Other people have already suggested my initial idea, the matrix method, but in addition to consolidating the if statements you can avoid some of what you have by making sure the arguments supplied are in the expected range and by using in-place returns (some coding standards I've seen enforce one-point-of-exit for functions, but I've found that multiple returns are very useful for avoiding arrow coding and with the prevalence of exceptions in Java there's not much point in strictly enforcing such a rule anyway as any uncaught exception thrown inside the method is a possible point of exit anyway). Nesting switch statements is a possibility, but for the small range of values you're checking here I find if statements to be more compact and not likely to result in much of a performance difference, especially if your program is turn-based rather than real-time.

public int fightMath(int one, int two) {
    if (one > 3 || one < 0 || two > 3 || two < 0) {
        throw new IllegalArgumentException("Result is undefined for arguments outside the range [0, 3]");
    }

    if (one <= 1) {
        if (two <= 1) return 0;
        if (two - one == 2) return 1;
        return 2; // two can only be 3 here, no need for an explicit conditional
    }

    // one >= 2
    if (two >= 2) return 3;
    if (two == 1) return 1;
    return 2; // two can only be 0 here
}

由于部分输入->结果映射的不规则性，这最终会导致可读性较差。我更喜欢矩阵风格，因为它的简单性和你如何设置矩阵在视觉上有意义(尽管这在一定程度上受到我对Karnaugh地图的记忆的影响):

int[][] results = {{0, 0, 1, 2},
                   {0, 0, 2, 1},
                   {2, 1, 3, 3},
                   {2, 1, 3, 3}};

更新:鉴于您提到了阻塞/命中，这里对函数进行了更彻底的更改，将属性/属性保存枚举类型用于输入和结果，并对结果进行了一些修改以考虑阻塞，这应该会产生一个更可读的函数。

enum MoveType {
    ATTACK,
    BLOCK;
}

enum MoveHeight {
    HIGH,
    LOW;
}

enum Move {
    // Enum members can have properties/attributes/data members of their own
    ATTACK_HIGH(MoveType.ATTACK, MoveHeight.HIGH),
    ATTACK_LOW(MoveType.ATTACK, MoveHeight.LOW),
    BLOCK_HIGH(MoveType.BLOCK, MoveHeight.HIGH),
    BLOCK_LOW(MoveType.BLOCK, MoveHeight.LOW);

    public final MoveType type;
    public final MoveHeight height;

    private Move(MoveType type, MoveHeight height) {
        this.type = type;
        this.height = height;
    }

    /** Makes the attack checks later on simpler. */
    public boolean isAttack() {
        return this.type == MoveType.ATTACK;
    }
}

enum LandedHit {
    NEITHER,
    PLAYER_ONE,
    PLAYER_TWO,
    BOTH;
}

LandedHit fightMath(Move one, Move two) {
    // One is an attack, the other is a block
    if (one.type != two.type) {
        // attack at some height gets blocked by block at same height
        if (one.height == two.height) return LandedHit.NEITHER;

        // Either player 1 attacked or player 2 attacked; whoever did
        // lands a hit
        if (one.isAttack()) return LandedHit.PLAYER_ONE;
        return LandedHit.PLAYER_TWO;
    }

    // both attack
    if (one.isAttack()) return LandedHit.BOTH;

    // both block
    return LandedHit.NEITHER;
}

You don't even have to change the function itself if you want to add blocks/attacks of more heights, just the enums; adding additional types of moves will probably require modification of the function, though. Also, EnumSets might be more extensible than using extra enums as properties of the main enum, e.g. EnumSet<Move> attacks = EnumSet.of(Move.ATTACK_HIGH, Move.ATTACK_LOW, ...); and then attacks.contains(move) rather than move.type == MoveType.ATTACK, though using EnumSets will probably be slightly slower than direct equals checks.

对于一个成功的块产生一个计数器的情况，您可以替换if (one。height == 2 .height)返回LandedHit.NEITHER;与

if (one.height == two.height) {
    // Successful block results in a counter against the attacker
    if (one.isAttack()) return LandedHit.PLAYER_TWO;
    return LandedHit.PLAYER_ONE;
}

另外，使用三元运算符(boolean_expression ?result_if_true: result_if_false)可以使代码更紧凑(例如，前面块中的代码将变成return one.isAttack() ?LandedHit。PLAYER_TWO: LandedHit.PLAYER_ONE;)，但这可能会导致更难阅读的一行代码，所以我不建议在更复杂的分支中使用它。

您可以创建包含结果的矩阵

int[][] results = {{0, 0, 1, 2}, {0, 0, 2, 1},{2, 1, 3, 3},{2, 1, 3, 3}};

当你想要获得价值时，你就会使用

public int fightMath(int one, int two) {
  return this.results[one][two]; 
}

我会使用Map, HashMap或TreeMap

特别是当参数不是0 <= X < N形式时

就像一组随机的正整数。

Code

public class MyMap
{
    private TreeMap<String,Integer> map;

    public MyMap ()
    {
        map = new TreeMap<String,Integer> ();
    }

    public void put (int key1, int key2, Integer value)
    {
        String key = (key1+":"+key2);

        map.put(key, new Integer(value));
    }

    public Integer get (int key1, int key2)
    {
        String key = (key1+":"+key2);

        return map.get(key);
    }
}

我希望我正确理解了逻辑。比如:

public int fightMath (int one, int two)
{
    int oneHit = ((one == 3 && two != 1) || (one == 2 && two != 0)) ? 1 : 0;
    int twoHit = ((two == 3 && one != 1) || (two == 2 && one != 0)) ? 2 : 0;

    return oneHit+twoHit;
}

检查一个击中高或一个击中低不被阻止，同样的球员二。

编辑:算法不完全理解，“命中”奖励时，我没有意识到(谢谢elias):

public int fightMath (int one, int two)
{
    int oneAttack = ((one == 3 && two != 1) || (one == 2 && two != 0)) ? 1 : (one >= 2) ? 2 : 0;
    int twoAttack = ((two == 3 && one != 1) || (two == 2 && one != 0)) ? 2 : (two >= 2) ? 1 : 0;

    return oneAttack | twoAttack;
}

您可以使用一个开关大小写，而不是多个if

还要提到的是，因为你有两个变量，那么你必须合并这两个变量，以使用它们在开关

检查这个Java开关语句来处理两个变量?

我想到的第一件事基本上与Francisco Presencia给出的答案相同，但有所优化:

public int fightMath(int one, int two)
{
    switch (one*10 + two)
    {
    case  0:
    case  1:
    case 10:
    case 11:
        return 0;
    case  2:
    case 13:
    case 21:
    case 30:
        return 1;
    case  3:
    case 12:
    case 20:
    case 31:
        return 2;
    case 22:
    case 23:
    case 32:
    case 33:
        return 3;
    }
}

你可以进一步优化它，使最后的情况(3)为默认情况:

    //case 22:
    //case 23:
    //case 32:
    //case 33:
    default:
        return 3;

此方法的优点是，与其他一些建议的方法相比，更容易看到1和2的哪个值对应于哪个返回值。