TicTacToe 的玩法 java
Play method for TicTacToe java
所以我们得到了一个练习，需要写一个 tictactoe class，我们需要使用一个 2d int 数组作为棋盘和 2 个玩家，其中一个有一个“1”作为 "X" 而另一个是“2”作为 "O"，“0”表示空字段。现在我们需要为玩家编写一个方法来实际在场上设置一些东西，但我能想到的只是与字符串或字符板有关的东西，而与 int 板无关。一个int板上的一个数字的问题这个设置是怎么实现的呢？感谢您的帮助！
我已经有了一个方法来检查板上是否有任何空闲位置，这应该是正确的。
public class TicTacToe extends BasicBoard implements Game {
int PlayerA = 1;
int PlayerB = 2;
int empty = 0;
private int row, column;
private int[][] board = new int[column][row];
public boolean isboardfull(int row, int column) {
for (int i = 0; i <= column; i++)
{
for (int j = 0; j <= row; j++)
{
if (board[i][j] == PlayerA || board[i][j] == PlayerB)
return true;
}
}
return false;
}
public void playerturn(){
if (!isboardfull(row, column))
}
}
您的 TicTacToe class 从 BasicBoard 和 Game class 扩展而来，但您没有提供它们。我假设这些 classes 为您提供了渲染棋盘和控制游戏演变的方法，但由于我们没有它们，所以我在我的示例中包含了类似（和简单）的东西（以演示它是如何工作的）。如果 game、printBoard、resetBoard 和 endGame 方法由 BasicBoard 和 Game [=61= 提供，您可以跳过这些方法]es.
这是我所做的一系列假设：
- 玩家被问到玩的坐标
- 棋盘满时游戏结束。一个更复杂的版本应该检查游戏的每次迭代是否有玩家获胜。
这里是我的方法的一般解释：
- X 和 O 到 1 和 2 之间的映射设置为
static constant，因为这永远不会改变。
- 行数和列数可能因执行和
TicTacToe 构造函数中的参数化而异。
- 玩家根据提示通过标准输入填写信息。
game 函数要求玩家移动并渲染棋盘（在标准输出上），直到棋盘完全填满。isBoardFull函数检查棋盘上是否有空位。因此，如果我们找到一个空槽，我们就知道它没有满，否则我们需要继续寻找空槽。如果我们搜索所有板并且没有空槽，那么它是满的。 （在我看来，这部分在您提供的代码中写错了）playerTurn 函数询问玩家想要玩的坐标并填充棋盘。为此，我们扫描标准输入的 2 行，将它们转换为 int 并检查该位置是否为空且在边界内。如果是这样，我们用玩家编号标记位置。
代码：
public class TicTacToe {
private static final int PLAYER_A = 1;
private static final int PLAYER_B = 2;
private static final int EMPTY = 0;
private final int numRows;
private final int numColumns;
private final int[][] board;
private final Scanner inputScanner;
public TicTacToe(int numRows, int numColumns) {
// Retrieve board sizes
this.numRows = numRows;
this.numColumns = numColumns;
// Instantiate board
this.board = new int[numRows][numColumns];
// Initialize board
resetBoard();
// Initialize the input scanner (for player choices)
this.inputScanner = new Scanner(System.in);
}
public void game() {
// Initialize the game
int numMoves = 0;
printBoard(numMoves);
// Play until the game is over
while (!isBoardFull() && !hasPlayerWon()) {
// A or B player should move
int currentPlayer = (numMoves % 2 == 0) ? PLAYER_A : PLAYER_B;
playerTurn(currentPlayer);
// We increase the number of moves
numMoves += 1;
// We render the board
printBoard(numMoves);
}
// Check winner and close game
endGame();
}
private void resetBoard() {
for (int i = 0; i < this.numRows; ++i) {
for (int j = 0; j < this.numColumns; ++j) {
this.board[i][j] = EMPTY;
}
}
}
private void printBoard(int currentMove) {
System.out.println("Move: " + currentMove);
for (int i = 0; i < this.numRows; ++i) {
for (int j = 0; j < this.numColumns; ++j) {
System.out.print(this.board[i][j] + " ");
}
System.out.println();
}
// A new line to split moves
System.out.println();
}
private boolean isBoardFull() {
for (int i = 0; i < this.numRows; ++i) {
for (int j = 0; j < this.numColumns; ++j) {
if (this.board[i][j] == EMPTY) {
// If there is an empty cell, the board is not full
return false;
}
}
}
// If there are no empty cells, the board is full
return true;
}
private boolean hasPlayerWon() {
// TODO: Return whether a player has won the game or not
return false;
}
private void playerTurn(int currentPlayer) {
// Log player information
System.out.println("Turn for player: " + currentPlayer);
// Ask the player to pick a position
boolean validPosition = false;
while (!validPosition) {
// Ask for X position
int posX = askForPosition("row", this.numRows);
// Ask for Y position
int posY = askForPosition("column", this.numColumns);
// Check position
if (posX >= 0 && posX < this.numRows) {
if (posY >= 0 && posY < this.numColumns) {
if (this.board[posX][posY] == EMPTY) {
// Mark as valid
validPosition = true;
// Mark the position
this.board[posX][posY] = currentPlayer;
} else {
System.out.println("Position is not empty. Please choose another one.");
}
} else {
System.out.println("Column position is not within bounds. Please choose another one.");
}
} else {
System.out.println("Row position is not within bounds. Please choose another one.");
}
}
}
private int askForPosition(String rc, int dimensionLimit) {
System.out.println("Select a " + rc + " position between 0 and " + dimensionLimit);
return Integer.valueOf(this.inputScanner.nextLine());
}
private void endGame() {
// Close input scanner
this.inputScanner.close();
// Log game end
System.out.println("GAME ENDED!");
// TODO: Check the board status
System.out.println("Draw");
}
public static void main(String[] args) {
TicTacToe ttt = new TicTacToe(3, 4);
ttt.game();
}
}
示例输出：
Move: 0
0 0 0 0
0 0 0 0
0 0 0 0
Turn for player: 1
Select a row position between 0 and 3
4
Select a column position between 0 and 4
1
Row position is not within bounds. Please choose another one.
Select a row position between 0 and 3
1
Select a column position between 0 and 4
1
Move: 1
0 0 0 0
0 1 0 0
0 0 0 0
Turn for player: 2
.
.
.
Move: 12
2 2 1 2
1 1 2 1
1 2 1 2
GAME ENDED!
Draw
所以我们得到了一个练习，需要写一个 tictactoe class，我们需要使用一个 2d int 数组作为棋盘和 2 个玩家，其中一个有一个“1”作为 "X" 而另一个是“2”作为 "O"，“0”表示空字段。现在我们需要为玩家编写一个方法来实际在场上设置一些东西，但我能想到的只是与字符串或字符板有关的东西，而与 int 板无关。一个int板上的一个数字的问题这个设置是怎么实现的呢？感谢您的帮助！
我已经有了一个方法来检查板上是否有任何空闲位置，这应该是正确的。
public class TicTacToe extends BasicBoard implements Game {
int PlayerA = 1;
int PlayerB = 2;
int empty = 0;
private int row, column;
private int[][] board = new int[column][row];
public boolean isboardfull(int row, int column) {
for (int i = 0; i <= column; i++)
{
for (int j = 0; j <= row; j++)
{
if (board[i][j] == PlayerA || board[i][j] == PlayerB)
return true;
}
}
return false;
}
public void playerturn(){
if (!isboardfull(row, column))
}
}
您的 TicTacToe class 从 BasicBoard 和 Game class 扩展而来，但您没有提供它们。我假设这些 classes 为您提供了渲染棋盘和控制游戏演变的方法，但由于我们没有它们，所以我在我的示例中包含了类似（和简单）的东西（以演示它是如何工作的）。如果 game、printBoard、resetBoard 和 endGame 方法由 BasicBoard 和 Game [=61= 提供，您可以跳过这些方法]es.
这是我所做的一系列假设：
- 玩家被问到玩的坐标
- 棋盘满时游戏结束。一个更复杂的版本应该检查游戏的每次迭代是否有玩家获胜。
这里是我的方法的一般解释：
- X 和 O 到 1 和 2 之间的映射设置为
static constant，因为这永远不会改变。 - 行数和列数可能因执行和
TicTacToe构造函数中的参数化而异。 - 玩家根据提示通过标准输入填写信息。
game函数要求玩家移动并渲染棋盘（在标准输出上），直到棋盘完全填满。isBoardFull函数检查棋盘上是否有空位。因此，如果我们找到一个空槽，我们就知道它没有满，否则我们需要继续寻找空槽。如果我们搜索所有板并且没有空槽，那么它是满的。 （在我看来，这部分在您提供的代码中写错了）playerTurn函数询问玩家想要玩的坐标并填充棋盘。为此，我们扫描标准输入的 2 行，将它们转换为int并检查该位置是否为空且在边界内。如果是这样，我们用玩家编号标记位置。
代码：
public class TicTacToe {
private static final int PLAYER_A = 1;
private static final int PLAYER_B = 2;
private static final int EMPTY = 0;
private final int numRows;
private final int numColumns;
private final int[][] board;
private final Scanner inputScanner;
public TicTacToe(int numRows, int numColumns) {
// Retrieve board sizes
this.numRows = numRows;
this.numColumns = numColumns;
// Instantiate board
this.board = new int[numRows][numColumns];
// Initialize board
resetBoard();
// Initialize the input scanner (for player choices)
this.inputScanner = new Scanner(System.in);
}
public void game() {
// Initialize the game
int numMoves = 0;
printBoard(numMoves);
// Play until the game is over
while (!isBoardFull() && !hasPlayerWon()) {
// A or B player should move
int currentPlayer = (numMoves % 2 == 0) ? PLAYER_A : PLAYER_B;
playerTurn(currentPlayer);
// We increase the number of moves
numMoves += 1;
// We render the board
printBoard(numMoves);
}
// Check winner and close game
endGame();
}
private void resetBoard() {
for (int i = 0; i < this.numRows; ++i) {
for (int j = 0; j < this.numColumns; ++j) {
this.board[i][j] = EMPTY;
}
}
}
private void printBoard(int currentMove) {
System.out.println("Move: " + currentMove);
for (int i = 0; i < this.numRows; ++i) {
for (int j = 0; j < this.numColumns; ++j) {
System.out.print(this.board[i][j] + " ");
}
System.out.println();
}
// A new line to split moves
System.out.println();
}
private boolean isBoardFull() {
for (int i = 0; i < this.numRows; ++i) {
for (int j = 0; j < this.numColumns; ++j) {
if (this.board[i][j] == EMPTY) {
// If there is an empty cell, the board is not full
return false;
}
}
}
// If there are no empty cells, the board is full
return true;
}
private boolean hasPlayerWon() {
// TODO: Return whether a player has won the game or not
return false;
}
private void playerTurn(int currentPlayer) {
// Log player information
System.out.println("Turn for player: " + currentPlayer);
// Ask the player to pick a position
boolean validPosition = false;
while (!validPosition) {
// Ask for X position
int posX = askForPosition("row", this.numRows);
// Ask for Y position
int posY = askForPosition("column", this.numColumns);
// Check position
if (posX >= 0 && posX < this.numRows) {
if (posY >= 0 && posY < this.numColumns) {
if (this.board[posX][posY] == EMPTY) {
// Mark as valid
validPosition = true;
// Mark the position
this.board[posX][posY] = currentPlayer;
} else {
System.out.println("Position is not empty. Please choose another one.");
}
} else {
System.out.println("Column position is not within bounds. Please choose another one.");
}
} else {
System.out.println("Row position is not within bounds. Please choose another one.");
}
}
}
private int askForPosition(String rc, int dimensionLimit) {
System.out.println("Select a " + rc + " position between 0 and " + dimensionLimit);
return Integer.valueOf(this.inputScanner.nextLine());
}
private void endGame() {
// Close input scanner
this.inputScanner.close();
// Log game end
System.out.println("GAME ENDED!");
// TODO: Check the board status
System.out.println("Draw");
}
public static void main(String[] args) {
TicTacToe ttt = new TicTacToe(3, 4);
ttt.game();
}
}
示例输出：
Move: 0
0 0 0 0
0 0 0 0
0 0 0 0
Turn for player: 1
Select a row position between 0 and 3
4
Select a column position between 0 and 4
1
Row position is not within bounds. Please choose another one.
Select a row position between 0 and 3
1
Select a column position between 0 and 4
1
Move: 1
0 0 0 0
0 1 0 0
0 0 0 0
Turn for player: 2
.
.
.
Move: 12
2 2 1 2
1 1 2 1
1 2 1 2
GAME ENDED!
Draw