我正在我的计算机科学课上使用处理井字游戏。在我的程序中,我正在对3行检查器进行编码,并且尝试使用2D数组代码对此进行编码,因为在对X / O进行编码时,我的数组已初始化为2D数组。此外,老师还给了我们一个该示例的代码但是是在1D数组中进行的,因此我不知道如何将他关于FOR循环的示例转换为2D,以便它可以运行到我的程序中。我已经将数组代码转换为适用于我的程序,并且该程序正在运行,但是有关有人是否获胜的代码(FOR循环)不起作用。有人能帮我吗?
void onlyX() { //views only X pieces, runs checker for onlyX (attempted to
convert to 2D array form)
for (int i=0; i<3; i++) {
for (int j=0; j<3; j++) {
if (boardPiece[i][j] == "X") {
onlyXPiece[i][j] = "X";
}
}
}
}
/*void onlyX() { //1D Form
for (int i=0; i<numberOfBoardPieces; i++) {
if (boardPiece[i] == "X") {
onlyXPiece[i] = "X";
}
}
}*/
void onlyO() { //views only O pieces, runs checker for onlyO (attempted to
convert to 2D array form)
for (int i=0; i<3; i++) {
for (int j=0; j<3; j++) {
if (boardPiece[i][j] == "O") {
onlyOPiece[i][j] = "O";
}
}
}
}
/*void onlyO() { //1D form
for (int i=0; i<numberOfBoardPieces; i++) {
if (boardPiece[i] == "O") {
onlyOPiece[i] = "O";
}
}
}*/
答案 0 :(得分:2)
您的2D数组FOR循环(通常称为double FOR LOOP)看起来不错,尽管我不太明白为什么您会将X和O放在单独的2D数组中。有多种检查获胜条件的方法,但这似乎过于复杂。
我在代码顶部改写了double for循环,以执行易于理解且可用于X和O的胜利检查。不要忘记您还必须检查垂直和对角线胜利。
String[][] boardPiece = {{"X", "", "O"},
{"O", "O","O"},
{"X", "", "X"}};
void setup() {
println("X wins = "+ str(checkRows("X")));
println("O wins = "+ str(checkRows("O")));
}
boolean checkRows(String XorO) {
for (int i=0; i<3; i++) { //for every row
boolean win = true; //set win to true
for (int j=0; j<3; j++) { //for every column
if (boardPiece[i][j] != XorO) { //if the column does not equal the sign that you are checking
win = false; // meaning its the other, or it's empty, then set win to false
}
}
if (win) {
return true; //if win is true, then there are 3 X's or 3 O's in this row, so return true
}
}
return false; // if none of the rows contain 3 equal, return false
}
答案 1 :(得分:2)
要检查是否有人赢得了比赛,您必须检查列,行和对角线。用一个参数编写一个checkBoard
函数,您可以在其中指定要检查的片段类型("X"
pr "O"
)
boolean checkBoard(String p) {
boolean won = false;
// check rows and columns
for (int i=0; !won && i<3; i++) {
boolean eqRow = true;
boolean eqCol = true;
for (int j=0; j<3; j++) {
eqRow = eqRow && boardPiece[i][j] == p;
eqCol = eqCol && boardPiece[j][i] == p;
}
won = eqRow || eqCol;
}
// check diagonals
if ( !won )
{
boolean d1 = true;
boolean d2 = true;
for (int i=0; !won && i<3; i++) {
d1 = d1 && boardPiece[i][i] == p;
d2 = d2 && boardPiece[2-i][i] == p;
}
won = d1 || d2;
}
return won;
}