794. 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:

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”}.

题意：
检测当前局面是否为合法局面。

思路：
1. 由规则可知，”X”一定最先开始，所以当前局面存在”O”的个数大于”X”的个数为非法。
2. 其次，由于”X”和”O”轮流，因此，当前局面中”X”的个数要么和”O”相等，要么比”O”多一。
3. “O”在当前局面赢得比赛的情况下，上一轮的”X”一定不能赢得局面。
4. “O”在当前局面赢得比赛的情况下，上一轮的”X”没有赢得局面时，合法局面必须满足”O”的个数等于”X”的个数。
5. “X”在当前局面赢得比赛的情况下，意味着上一轮”O”没有赢得局面，合法局面下，”X”的个数正好比”O”的个数多一。

class Solution {
    public boolean validTicTacToe(String[] board) {
        int x_count = 0;
        int o_count = 0;
        for (int i = 0; i < 3; ++i) {
            for (int j = 0; j < 3; ++j) {
                if (board[i].charAt(j) == 'X') x_count++;
                else if (board[i].charAt(j) == 'O') o_count++;
            }
        }
        char[][] map = new char[3][3];
        for (int i = 0; i < 3; ++i) {
            map[i] = board[i].toCharArray();
        }
        if (o_count > x_count || x_count - o_count > 1) return false;

        if (check_win_positions(map, 'O')) {
            if (check_win_positions(map, 'X')) {
                return false;
            }
            return o_count == x_count;
        }

        if (check_win_positions(map, 'X') && x_count != o_count + 1) return false;

        return true;
    }

    boolean check_win_positions(char[][] map, char player) {
        for (int i = 0; i < 3; ++i) {
            if (map[i][0] == map[i][1] && map[i][1] == map[i][2]
                    && map[i][2] == player) {
                return true;
            }
        }

        for (int j = 0; j < 3; ++j) {
            if (map[0][j] == map[1][j] && map[1][j] == map[2][j]
                    && map[2][j] == player) {
                return true;
            }
        }


        if (map[0][0] == map[1][1] && map[1][1] == map[2][2] && map[2][2]  == player ||
                map[0][2] == map[1][1] && map[1][1] == map[2][0] && map[2][0] == player)
            return true;

        return false;
    }
}