如何在特定范围内生成随机int值？
以下方法存在与整数溢出相关的错误：
randomNum = minimum + (int)(Math.random() * maximum);
// Bug: `randomNum` can be bigger than `maximum`.
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum = minimum + i;
// Bug: `randomNum` can be smaller than `minimum`.
当前回答
private static Random random = new Random();
public static int getRandomInt(int min, int max){
return random.nextInt(max - min + 1) + min;
}
OR
public static int getRandomInt(Random random, int min, int max)
{
return random.nextInt(max - min + 1) + min;
}
其他回答
Random rng = new Random();
int min = 3;
int max = 11;
int upperBound = max - min + 1; // upper bound is exclusive, so +1
int num = min + rng.nextInt(upperBound);
System.out.println(num);
下面是一个函数，它按照用户42155的请求，在lowerBoundIncluded和upperBoundIncluded定义的范围内返回一个整数随机数
SplitableRandom splitableRandom=新的Splitablerandom（）；
BiFunction<Integer,Integer,Integer> randomInt = (lowerBoundIncluded, upperBoundIncluded)
-> splittableRandom.nextInt(lowerBoundIncluded, upperBoundIncluded + 1);
randomInt.apply（…，…）；//获取随机数
…或更短，用于一次性生成随机数
new SplittableRandom().nextInt(lowerBoundIncluded, upperBoundIncluded + 1);
以前的大多数建议都不考虑“溢出”。例如：min=整数.min_VALUE，max=100。到目前为止，我采用的正确方法之一是：
final long mod = max- min + 1L;
final int next = (int) (Math.abs(rand.nextLong() % mod) + min);
这里有一个有用的类，可以在包含/排除边界的任意组合范围内生成随机整数：
import java.util.Random;
public class RandomRange extends Random {
public int nextIncInc(int min, int max) {
return nextInt(max - min + 1) + min;
}
public int nextExcInc(int min, int max) {
return nextInt(max - min) + 1 + min;
}
public int nextExcExc(int min, int max) {
return nextInt(max - min - 1) + 1 + min;
}
public int nextIncExc(int min, int max) {
return nextInt(max - min) + min;
}
}
int randomNum = 5 + (int)(Math.random()*5);
范围5-10
