/**
1. 判断井字棋是否合法, 那么把每个条件都细化为逻辑来判断
2. 可以根据 X赢 或 O赢分为4中情况讨论
    2.1 X赢了, O没赢: x先手, 所以此时合法情况是 x 比 o 多一个
    2.1 X没赢, O赢了: x先手, 所以此时合法情况是 x 与 o 相等
    2.1 X没赢, O没赢: 中间情况 0 <= count(x) - count(o) <= 1
    2.1 X赢了, O赢了: 必定非法

3. 怎么判断有没有人赢: 当有 3 个相同（且非空）的字符填充任何行、列或对角线时，游戏结束。
*/


class Solution {

    final static int n = 3;
    public boolean validTicTacToe(String[] board) {
        int cx = 0 , co = 0;
        char[][] plant = new char[n][n];
        for (int i = 0 ; i < n; i ++) plant[i] = board[i].toCharArray();
        for (int i = 0 ; i < n ;i ++){
            for(int j = 0; j < n ; j++ ){
                if (plant[i][j] == 'X') cx ++;
                else if (plant[i][j] == 'O') co ++;
            }
        }
        int diff = cx - co;
        boolean xIsWiner = isWiner('X', plant);
        boolean oIsWiner = isWiner('O', plant);

        if (xIsWiner && !oIsWiner)  return diff == 1;
        if (oIsWiner && !xIsWiner)  return diff == 0;
        if (!xIsWiner && !oIsWiner) return 0 <= diff && diff <= 1;
        return false;
    }

    public boolean isWiner(char player, char[][] plant){
        int[] row = new int[n];
        int[] col = new int[n];
        int ldig = 0, rdig = 0;
        for (int i = 0 ; i < n; i++){
            for (int j = 0; j < n ; j++){
                if (plant[i][j] != player) continue;
                row[i]++;
                col[j]++;
                if (i == j) ldig ++;
                if (i + j == n-1) rdig ++;
                if (row[i] == 3 || col[j] == 3 || ldig == 3 || rdig == 3) return true;
            }
        }
        return false;
    }
}