戏，我的Connect4没有赢得比赛时，它是对角向右连接。它只适用于一些组合，当它对角连接到左边时。坐标:左上角：(0,0)，左下角：(5,0)，右上角：(0,6)，右下角：(5,6)。Connect4板是6乘7。
问题:向左对角连接的只适用于某些组合。但是，没有一个连接是对角连接到正确的工作。
/** A method of winning diagonally towards the left side when playing connect 4 two player*/
/** Giving the new method with all the possible possibilities for a user to win diagonally-left */
public static void diagWinningLeft() {
for (int x = 5; x > 2; x--) { // Checks to see if same colored pegs are lining diagonally to the left
for (int y = 6; y > 3; y--) {
if (board[x][y] == 1 && board[x - 1][y - 1] == 1 && board[x - 2][y - 2] == 1 && board[x - 3][y - 3] == 1) {
JOptionPane.showMessageDialog(null, playerNames[0]+" has connected four diagonally-left in a row in " +(countForRed)+ " turns!");
b.drawLine(x,y,x-3,y-3);
}
if (board[x - 1][y - 1] == 1 && board[x - 2][y - 2] == 1 && board[x - 3][y - 3] == 1 && board[x - 4][y - 4] == 1) {
JOptionPane.showMessageDialog(null, playerNames[0]+" has connected four diagonally-left in a row in " +(countForRed)+ " turns!");
b.drawLine(x,y,x-3,y-3);
}
if (board[x][y] == 2 && board[x - 1][y - 1] == 2 && board[x - 2][y - 2] == 2 && board[x - 3][y - 3] == 2) {
JOptionPane.showMessageDialog(null, playerNames[1]+ " has connected four diagonally-left in a row in " +(countForYellow)+ " turns!");
b.drawLine(x,y,x-3,y-3);
}
}
}
}
/** Another method of winning diagonally towards the right side when playing connect 4 two player*/
/** Giving the new method with all the possible possibilities for a user to win diagonally-right*/
public static void diagWinningRight() {
for (int x = 0; x < 2; x++) { // Check to see if same colored pegs are lining diagonally to the right
for (int y = 0; y < 3; y++) {
if (board[x][y] == 1 && board[x + 1][y + 1] == 1 && board[x + 2][y + 2] == 1 && board[x + 3][y + 3] == 1) {
JOptionPane.showMessageDialog(null, playerNames[0]+" has connected four diagonally-right in a row in " +(countForRed)+ " turns!");
}
if (board[x][y] == 2 && board[x + 1][y + 1] == 2 && board[x + 2][y + 2] == 2 && board[x + 3][y + 3] == 2) {
JOptionPane.showMessageDialog(null, playerNames[1]+" has connected four diagonally-right in a row in " +(countForYellow)+ " turns!");
}
}
}
}回答 2
Stack Overflow用户
发布于 2015-06-19 09:31:55
请原谅我没有直接回答这个问题，但这将帮助您解决这个问题，并最终获得更好的代码和将来编写代码的能力。
将"if“条件的逻辑提取到单独的方法中，可以更容易地独立考虑该逻辑，并允许您独立于程序的其余部分对其进行测试。
因此，与其：
if (board[x][y] == 1 && board[x - 1][y - 1] == 1 && board[x - 2][y - 2] == 1 && board[x - 3][y - 3] == 1) {
JOptionPane.showMessageDialog(...)
}..。用途：
if(isDiagonalLeft(x,y,1) { ... }..。还有..。
boolean isDiagonalLeft(int x, int y, int player) {
return board[x][y] == player &&
board[x - 1][y - 1] == player &&
board[x - 2][y - 2] == player &&
board[x - 3][y - 3] == player
}现在，您可以在()上运行单元测试，以确保其工作正常。也就是说，一个小程序设置一个板，只运行isDiagonalLeft()，以确保它在各种情况下给出了正确的答案。这感觉像是额外的工作，但是大多数尝试过它的人都知道，它可以通过早期捕获bug来节省精力。
您所做的是在某种程度上将游戏逻辑与演示代码(JOptionPane)分开，这样当您只想使用游戏逻辑时，表示代码就不会妨碍您。在以后的编程研究中，您将遇到更多的方法来分离它们，比如模型。
如果你需要问一些关于堆栈溢出的问题，像这样的逻辑也是有帮助的--通过把游戏逻辑和Swing分开，你可以向那些对Swing一无所知的潜在回答者打开这个问题。
而且，您可以对每个玩家重复使用此方法一次，而不是像您所拥有的那样将逻辑复制到两个位置。
如果它不起作用，请使用IDE中的调试器逐步完成它。
现在，您已经这样做了，您可以改进该方法，以便计算机执行递减，而不是程序员.
boolean isDiagonalLeft(int x, int y, int player) {
for(int i = 0; i<4; i++) {
if(board[x-i][y-i] != player) {
return false;
}
}
return true;
}...and，您可以将其概括为涵盖对角线的两个方向：
boolean isDiagonal(int x, int y, int player, boolean direction) {
int dirUnit = direction ? -1 : 1;
for(int i = 0; i<4; i++) {
if(board[x-i][y + dirUnit] != player) {
return false;
}
}
return true;
}..。所以现在你可以在四个地方重复使用这个方法:每个玩家和每个方向。
当您遇到在GUI中不工作的情况时，进行单元测试，按照GUI中的方式设置板，并在其上运行isDiagonal()。如果测试通过了，你就知道问题出在别的地方。如果测试失败，您可以使用调试器和方法的代码，以使其通过。
Stack Overflow用户
发布于 2015-06-19 05:41:07
以下代码应该有效：
//Winning diagonally towards left
public static void diagWinningLeft() {
for (int x = 5; x > 2; x--) { // Checks to see if same colored pegs are lining diagonally to the left
for (int y = 6; y > 2; y--) {
if (board[x][y] == 1 && board[x - 1][y - 1] == 1 && board[x - 2][y - 2] == 1 && board[x - 3][y - 3] == 1) {
JOptionPane.showMessageDialog(null, playerNames[0]+" has connected four diagonally-left in a row in " +(countForRed)+ " turns!");
b.drawLine(x,y,x-3,y-3);
}
if (board[x][y] == 2 && board[x - 1][y - 1] == 2 && board[x - 2][y - 2] == 2 && board[x - 3][y - 3] == 2) {
JOptionPane.showMessageDialog(null, playerNames[1]+ " has connected four diagonally-left in a row in " +(countForYellow)+ " turns!");
b.drawLine(x,y,x-3,y-3);
}
}
}
}
//Winning diagonally towards right
public static void diagWinningRight() {
for (int x = 5; x > 2; x--) { // Check to see if same colored pegs are lining diagonally to the right
for (int y = 0; y < 4; y++) {
if (board[x][y] == 1 && board[x - 1][y + 1] == 1 && board[x - 2][y + 2] == 1 && board[x - 3][y + 3] == 1) {
JOptionPane.showMessageDialog(null, playerNames[0]+" has connected four diagonally-right in a row in " +(countForRed)+ " turns!");
}
if (board[x][y] == 2 && board[x - 1][y + 1] == 2 && board[x - 2][y + 2] == 2 && board[x - 3][y + 3] == 2) {
JOptionPane.showMessageDialog(null, playerNames[1]+" has connected four diagonally-right in a row in " +(countForYellow)+ " turns!");
}
}
}
}
代码的问题是，在向右检查时，您的坐标(索引)继续检查左方向(因为ax中有4条对角线)。ax+1在NW-SE方向(或左对角线)构成对角线，ax+1在NE-SW方向(或右对角线)构成对角线.
此外，您的内部for循环没有运行到limits.Hence，diagWinningLeft()有时不工作。
https://stackoverflow.com/questions/30905138
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