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;
}
下面是一个相当简洁的版本，类似于JAB的回应。这使用地图来存储哪个移动战胜其他移动。
public enum Result {
P1Win, P2Win, BothWin, NeitherWin;
}
public enum Move {
BLOCK_HIGH, BLOCK_LOW, ATTACK_HIGH, ATTACK_LOW;
static final Map<Move, List<Move>> beats = new EnumMap<Move, List<Move>>(
Move.class);
static {
beats.put(BLOCK_HIGH, new ArrayList<Move>());
beats.put(BLOCK_LOW, new ArrayList<Move>());
beats.put(ATTACK_HIGH, Arrays.asList(ATTACK_LOW, BLOCK_LOW));
beats.put(ATTACK_LOW, Arrays.asList(ATTACK_HIGH, BLOCK_HIGH));
}
public static Result compare(Move p1Move, Move p2Move) {
boolean p1Wins = beats.get(p1Move).contains(p2Move);
boolean p2Wins = beats.get(p2Move).contains(p1Move);
if (p1Wins) {
return (p2Wins) ? Result.BothWin : Result.P1Win;
}
if (p2Wins) {
return (p1Wins) ? Result.BothWin : Result.P2Win;
}
return Result.NeitherWin;
}
}
例子:
System.out.println(Move.compare(Move.ATTACK_HIGH, Move.BLOCK_LOW));
打印:
P1Win
下面是一个相当简洁的版本，类似于JAB的回应。这使用地图来存储哪个移动战胜其他移动。
public enum Result {
P1Win, P2Win, BothWin, NeitherWin;
}
public enum Move {
BLOCK_HIGH, BLOCK_LOW, ATTACK_HIGH, ATTACK_LOW;
static final Map<Move, List<Move>> beats = new EnumMap<Move, List<Move>>(
Move.class);
static {
beats.put(BLOCK_HIGH, new ArrayList<Move>());
beats.put(BLOCK_LOW, new ArrayList<Move>());
beats.put(ATTACK_HIGH, Arrays.asList(ATTACK_LOW, BLOCK_LOW));
beats.put(ATTACK_LOW, Arrays.asList(ATTACK_HIGH, BLOCK_HIGH));
}
public static Result compare(Move p1Move, Move p2Move) {
boolean p1Wins = beats.get(p1Move).contains(p2Move);
boolean p2Wins = beats.get(p2Move).contains(p1Move);
if (p1Wins) {
return (p2Wins) ? Result.BothWin : Result.P1Win;
}
if (p2Wins) {
return (p1Wins) ? Result.BothWin : Result.P2Win;
}
return Result.NeitherWin;
}
}
例子:
System.out.println(Move.compare(Move.ATTACK_HIGH, Move.BLOCK_LOW));
打印:
P1Win
这使用了一点bitmagic(你已经通过在一个整数中保存两个比特信息(low/high & attack/block)来做到这一点):
我没有运行，只是在这里输入，请仔细检查。这个想法肯定有效。
编辑:现在对每个输入进行测试，工作正常。
public int fightMath(int one, int two) {
if(one<2 && two<2){ //both players blocking
return 0; // nobody hits
}else if(one>1 && two>1){ //both players attacking
return 3; // both hit
}else{ // some of them attack, other one blocks
int different_height = (one ^ two) & 1; // is 0 if they are both going for the same height - i.e. blocker wins, and 1 if height is different, thus attacker wins
int attacker = one>1?1:0; // is 1 if one is the attacker, two is the blocker, and 0 if one is the blocker, two is the attacker
return (attacker ^ different_height) + 1;
}
}
或者我应该建议把这两个信息分离成单独的变量?
主要基于上面这样的位操作的代码通常很难维护。
