Question

我必须制作一个单人游戏。我们有4-4个玩家X和O，还有一个空格（从0到8）。看起来像这样：

|X|X|X|X| |O|O|O|O|

我必须交换X和O：

|O|O|O|O| |X|X|X|X|

规则是：

任何符号都可以移动到相邻的空格。
X旁边的空白为空白，则X可以跳过O，反之亦然。例如：
```
|X|X|X| |X|O|O|O|O| -> |X|X|X|O|X| |O|O|O|
```
没有其他方法可以移动它们。

我坚持第二条规则大约2天。

public static final int charX = 88;
public static final int charO = 79;
public static final int charSpace = 32;
    //steps[0] = from;
    //steps[1] = to
    // X = 88 ; O = 79 ; Space = 32;
    //char[] table = {'X', 'X', 'X', 'X', ' ', 'O', 'O', 'O', 'O'};
    //int[] table = {88, 88, 88, 88, 32, 79, 79, 79, 79};    



public static int[] ask() {
    Scanner sc = new Scanner(System.in);
    int from;
    int to;
    System.out.println("From where? (0-8)");
    from = sc.nextInt();
    System.out.println("To where? (0-8)");
    to = sc.nextInt();
    int[] step = {from, to};
    return step;
}

public static boolean isValidMove(int[] steps, int[] table) {

    int from = steps[0];
    int to = steps[1];

    boolean checker = (Math.abs(from - to) != 1 && table[to] != charSpace);
    if (checker == true) {
        return true;
    } else 
        return false;

}

public static int[] validMove(Scanner sc, int to, int from, int[] table) {

    int[] steps = ask();

    from = steps[0];
    to = steps[1];

    while (isValidMove(steps, table)) {
        System.out.println("Invalid move!");
        steps = ask();
        from = steps[0];
        to = steps[1];
    }

    return steps;
}

Answer 1

类似的东西：

for (int i=Math.min(steps[0],steps[1]); i<Math.max(steps[0],steps[1])-1; i++){
    // if table[i] is opposite of table[step[0]]
        // move not valid
}

河内塔之类的东西

1 个答案: