// 判斷是否具備消除的基本條件：兩個方塊不能是同一個坐標(biāo)；兩個方塊必須是同種類型；兩個方塊中不能有任何一個已經(jīng)消除過的
public static boolean basicCondition(Point a, Point b) {
    return !a.equals(b) && board[a.x][a.y] == board[b.x][b.y] && !isNull(a) && !isNull(b);
}

// 判斷格子是否為空或已經(jīng)被消除
public static boolean isNull(Point c) {
    return board[c.x][c.y] == 0 || board[c.x][c.y] == mark;
}

// 判斷同一直線能否相連
public static boolean matchLine(Point a, Point b) {
    // 水平
    if (a.x == b.x) {
        int minY = Math.min(a.y, b.y), maxY = Math.max(a.y, b.y);
        for (int i = minY + 1; i < maxY; i++) {
            if (!isNull(new Point(a.x, i))) return false;
        }
        return true;
    }
    // 垂直
    else if (a.y == b.y) {
        int minX = Math.min(a.x, b.x), maxX = Math.max(a.x, b.x);
        for (int i = minX + 1; i < maxX; i++) {
            if (!isNull(new Point(i, a.y))) return false;
        }
        return true;
    }
    // 不在水平或垂直上
    return false;
}

// 判斷 1 折能否相連：拐角點(diǎn) c1 和 c2 與 a b 點(diǎn)能相連并且拐角點(diǎn)為空
public static boolean matchOneTurn(Point a, Point b) {
    Point c1 = new Point(a.x, b.y);
    Point c2 = new Point(b.x, a.y);
    return matchLine(a, c1) && matchLine(b, c1) && isNull(c1)
            || matchLine(a, c2) && matchLine(b, c2) && isNull(c2);
}

// 判斷 2 折能否相連：掃描 a 所在的行和列，找一點(diǎn) c 使之與 a 直線匹配，與 b 1 折匹配；掃描 b 所在的行和列，找一點(diǎn) c 使之與 b 直線匹配，與 a 1 折匹配
public static boolean matchTwoTurn(Point a, Point b) {
    // 掃描 a b 所在的行
    for (int i = 0; i < c; i++) {
        Point c1 = new Point(a.x, i);
        Point c2 = new Point(b.x, i);
        if (i != a.y && matchLine(c1, a) && matchOneTurn(c1, b) && isNull(c1)
                || i != b.y && matchLine(c2, b) && matchOneTurn(c2, a) && isNull(c2))
            return true;
    }
    // 掃描 a b 所在的列
    for (int i = 0; i < r; i++) {
        Point c1 = new Point(i, a.y);
        Point c2 = new Point(i, b.y);
        if (i != a.x && matchLine(c1, a) && matchOneTurn(c1, b) && isNull(c1)
                || i != b.x && matchLine(c2, b) && matchOneTurn(c2, a) && isNull(c2))
            return true;
    }
    // 不存在這樣的 c 點(diǎn)
    return false;
}

// 整合判斷
public static boolean match(Point a, Point b) {
    return basicCondition(a, b) && (matchLine(a, b) || matchOneTurn(a, b) || matchTwoTurn(a, b));
}