【BISHI15】小红的夹吃棋

求解代码

public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer in = new StringTokenizer(br.readLine());
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

        int t = Integer.parseInt(in.nextToken());

        for (int i = 0; i < t; i++) {
            char[][] board = new char[3][3];

            for (int j = 0; j < 3; j++) {
                String s = br.readLine().trim();
                board[j] = s.toCharArray();
            }

            boolean redEat = hasEat(board, '*');
            boolean purpleEat = hasEat(board, 'o');

            if (redEat && !purpleEat) {
                out.println("yukari");
            } else if (!redEat && purpleEat) {
                out.println("kou");
            } else {
                out.println("draw");
            }
        }

        out.flush();
        out.close();
        br.close();
    }

    private static boolean hasEat(char[][] board, char piece) {
        char opponent = (piece == '*') ? 'o' : '*';

        //遍历 3 行（行索引 0-2），
        // 仅检查 "该行中间列（索引 1）是己方棋子，
        // 左右列（索引 0、2）全是对方棋子"
        for (int row = 0; row < 3; row++) {
            if (board[row][1] == piece &&
                    board[row][0] == opponent &&
                    board[row][2] == opponent) {
                return true;
            }
        }

        //遍历 3 列（列索引 0-2），
        // 仅检查 "该列中间行（索引 1）是己方棋子，
        // 上下行（索引 0、2）全是对方棋子"
        for (int col = 0; col < 3; col++) {
            if (board[1][col] == piece &&
                    board[0][col] == opponent &&
                    board[2][col] == opponent) {
                return true;
            }
        }

        return false;
    }

小贴士

逐行读取 3 行字符串时需要用br.readLine().trim()，可以避免StringTokenizer的分割符问题，其实这题用in.nextToken()还过不了。