所以我是java的新手，我有一个创建连接4游戏的任务。我创建了一个6行和7列充满字符的电路板，如下所示 - ＆gt; ' - '，当用户输入欲望列时，它用B或R（红色或黑色检查器）替换' - '，无论这只是一个背景。一切都有效，除了我检查对角线的代码很长的部分，我无法找到一种方法来通过所有可能的4组对角线，其中一个玩家可以赢，除了在不同的for循环中做每一个。 ..我知道你可以帮助我缩短它的可怕希望:(
这是对角线检查的代码:(大声笑只看着它让我感到难过）
public class Connect4 {
public static void main(String[] args) {
//Create board
Scanner input = new Scanner(System.in);
char[][] grid = new char[6][7];
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
grid[i][j] = '-';
}
}
public static void checkWinner(char[][] grid) {
try{
//A LOT OF FOR LOOPS FOR DIAGONAL CHECKS
for (int i = 5; i > 1; i-- ) {
for(int j = 0; j < 4; j++) {
if ( grid[i][j] == 'R' &&
grid[i-1][j+1] == 'R' &&
grid[i-2][j+2] == 'R' &&
grid[i-3][j+3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if ( grid[i][j] == 'B' &&
grid[i-1][j+1] == 'B' &&
grid[i-2][j+2] == 'B' &&
grid[i-3][j+3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for (int i = 5; i > 1; i--) {
for (int j = 1; j < 4; j++) {
if (grid[i][j] == 'R' &&
grid[i-1][j+1] == 'R' &&
grid[i-2][j+2] == 'R' &&
grid[i-3][j+3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i-1][j+1] == 'B' &&
grid[i-2][j+2] == 'B' &&
grid[i-3][j+3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for (int i = 0; i < 4; i++) {
for (int j = 4; j < 7; j++) {
if ( grid[i][j] == 'R' &&
grid[i+1][j-1] == 'R' &&
grid[i+2][j-2] == 'R' &&
grid[i+3][j-3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i+1][j-1] == 'B' &&
grid[i+2][j-2] == 'B' &&
grid[i+3][j-3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for (int i = 1; i < 2; i++) {
for (int j = 6; j > 5; j--) {
if ( grid[i][j] == 'R' &&
grid[i+1][j-1] == 'R' &&
grid[i+2][j-2] == 'R' &&
grid[i+3][j-3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i+1][j-1] == 'B' &&
grid[i+2][j-2] == 'B' &&
grid[i+3][j-3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for (int i = 4; i < 5; i++){
for (int j = 2; j < 3; j++){
if (grid[i][j] == 'R' &&
grid[i-1][j+1] == 'R' &&
grid[i-2][j+2] == 'R' &&
grid[i-3][j+3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i-1][j+1] == 'B' &&
grid[i-2][j+2] == 'B' &&
grid[i-3][j+3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for (int i = 0; i < 4; i++) {
for (int j = 3; j > 0; j--) {
if (grid[i][j] == 'R' &&
grid[i+1][j+1] == 'R' &&
grid[i+2][j+2] == 'R' &&
grid[i+3][j+3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i+1][j+1] == 'B' &&
grid[i+2][j+2] == 'B' &&
grid[i+3][j+3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for(int i =0; i < 1; i++) {
for (int j = 0; j <1; j++) {
if (grid[i][j] == 'R' &&
grid[i+1][j+1] == 'R' &&
grid[i+2][j+2] == 'R' &&
grid[i+3][j+3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i+1][j+1] == 'B' &&
grid[i+2][j+2] == 'B' &&
grid[i+3][j+3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for (int j = 0; j < 1; j++) {
for(int i =1 ; i < 3; i++) {
if (grid[i][j] == 'R' &&
grid[i+1][j+1] == 'R' &&
grid[i+2][j+2] == 'R' &&
grid[i+3][j+3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i+1][j+1] == 'B' &&
grid[i+2][j+2] == 'B' &&
grid[i+3][j+3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for (int j = 3; j < 4; j++) {
for (int i = 0; i < 3; i++) {
if (grid[i][j] == 'R' &&
grid[i+1][j+1] == 'R' &&
grid[i+2][j+2] == 'R' &&
grid[i+3][j+3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i+1][j+1] == 'B' &&
grid[i+2][j+2] == 'B' &&
grid[i+3][j+3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for (int i = 2; i < 3; i++) {
for (int j = 2; j > 0; j--) {
if (grid[i][j] == 'R' &&
grid[i+1][j+1] == 'R' &&
grid[i+2][j+2] == 'R' &&
grid[i+3][j+3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i+1][j+1] == 'B' &&
grid[i+2][j+2] == 'B' &&
grid[i+3][j+3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for (int i = 1; i < 2; i++) {
for (int j = 2; j < 3; j++){
if (grid[i][j] == 'R' &&
grid[i+1][j+1] == 'R' &&
grid[i+2][j+2] == 'R' &&
grid[i+3][j+3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i+1][j+1] == 'B' &&
grid[i+2][j+2] == 'B' &&
grid[i+3][j+3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
for (int i = 1; i < 2; i++) {
for (int j = 1; j < 2; j++){
if (grid[i][j] == 'R' &&
grid[i+1][j+1] == 'R' &&
grid[i+2][j+2] == 'R' &&
grid[i+3][j+3] == 'R') {
System.out.println("Player 1 Wins!");
System.exit(0);
}
else if (grid[i][j] == 'B' &&
grid[i+1][j+1] == 'B' &&
grid[i+2][j+2] == 'B' &&
grid[i+3][j+3] == 'B') {
System.out.println("Player 2 Wins!");
System.exit(0);
}
}
}
}
catch(ArrayIndexOutOfBoundsException e){
System.out.println("Exception thrown :" + e);
}
}
答案 0 :(得分：3)
不是检查棋盘上的每一个可能位，而是根据人类如何玩它来玩游戏 - 只有当有人掉入一块然后形成一条四条瓷砖的线条时才会连接4。所以：不要检查每个可能的磁贴，使用刚刚放入的磁贴，只检查涉及该磁贴的行：
你已经知道了瓷砖的“颜色”，所以你的支票（假代码）只是形式：
int stretch = 0;
if( <THE CHECK TILE COLOR>.equals(<DROPPED TILE COLOR>)) {
// check the next possible tile
stretch++;
} else { stretch = 0; }
如果您找到的具有相同颜色的最大一块瓷砖是4，那么就完成了。如果没有，没有连接-4。
那就是说，这是一项家庭作业：S.O。当你在编程时遇到问题时，就在这里，但我们不是为你做功课。如果你遇到困难，请问你的同学甚至你的老师。当你上一门课程时，互联网并不是唯一可以提供帮助的地方，尤其是。
答案 1 :(得分：1)
而不是在if语句中编码，我会例如对于\对角只是从硬币的位置开始并向右下方计数，在颜色变化之前有多少相同的颜色，或者到达边界。在左上方向相同。
最后，我只是检查两个计数加一的总和是否大于或等于4.
/ diagonal的方法相同。
警告：您的实施 - 和|还没有数组边界检查。我也会为他们推荐计数方法。
P.S。为了使您的代码更清晰，请将每张支票放在一个单独的方法中：
isHorizonallWin(x,y,color,grid)
isVerticalWin(x,y,color,grid)
isLdiagonalLeftUpper2RightLowerWin(x,y,color,grid)
isLdiagonalRightUpper2leftLowerWin(x,y,color,grid)