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

如果你不想重新发明轮子,这里有一个简单明了的解决方案:

RandomStringUtils.randomNumeric(count); // Where count is the number of digits you want in the random number.

Apache Commons,例如组件Lang中的RandomUtils,提供了许多选项来生成任意格式的随机数。

其他回答

在尝试1中进行以下更改应该可以完成工作-

randomNum = minimum + (int)(Math.random() * (maximum - minimum) );

检查此项以获取工作代码。

范围[最小值最大值](含)内的随机数:

int randomFromMinToMaxInclusive = ThreadLocalRandom.current()
        .nextInt(min, max + 1);
int func(int max, int min){

      int range = max - min + 1;
      
      // Math.random() function will return a random no between [0.0,1.0).
      int res = (int) ( Math.random()*range)+min;

      return res;
}

有一个图书馆在https://sourceforge.net/projects/stochunit/用于处理范围的选择。

StochIntegerSelector randomIntegerSelector = new StochIntegerSelector();
randomIntegerSelector.setMin(-1);
randomIntegerSelector.setMax(1);
Integer selectInteger = randomIntegerSelector.selectInteger();

它具有边缘包含/排除。

import java.util.Random; 

public class RandomUtil {
    // Declare as class variable so that it is not re-seeded every call
    private static Random random = new Random();

    /**
     * Returns a psuedo-random number between min and max (both inclusive)
     * @param min Minimim value
     * @param max Maximim value. Must be greater than min.
     * @return Integer between min and max (both inclusive)
     * @see java.util.Random#nextInt(int)
     */
    public static int nextInt(int min, int max) {
        // nextInt is normally exclusive of the top value,
        // so add 1 to make it inclusive
        return random.nextInt((max - min) + 1) + min;
    }
}