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
让我们举个例子。
假设我希望生成5-10之间的数字：
int max = 10;
int min = 5;
int diff = max - min;
Random rn = new Random();
int i = rn.nextInt(diff + 1);
i += min;
System.out.print("The Random Number is " + i);
让我们了解这一点。。。
用最高值初始化max，用最低值初始化min。现在，我们需要确定可以获得多少可能的值。在本例中，应为：5, 6, 7, 8, 9, 10所以，这个计数应该是max-min+1。即10-5+1=6随机数将生成0-5之间的数字。即0、1、2、3、4、5将最小值添加到随机数将产生：5, 6, 7, 8, 9, 10 因此，我们获得了所需的范围。
其他回答
rand.nextInt((max+1) - min) + min;
这很好。
我已经创建了一个方法来获取给定范围内的唯一整数。
/*
* minNum is the minimum possible random number
* maxNum is the maximum possible random number
* numbersNeeded is the quantity of random number required
* the give method provides you with unique random number between min & max range
*/
public static Set<Integer> getUniqueRandomNumbers( int minNum , int maxNum ,int numbersNeeded ){
if(minNum >= maxNum)
throw new IllegalArgumentException("maxNum must be greater than minNum");
if(! (numbersNeeded > (maxNum - minNum + 1) ))
throw new IllegalArgumentException("numberNeeded must be greater then difference b/w (max- min +1)");
Random rng = new Random(); // Ideally just create one instance globally
// Note: use LinkedHashSet to maintain insertion order
Set<Integer> generated = new LinkedHashSet<Integer>();
while (generated.size() < numbersNeeded)
{
Integer next = rng.nextInt((maxNum - minNum) + 1) + minNum;
// As we're adding to a set, this will automatically do a containment check
generated.add(next);
}
return generated;
}
使用nexttint（n）方法为最小值和最大值的差值生成一个随机数，然后将最小值添加到结果中：
Random rn = new Random();
int result = rn.nextInt(max - min + 1) + min;
System.out.println(result);
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);
我想知道Apache Commons Math库提供的任何随机数生成方法是否符合要求。
例如：RandomDataGenerator.nextInt或RandomDataGenerator.nextLong
