下面的代码确实按照我需要的方式工作,但它很丑,过多或其他一些事情。我已经看了公式,并试图写一些解决方案,但我最终得到了类似数量的语句。

在这种情况下,是否有一种数学公式对我有益,或者是否可以接受16个if语句?

为了解释代码,这是一款基于同时回合制的游戏。两名玩家各有四个操作按钮,结果来自一个数组(0-3),但变量“1”和“2”可以赋值任何东西,如果这有帮助的话。结果是,0 =双方都不赢,1 = p1赢,2 = p2赢,3 =双方都赢。

public int fightMath(int one, int two) {

    if(one == 0 && two == 0) { result = 0; }
    else if(one == 0 && two == 1) { result = 0; }
    else if(one == 0 && two == 2) { result = 1; }
    else if(one == 0 && two == 3) { result = 2; }
    else if(one == 1 && two == 0) { result = 0; }
    else if(one == 1 && two == 1) { result = 0; }
    else if(one == 1 && two == 2) { result = 2; }
    else if(one == 1 && two == 3) { result = 1; }
    else if(one == 2 && two == 0) { result = 2; }
    else if(one == 2 && two == 1) { result = 1; }
    else if(one == 2 && two == 2) { result = 3; }
    else if(one == 2 && two == 3) { result = 3; }
    else if(one == 3 && two == 0) { result = 1; }
    else if(one == 3 && two == 1) { result = 2; }
    else if(one == 3 && two == 2) { result = 3; }
    else if(one == 3 && two == 3) { result = 3; }

    return result;
}

当前回答

我没有使用Java的经验,所以可能会有一些错字。请将此代码视为伪代码。

我会选择一个简单的开关。为此,您需要一个单一的数字求值。然而,对于这种情况,由于0 <= one < 4 <= 9和0 <= two < 4 <= 9,我们可以将这两个整型都转换为简单的整型,方法是1乘以10再加上2。然后在结果数字中使用一个开关,就像这样:

public int fightMath(int one, int two) {
    // Convert one and two to a single variable in base 10
    int evaluate = one * 10 + two;

    switch(evaluate) {
        // I'd consider a comment in each line here and in the original code
        // for clarity
        case 0: result = 0; break;
        case 1: result = 0; break;
        case 1: result = 0; break;
        case 2: result = 1; break;
        case 3: result = 2; break;
        case 10: result = 0; break;
        case 11: result = 0; break;
        case 12: result = 2; break;
        case 13: result = 1; break;
        case 20: result = 2; break;
        case 21: result = 1; break;
        case 22: result = 3; break;
        case 23: result = 3; break;
        case 30: result = 1; break;
        case 31: result = 2; break;
        case 32: result = 3; break;
        case 33: result = 3; break;
    }

    return result;
}

还有另一个简单的方法,我只是想指出一个理论代码。然而,我不会使用它,因为它有一些额外的复杂性,你通常不想处理。额外的复杂性来自于以4为基数,因为计数是0、1、2、3、10、11、12、13、20,……

public int fightMath(int one, int two) {
    // Convert one and two to a single variable in base 4
    int evaluate = one * 4 + two;

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

    return allresults[evaluate];
}

这只是额外的说明,以防我在Java中遗漏了一些东西。在PHP中我会这样做:

function fightMath($one, $two) {
    // Convert one and two to a single variable in base 4
    $evaluate = $one * 10 + $two;

    $allresults = array(
         0 => 0,  1 => 0,  2 => 1,  3 => 2,
        10 => 0, 11 => 0, 12 => 2, 13 => 1,
        20 => 2, 21 => 1, 22 => 3, 23 => 3,
        30 => 1, 31 => 2, 32 => 3, 33 => 3 );

    return $allresults[$evaluate];
}

其他回答

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。

试试开关外壳……

看看这里或这里关于它的更多信息

switch (expression)
{ 
  case constant:
        statements;
        break;
  [ case constant-2:
        statements;
        break;  ] ...
  [ default:
        statements;
        break;  ] ...
}

您可以向它添加多个条件(不是同时添加),甚至可以在没有满足其他情况的情况下添加默认选项。

PS:只有在满足一个条件的情况下。

如果两种情况同时出现。我认为开关不能用。 但是您可以减少这里的代码。

Java开关语句多种情况

看看我们都知道些什么

1:你的答案对于参与人1 P1和参与人2 P2是对称的。这对于格斗游戏来说很有意义,但你也可以利用它来完善你的逻辑。

2:3拍0拍2拍1拍3。这些情况中唯一不包括的情况是0对1和2对3的组合。换句话说,唯一的胜利表是这样的:0击败2,1击败3,2击败1,3击败0。

3:如果0/1人对位,则平局无命中,但如果2/3人对位,则双方均命中

首先,让我们构建一个单向函数,告诉我们是否赢了:

// returns whether we beat our opponent
public boolean doesBeat(int attacker, int defender) {
  int[] beats = {2, 3, 1, 0};
  return defender == beats[attacker];
}

然后我们可以使用这个函数来组合最终的结果:

// returns the overall fight result
// bit 0 = one hits
// bit 1 = two hits
public int fightMath(int one, int two)
{
  // Check to see whether either has an outright winning combo
  if (doesBeat(one, two))
    return 1;

  if (doesBeat(two, one))
    return 2;

  // If both have 0/1 then its hitless draw but if both have 2/3 then they both hit.
  // We can check this by seeing whether the second bit is set and we need only check
  // one's value as combinations where they don't both have 0/1 or 2/3 have already
  // been dealt with 
  return (one & 2) ? 3 : 0;
}

虽然这可以说比许多答案中提供的查找表更复杂,而且可能更慢,但我相信这是一种更好的方法,因为它实际上封装了代码的逻辑,并向阅读您代码的任何人描述它。我认为这是一个更好的实现。

(这是一段时间以来,我做任何Java,所以抱歉,如果语法错误,希望它仍然是可理解的,如果我有一点错误)

顺便说一下,0-3显然意味着什么;它们不是任意的值,所以给它们命名会有帮助。

说实话,每个人都有自己的代码风格。我没想到性能会受到太大影响。如果您比使用开关箱版本更能理解这一点,那么请继续使用此版本。

您可以嵌套if语句,因此最后的if检查可能会略微提高性能,因为它不会经过那么多if语句。但是在你的java基础课程中,它可能不会有什么好处。

else if(one == 3 && two == 3) { result = 3; }

所以,与其…

if(one == 0 && two == 0) { result = 0; }
else if(one == 0 && two == 1) { result = 0; }
else if(one == 0 && two == 2) { result = 1; }
else if(one == 0 && two == 3) { result = 2; }

你会做……

if(one == 0) 
{ 
    if(two == 0) { result = 0; }
    else if(two == 1) { result = 0; }
    else if(two == 2) { result = 1; }
    else if(two == 3) { result = 2; }
}

按照你的喜好重新格式化。

这并没有使代码看起来更好,但我相信它可能会加快一点速度。

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?