794. Valid Tic-Tac-Toe State (Medium)

https://leetcode.com/problems/valid-tic-tac-toe-state/

A Tic-Tac-Toe board is given as a string array board. Return True if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.

The board is a 3 x 3 array, and consists of characters " ", "X", and "O".  The " " character represents an empty square.

Here are the rules of Tic-Tac-Toe:

  • Players take turns placing characters into empty squares (" ").
  • The first player always places "X" characters, while the second player always places "O" characters.
  • "X" and "O" characters are always placed into empty squares, never filled ones.
  • The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
  • The game also ends if all squares are non-empty.
  • No more moves can be played if the game is over.
Example 1:
Input: board = ["O  ", "   ", "   "]
Output: false
Explanation: The first player always plays "X".

Example 2:
Input: board = ["XOX", " X ", "   "]
Output: false
Explanation: Players take turns making moves.

Example 3:
Input: board = ["XXX", "   ", "OOO"]
Output: false

Example 4:
Input: board = ["XOX", "O O", "XOX"]
Output: true

Note:

  • board is a length-3 array of strings, where each string board[i] has length 3.
  • Each board[i][j] is a character in the set {" ", "X", "O"}.

Hints

Solutions

class Solution {
    public boolean validTicTacToe(String[] board) {
        if (hasWon(board, 'X')) {
            return getCount(board, 'X') - getCount(board, 'O') == 1;
        }

        if (hasWon(board, 'O')) {
            return getCount(board, 'X') - getCount(board, 'O') == 0;
        }

        int diff = getCount(board, 'O') - getCount(board, 'X');

        return diff == 0 || diff == -1;
    }

    private boolean hasWon(String[] board, char player) {
        boolean b1 = (board[0].charAt(0) == player && board[0].charAt(1) == player && board[0].charAt(2) == player);
        boolean b2 = (board[0].charAt(0) == player && board[1].charAt(0) == player && board[2].charAt(0) == player);
        boolean b3 = (board[0].charAt(1) == player && board[1].charAt(1) == player && board[2].charAt(1) == player);
        boolean b4 = (board[0].charAt(2) == player && board[1].charAt(2) == player && board[2].charAt(2) == player);
        boolean b5 = (board[0].charAt(0) == player && board[1].charAt(1) == player && board[2].charAt(2) == player);
        boolean b6 = (board[0].charAt(2) == player && board[1].charAt(1) == player && board[2].charAt(0) == player);
        boolean b7 = (board[1].charAt(0) == player && board[1].charAt(1) == player && board[1].charAt(2) == player);
        boolean b8 = (board[2].charAt(0) == player && board[2].charAt(1) == player && board[2].charAt(2) == player);

        return b1 || b2 || b3 || b4 || b5|| b6 || b7 || b8;
    }

    private int getCount(String[] board, char player) {
        int count = 0;

        for (String row : board) {
            for (char c : row.toCharArray()) {
                if (c == player) {
                    count++;
                }
            }
        }

        return count;
    }
}
Copyright © iovi.com 2017 all right reserved,powered by GitbookLast Modification: 2019-04-15 19:13:58

results matching ""

    No results matching ""