在TictoeJava应用程序中添加计算机播放器?
如何将播放器2转换为计算机生成的答案,而不是另一个输入。只要遵循游戏规则并选择一个开放空间,计算机的选择就可以是完全随机的 此外,我想保留我现有代码的基本结构…只需要将一个播放器转换为计算机生成的选择。选择不必是最佳选择,只要是有效的选择即可 我不希望答案只是一个开始,而不破坏我所拥有的…请任何帮助在TictoeJava应用程序中添加计算机播放器?,java,Java,如何将播放器2转换为计算机生成的答案,而不是另一个输入。只要遵循游戏规则并选择一个开放空间,计算机的选择就可以是完全随机的 此外,我想保留我现有代码的基本结构…只需要将一个播放器转换为计算机生成的选择。选择不必是最佳选择,只要是有效的选择即可 我不希望答案只是一个开始,而不破坏我所拥有的…请任何帮助 package tictactoe; import java.util.Scanner; public class TicTacToe { private final int BOARDSIZ
package tictactoe;
import java.util.Scanner;
public class TicTacToe
{
private final int BOARDSIZE = 3; // size of the board
private enum Status { WIN, DRAW, CONTINUE }; // game states
private char[][] board; // board representation
private boolean firstPlayer; // whether it's player 1's move
private boolean gameOver; // whether game is over
// Constructor
public TicTacToe()
{
board = new char[ BOARDSIZE ][ BOARDSIZE ];
firstPlayer = true;
gameOver = false;
} // end Constructor
// start game
public void play()
{
Scanner input = new Scanner( System.in );
int row; // row for next move
int column; // column for next move
System.out.println( "Player X's turn." );
while ( !gameOver )
{
char player = ( firstPlayer ? 'X' : 'O' );
// player's turn
do
{
System.out.printf(
"Player %c: Enter row ( 0, 1 or 2 ): ", player );
row = input.nextInt();
System.out.printf(
"Player %c: Enter column ( 0, 1 or 2 ): ", player );
column = input.nextInt();
} while ( !validMove( row, column ) );
board[ row ][ column ] = player;
firstPlayer = !firstPlayer;
printBoard();
printStatus( player );
} // end while
} // end method play
// show game status in status bar
private void printStatus( int player )
{
Status status = gameStatus();
// check game status
switch ( status )
{
case WIN:
System.out.printf( "Player %c wins.", player );
gameOver = true;
break;
case DRAW:
System.out.println( "Game is a draw." );
gameOver = true;
break;
case CONTINUE:
if ( player == 'X' )
System.out.println( "Player O's turn." );
else
System.out.println( "Player X's turn." );
break;
} // end switch
} // end method printStatus
// get game status
private Status gameStatus()
{
int a;
// check for a win on diagonals
if ( board[ 0 ][ 0 ] != 0 && board[ 0 ][ 0 ] == board[ 1 ][ 1 ] &&
board[ 0 ][ 0 ] == board[ 2 ][ 2 ] )
return Status.WIN;
else if ( board[ 2 ][ 0 ] != 0 && board[ 2 ][ 0 ] ==
board[ 1 ][ 1 ] && board[ 2 ][ 0 ] == board[ 0 ][ 2 ] )
return Status.WIN;
// check for win in rows
for ( a = 0; a < 3; a++ )
if ( board[ a ][ 0 ] != 0 && board[ a ][ 0 ] ==
board[ a ][ 1 ] && board[ a ][ 0 ] == board[ a ][ 2 ] )
return Status.WIN;
// check for win in columns
for ( a = 0; a < 3; a++ )
if ( board[ 0 ][ a ] != 0 && board[ 0 ][ a ] ==
board[ 1 ][ a ] && board[ 0 ][ a ] == board[ 2 ][ a ] )
return Status.WIN;
// check for a completed game
for ( int r = 0; r < 3; r++ )
for ( int c = 0; c < 3; c++ )
if ( board[ r ][ c ] == 0 )
return Status.CONTINUE; // game is not finished
return Status.DRAW; // game is a draw
} // end method gameStatus
// display board
public void printBoard()
{
System.out.println( " _______________________ " );
for ( int row = 0; row < BOARDSIZE; row++ )
{
System.out.println( "| | | |" );
for ( int column = 0; column < BOARDSIZE; column++ )
printSymbol( column, board[ row ][ column ] );
System.out.println( "|_______|_______|_______|" );
} // end for
} // end method printBoard
// print moves
private void printSymbol( int column, char value )
{
System.out.printf( "| %c ", value );
if ( column == 2 )
System.out.println( "|" );
} // end method printSymbol
// validate move
private boolean validMove( int row, int column )
{
return row >= 0 && row < 3 && column >= 0 && column < 3 &&
board[ row ][ column ] == 0;
} // end method validMove
} // end class TicTacToe
首先,您将在游戏中修改do,在询问用户和为计算机选择移动之间切换 如果您只想随机移动,您会发现Java类random很有用。在循环外部创建一个实例
Random random = new Random();
现在,您可以要求在[0,2]范围内使用以下参数获得一个随机整数:
int row = random.nextInt(3);
int col = random.nextInt(3);
检查电路板,看那个点是否填满了。如果是,您需要选择另一个
如果你想找到最好的移动,你可以使用算法。它会搜索两名玩家可能采取的行动,并假定每个玩家都会发挥其最佳水平。在您向用户询问其行动的行和列时,将其替换为调用另一个表示某个玩家的对象的方法,在这种情况下,它只会提示输入,然后将其返回给调用者或计算机播放器,在这种情况下,它将根据您确定的任何方式计算移动,并返回 该方法需要将董事会传递给它。玩家类可以忽略它,但是计算机玩家需要它来决定它的移动
您还需要创建某种Move类,以便将行和列的移动作为单个返回值传递回去。如果这是我的代码,我会重新开始重构所有内容,并尝试创建各自负责的不同对象,包括游戏对象、玩家对象、计算机玩家对象。。。分而治之。我必须将这个基本结构用于我正在进行的项目。。。
public int[] autoPlay()
{
Random rnd = new Random();
int randomValue[] = new int[2];
do{
randomValue[0] = rnd.nextInt(3);
randomValue[1] = rnd.nextInt(3);
} while ( !validMove( randomValue[0], randomValue[1] ) );
return randomValue;
}
// start game
public void play()
{
Scanner input = new Scanner( System.in );
int row; // row for next move
int column; // column for next move
System.out.println( "Player X's turn." );
while ( !gameOver )
{
char player = ( firstPlayer ? 'X' : 'O' );
// player's turn
do
{
if(player == 'X') {
System.out.printf(
"Player %c: Enter row ( 0, 1 or 2 ): ", player);
row = input.nextInt();
System.out.printf(
"Player %c: Enter column ( 0, 1 or 2 ): ", player);
column = input.nextInt();
}else
{
int array[] = autoPlay();
row = array[0];
column = array[1];
}
} while ( !validMove( row, column ) );
board[ row ][ column ] = player;
firstPlayer = !firstPlayer;
printBoard();
printStatus( player );
} // end while
} // end method play