太多的“如果”语句?

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];
}

2014-03-19 17:19:51

其他回答

感谢@Joe Harper，因为我最终使用了他的答案的变体。为了进一步瘦身，每4个结果中有2个是相同的，我进一步瘦身。

我可能会在某个时候回到这个问题上，但如果没有由多个if语句引起的主要阻力，那么我现在就保留这个问题。我将进一步研究表格矩阵和开关语句解决方案。

public int fightMath(int one, int two) {
  if (one === 0) {
    if (two === 2) { return 1; }
    else if(two === 3) { return 2; }
    else { return 0; }
  } else if (one === 1) {
    if (two === 2) { return 2; }
    else if (two === 3) { return 1; }
    else { return 0; }
  } else if (one === 2) {
    if (two === 0) { return 2; }
    else if (two === 1) { return 1; }
    else { return 3; }
  } else if (one === 3) {
    if (two === 0) { return 1; }
    else if (two === 1) { return 2; }
    else { return 3; }
  }
}

2014-03-19 12:45:58

使用常量或枚举使代码更具可读性尝试将代码拆分为更多的函数试着利用问题的对称性

这里是一个建议，但在这里使用int型仍然有点难看:

static final int BLOCK_HIGH = 0;
static final int BLOCK_LOW = 1;
static final int ATTACK_HIGH = 2;
static final int ATTACK_LOW = 3;

public static int fightMath(int one, int two) {
    boolean player1Wins = handleAttack(one, two);
    boolean player2Wins = handleAttack(two, one);
    return encodeResult(player1Wins, player2Wins); 
}



private static boolean handleAttack(int one, int two) {
     return one == ATTACK_HIGH && two != BLOCK_HIGH
        || one == ATTACK_LOW && two != BLOCK_LOW
        || one == BLOCK_HIGH && two == ATTACK_HIGH
        || one == BLOCK_LOW && two == ATTACK_LOW;

}

private static int encodeResult(boolean player1Wins, boolean player2Wins) {
    return (player1Wins ? 1 : 0) + (player2Wins ? 2 : 0);
}

使用结构化类型作为输入和输出会更好。输入实际上有两个字段:位置和类型(阻挡或攻击)。输出也有两个字段:player1Wins和player2Wins。将其编码为单个整数会使代码更难阅读。

class PlayerMove {
    PlayerMovePosition pos;
    PlayerMoveType type;
}

enum PlayerMovePosition {
    HIGH,LOW
}

enum PlayerMoveType {
    BLOCK,ATTACK
}

class AttackResult {
    boolean player1Wins;
    boolean player2Wins;

    public AttackResult(boolean player1Wins, boolean player2Wins) {
        this.player1Wins = player1Wins;
        this.player2Wins = player2Wins;
    }
}

AttackResult fightMath(PlayerMove a, PlayerMove b) {
    return new AttackResult(isWinningMove(a, b), isWinningMove(b, a));
}

boolean isWinningMove(PlayerMove a, PlayerMove b) {
    return a.type == PlayerMoveType.ATTACK && !successfulBlock(b, a)
            || successfulBlock(a, b);
}

boolean successfulBlock(PlayerMove a, PlayerMove b) {
    return a.type == PlayerMoveType.BLOCK 
            && b.type == PlayerMoveType.ATTACK 
            && a.pos == b.pos;
}

不幸的是，Java并不擅长表达这类数据类型。

2014-03-29 18:03:48

如果你无法想出一个公式，你可以使用一个表格来描述有限的结果:

final int[][] result = new int[][] {
  { 0, 0, 1, 2 },
  { 0, 0, 2, 1 },
  { 2, 1, 3, 3 },
  { 1, 2, 3, 3 }
};
return result[one][two];

2014-03-19 09:32:15

相反，你可以这样做

   public int fightMath(int one, int two) {
    return Calculate(one,two)

    }


    private int Calculate(int one,int two){

    if (one==0){
        if(two==0){
     //return value}
    }else if (one==1){
   // return value as per condtiion
    }

    }

2014-03-19 09:32:12

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。

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;
}

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];
}

2014-03-19 17:19:51

太多的“如果”语句?

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