英文:
How to randomly choose what code should be run?
问题
除了我这里的写法，是否有一种随机选择要运行的代码的方法？肯定有一种更简单的方法。我知道有关switch和case，但不知道是否可以在case中随机执行一个情况。
Random r = new Random();
int i = r.nextInt(4);
switch (i) {
case 0:
// 做某事
break;
case 1:
// 做其他事情
break;
case 2:
// 做其他事情
break;
case 3:
// 做其他事情
break;
default:
// 在没有匹配的情况下执行的代码
break;
}
英文:
Is there a way to randomly pick what code should run other than how I have it here? There's gotta be an easier way. I know about switch and case but didn't know if I could randomly have a case there.
Random r = new Random();
int i = r.nextInt(4);
if (i == 0) {
//dosomething
} else if (i == 1) {
// dosomethingelse
} else if(i == 2) {
//dosomethingelse
} else if(i == 3) {
//dosomethingelse
} else if(i == 4) {
//dosomethingelse
}
</details>
# 答案1
**得分**: 2
你当前的代码已经非常优化，至少对于在Java中执行类似操作而言是这样。作为一个指针，你可以重写逻辑以使用`switch`语句，这样至少可以使代码变得更易读：
```java
Random r = new Random();
int i = r.nextInt(5);
switch(i) {
case 0:
// 做一些事情
break;
case 1:
// 做另一些事情
break;
case 2:
// 做另一些事情
break;
case 3:
// 做另一些事情
break;
case 4:
// 做另一些事情
break;
}
注意：如果你想生成一个在0和4之间的随机整数（包括4），则使用nextInt(5)。使用nextInt(4)将永远不会生成值4。
英文:
Your current code is already pretty optimal, at least for how you would do something like this in Java. As a pointer, you could rewrite your logic to use a switch statement, which would at least make the code a bit easier to read:
Random r = new Random();
int i = r.nextInt(5);
switch(i) {
case 0:
// do something
break;
case 1:
// do something else
break;
case 2:
// do something else
break;
case 3:
// do something else
break;
case 4:
// do something else
break;
}
Note: If you want to generate a random integer between 0 and 4 (inclusive of the 4), then use nextInt(5). Using nextInt(4) will never actually generate the value 4.
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