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;
}
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;
}
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];
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);
}
}
