我有以下代码块。基本上它决定了棒球运动员是否获得了多个基础，如果有人在第1 /第2 /第3位，会发生什么。我这样做了蛮力但有没有办法重写这个以大幅减少行数？我无法想出一个好办法。
int runsScored = 0;
switch (action) {
case 0:
break;
case 1:
if (third) {
if (second) {
if (first) {
runsScored = 2;
second = false;
} else {
runsScored = 2;
first = true;
second = false;
third = false;
}
} else {
if (first) {
runsScored = 1;
second = true;
third = false;
} else {
runsScored = 1;
first = true;
third = false;
}
}
} else {
if (second) {
if (first) {
third = true;
} else {
first = true;
}
} else {
if (first) {
second = true;
} else {
first = true;
}
}
}
break;
case 2:
if (third) {
if (second) {
if (first) {
runsScored = 2;
first = false;
} else {
runsScored = 2;
third = false;
}
} else {
if (first) {
runsScored = 1;
second = true;
} else {
runsScored = 1;
second = true;
third = false;
}
}
} else {
if (second) {
if (first) {
runsScored = 1;
third = true;
second = true;
first = false;
} else {
runsScored = 1;
}
} else {
if (first) {
first = false;
second = true;
third = true;
} else {
second = true;
}
}
}
break;
case 3:
if (third) {
if (second) {
if (first) {
runsScored = 3;
first = false;
second = false;
third = true;
} else {
runsScored = 2;
second = false;
}
} else {
if (first) {
runsScored = 2;
first = false;
} else {
runsScored = 1;
}
}
} else {
if (second) {
if (first) {
runsScored = 2;
third = true;
second = false;
first = false;
} else {
runsScored = 1;
second = false;
third = true;
}
} else {
if (first) {
runsScored = 1;
first = false;
second = false;
third = true;
} else {
third = true;
}
}
}
break;
case 4:
if (third) {
if (second) {
if (first) {
runsScored = 4;
first = false;
second = false;
third = false;
} else {
runsScored = 3;
second = false;
third = false;
}
} else {
if (first) {
runsScored = 3;
first = false;
third = false;
} else {
runsScored = 2;
third = false;
}
}
} else {
if (second) {
if (first) {
runsScored = 3;
third = false;
second = false;
first = false;
} else {
runsScored = 2;
second = false;
third = false;
}
} else {
if (first) {
runsScored = 2;
first = false;
second = false;
third = false;
} else {
runsScored = 1;
}
}
}
break;
default:
throw new AssertionError();
}
return runsScored;
答案 0 :(得分：1)
好吧，我会用课程来描述这个并让它自我管理，但这是一个很好的第一步。
if (third) {
if (second) {
if (first) {
runsScored = 2;
second = false;
} else {
runsScored = 2;
first = true;
second = false;
third = false;
}
if (third) {
if (second) {
runsScored = 2;
second = false;
if (!first) {
first = true;
third = false;
}
即。在else的两边分解相同的代码。一旦你清除了一些灌木丛，就会看到更简单的东西。
答案 1 :(得分：1)
我使用按位运算符（shift和bitwise和）
（见http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html）
int runsScored = 0;
int action=1;
boolean first = false;
boolean second = true;
boolean third = true;
// Encode the bases in a bitstring.
int onbases = 0;
onbases += third ? 1 : 0;
onbases <<= 1;
onbases += second ? 1 : 0;
onbases <<= 1;
onbases += first ? 1 : 0;
onbases <<= 1;
// Bitmasks for use in the loop.
int homeplate = 16;
int basesloaded = 2+4+8;
onbases += 1; // Represent the batter who hasn't yet left the plate.
for (int i=0; i<action; i++) {
onbases <<= 1;
if ((onbases & homeplate) > 0) {
runsScored += 1;
}
}
// Remove the "players" who crossed home plate.
onbases &= basesloaded;
// onbases now reflects who's on base *after* the run.
// Decode the bitstring.
first = (onbases & 2) > 0;
second = (onbases & 4) > 0;
third = (onbases & 8) > 0;
Integer.toBinaryString(int)对于调试非常有用，如果你使用这些运算符，我发现在括号上加重是有帮助的，因为操作顺序有时并不直观。
编辑：我之前发布的内容忘了将击球手放在底座上。我相信它是固定的。
答案 2 :(得分：0)
假设你的action变量是击球手最终的基础，first, second, third是基础击球员的布尔值：
可以将它们转换为单个数组，而不是检查每个布尔值。例如，第一个和第三个的跑步者看起来像[1,0,1]。第二名和第三名的跑步者看起来像[0,1,1]。然后你可以编写一个函数，它接收你的action变量和基数组，并返回运行次数和新的基数数组。
这当然可以从case语句中删除嵌套语句，但也许您可以想出一种方法来使这项工作比嵌套if更好？
答案 3 :(得分：0)
布尔人的简短版本（仅限）：
boolean new_first = (action == 1) || (action == 2) && (first &!second & third);
boolean new_second = (action == 2) || (action == 1
&& (first && !(second && third))
|| (!first && second && !third));
boolean new_third = (action == 3)
|| (action == 2 && first)
|| (action == 1 && (first && second));
runsScored部分更难。