Question

我正在尝试编写一个名为Mancala的电路板代码，我想更改winner（）函数。

如果一方在其任何一个坑中都没有任何一块石头，则游戏结束，然后确定获胜者是否PlayerOne的mancala比PlayerTwo更多的石头，反之亦然。

但是我不知道如何定位对象中的特定元素，所以我可以扩展win函数...任何帮助将不胜感激！

我想要达到的目标是：

if (sideOne.get(13).number > sideOne.get(14).number) {
    return...
}

我知道上面的代码是不正确的，但我想要的目标是数组列表，该列表的特定索引和对象中的整数元素。

我已经包含了游戏板的构造函数和当前的winner（）函数。

MancalaPit等级

public class MancalaPit {
    String player;
    Integer stones;
    Boolean pit;
    Integer number;
    MancalaPit next;

    public MancalaPit(String player, int stones, boolean pit, int number, MancalaPit next) {
        this.player = player;
        this.stones = stones;
        this.pit = pit;
        this.number = number;
        this.next = next;
    }
}

public class Board { // Creates the board

    private MancalaPit a;
    private MancalaPit b;
    private MancalaPit c;
    private MancalaPit d;
    private MancalaPit e;
    private MancalaPit f;
    private MancalaPit mancalaOne;
    private MancalaPit g;
    private MancalaPit h;
    private MancalaPit i;
    private MancalaPit j;
    private MancalaPit k;
    private MancalaPit l;
    private MancalaPit mancalaTwo;
    private ArrayList<MancalaPit> sideOne;
    private ArrayList<MancalaPit> sideTwo;

    public Board() { // Constructs the bad boy

        // Constructs the array lists
        sideOne = new ArrayList<>();
        sideTwo = new ArrayList<>();

        // Constructs the pits and Mancala's
        a = new MancalaPit("playerOne", 4, false, 1, null);
        b = new MancalaPit("playerOne", 4, false, 2, null);
        c = new MancalaPit("playerOne", 4, false, 3, null);
        d = new MancalaPit("playerOne", 4, false, 4, null);
        e = new MancalaPit("playerOne", 4, false, 5, null);
        f = new MancalaPit("playerOne", 4, false, 6, null);
        mancalaOne = new MancalaPit("playerOne", 0, true, 13, null);
        g = new MancalaPit("playerTwo", 4, false, 7, null);
        h = new MancalaPit("playerTwo", 4, false, 8, null);
        i = new MancalaPit("playerTwo", 4, false, 9, null);
        j = new MancalaPit("playerTwo", 4, false, 10, null);
        k = new MancalaPit("playerTwo", 4, false, 11, null);
        l = new MancalaPit("playerTwo", 4, false, 12, null);
        mancalaTwo = new MancalaPit("playerTwo", 0, true, 14, null);

        // Constructs the order of the pits
        a.next = b;
        b.next = c;
        c.next = d;
        d.next = e;
        e.next = f;
        f.next = mancalaOne;
        mancalaOne.next = g;
        g.next = h;
        h.next = i;
        i.next = j;
        j.next = k;
        k.next = l;
        l.next = mancalaTwo;
        mancalaTwo.next = a;

        // Constructs sides
        sideOne.add(a);
        sideOne.add(b);
        sideOne.add(c);
        sideOne.add(d);
        sideOne.add(e);
        sideOne.add(f);
        sideTwo.add(g);
        sideTwo.add(h);
        sideTwo.add(i);
        sideTwo.add(j);
        sideTwo.add(k);
        sideTwo.add(l);

    }

    public String winner() {
        Boolean one = true;
        Boolean two = true;
        for(MancalaPit pit : sideOne) {
            if(pit.stones == 0) {
                continue;
            } else {
                one = false;
                break;
            }
        }
        if(one == true) {
            return "Your boy player one is the champion";
        }
        for(MancalaPit pit : sideTwo) {
            if(pit.stones == 0) {
                continue;
            } else {
                two = false;
                break;
            }
        }
        if(two == true) {
            return "Your boy player two is the champion";
        }

        return "No winners so far";
    }

Answer 1

我认为您根本不需要定位索引（特别是因为这些坑不在您创建的列表中），您已经定义了变量以比较获胜条件。你应该能够将mancalaOne.stones与mancalaTwo.stones进行比较，如果其中一个＆＃34;一个＆＃34;是真的还是＆＃34;两个＆＃34;是真的。 BTW一和二应该是布尔而不是布尔。

即。做两个for循环，然后做 if（one || two）{ if（mancalaOne.stones＆gt; mancalaTwo.stones）{ 返回＆＃34;玩家1获胜！＆＃34; 如果...... }

如何定位特定列表索引（JAVA）

1 个答案: