Level Hard on Tic Tac Toe

[programmer1337]

I am making a Tic Tac Toe game, now I have the easy, and medium done, all I need is the hard, here is my code at the moment

code:

public static void hard(String phrase) {         int winner = 0;         board[1][1] = " X ";         int c = 2;         phrase = getPhrase();         while (winner == 0) {             if (c % 2 == 0) {                 rowColumn(phrase, "1", 2);             } else {                 System.out.println("hello");                 if (board[0][0].equals(" - ")) {                     board[0][0] = " X ";                 } else if (!board[0][0].equals(" - ") && board[0][2].equals(" - ")) {                     board[0][2] = " X ";                 } else if (!board[0][0].equals(" - ") && !board[0][2].equals(" - ") && board[2][0].equals(" - ")) {                     board[2][0] = " X ";                 } else if (!board[0][0].equals(" - ") && !board[0][2].equals(" - ") && !board[2][0].equals(" - ") && board[2][2].equals(" - ")) {                     board[2][2] = " X ";                 }             }             phrase = getPhrase();             winner = determineWin();             c++;         }         if (winner == 2) {             System.out.println("you win");         } else if (winner == 1) {             System.out.println("cpu wins");         }     }

Now I dont want help with coding or anything, but the code above will put an X according to which corners are open, and I wouldnt know which part of the if statement I would need to put another if statment to check if there is a spot open for a win, like for example if you have(below), which part of the if statement would you make the computer put an x where it
would win

Quote:

X - X
- X O
- - O

[programmer1337]

code:

public static void hard(String phrase) {         int winner = 0;         board[1][1] = " X ";         int c = 2;         int d = 0;         int xVal = 0;         phrase = getPhrase();         while (winner == 0) {             if (c % 2 == 0) {                 rowColumn(phrase, "1", 2);             } else {                 if (!board[0][0].equals(" - ") && !board[0][2].equals(" - ") && !board[2][0].equals(" - ") && !board[2][2].equals(" - ")) {                     if (board[0][0].equals(" X ") && board[0][2].equals(" X ")) {                         board[0][1] = " X ";                     } else if (board[0][0].equals(" X ") && board[2][0].equals(" X ")) {                         board[1][0] = " X ";                     } else if (board[2][2].equals(" X ") && board[0][2].equals(" X ")) {                         board[1][2] = " X ";                     } else if (board[2][2].equals(" X ") && board[2][0].equals(" X ")) {                         board[2][1] = " X ";                     }                 }                 if (board[0][0].equals(" - ")) {                     board[0][0] = " X ";                 } else if (!board[0][0].equals(" - ") && board[0][2].equals(" - ")) {                     board[0][2] = " X ";                 } else if (!board[0][0].equals(" - ") && !board[0][2].equals(" - ") && board[2][0].equals(" - ")) {                     board[2][0] = " X ";                 } else if (!board[0][0].equals(" - ") && !board[0][2].equals(" - ") && !board[2][0].equals(" - ") && board[2][2].equals(" - ")) {                     board[2][2] = " X ";                 }             }             phrase = getPhrase();             winner = determineWin();             c++;         }         if (winner == 2) {             JOptionPane.showMessageDialog(null, "You Win");         } else if (winner == 1) {             JOptionPane.showMessageDialog(null, "CPU Wins");         }     }

My new code, still not unbeatable

[Tony]

Tic Tac Toe is a solved game. For any board, there exists an optimal move. All such moves are here -- http://xkcd.com/832/