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

import java.util.Random;

public class RandomSSNTest {

    public static void main(String args[]) {
        generateDummySSNNumber();
    }


    //831-33-6049
    public static void generateDummySSNNumber() {
        Random random = new Random();

        int id1 = random.nextInt(1000);//3
        int id2 = random.nextInt(100);//2
        int id3 = random.nextInt(10000);//4

        System.out.print((id1+"-"+id2+"-"+id3));
    }

}

您也可以使用

import java.util.concurrent.ThreadLocalRandom;
Random random = ThreadLocalRandom.current();

然而,这个类在多线程环境中表现不佳。

其他回答

我发现这个例子生成随机数:


此示例生成特定范围内的随机整数。

import java.util.Random;

/** Generate random integers in a certain range. */
public final class RandomRange {

  public static final void main(String... aArgs){
    log("Generating random integers in the range 1..10.");

    int START = 1;
    int END = 10;
    Random random = new Random();
    for (int idx = 1; idx <= 10; ++idx){
      showRandomInteger(START, END, random);
    }

    log("Done.");
  }

  private static void showRandomInteger(int aStart, int aEnd, Random aRandom){
    if ( aStart > aEnd ) {
      throw new IllegalArgumentException("Start cannot exceed End.");
    }
    //get the range, casting to long to avoid overflow problems
    long range = (long)aEnd - (long)aStart + 1;
    // compute a fraction of the range, 0 <= frac < range
    long fraction = (long)(range * aRandom.nextDouble());
    int randomNumber =  (int)(fraction + aStart);    
    log("Generated : " + randomNumber);
  }

  private static void log(String aMessage){
    System.out.println(aMessage);
  }
} 

此类的示例运行:

Generating random integers in the range 1..10.
Generated : 9
Generated : 3
Generated : 3
Generated : 9
Generated : 4
Generated : 1
Generated : 3
Generated : 9
Generated : 10
Generated : 10
Done.

我正在考虑使用以下方法将生成的随机数线性归一化到所需范围。设x为随机数,设a和b为期望归一化数的最小和最大范围。

下面是一个非常简单的代码片段,用来测试线性映射产生的范围。

public static void main(String[] args) {
    int a = 100;
    int b = 1000;
    int lowest = b;
    int highest = a;
    int count = 100000;
    Random random = new Random();
    for (int i = 0; i < count; i++) {
        int nextNumber = (int) ((Math.abs(random.nextDouble()) * (b - a))) + a;
        if (nextNumber < a || nextNumber > b) {
            System.err.println("number not in range :" + nextNumber);
        }
        else {
            System.out.println(nextNumber);
        }
        if (nextNumber < lowest) {
            lowest = nextNumber;
        }
        if (nextNumber > highest) {
            highest = nextNumber;
        }
    }
    System.out.println("Produced " + count + " numbers from " + lowest
            + " to " + highest);
}
public static Random RANDOM = new Random(System.nanoTime());

public static final float random(final float pMin, final float pMax) {
    return pMin + RANDOM.nextFloat() * (pMax - pMin);
}

要生成“介于两个数字之间”的随机数,请使用以下代码:

Random r = new Random();
int lowerBound = 1;
int upperBound = 11;
int result = r.nextInt(upperBound-lowerBound) + lowerBound;

这将为您提供一个介于1(含)和11(不含)之间的随机数,因此通过添加1来初始化上限值。例如,如果要生成1到10之间的随机数,则使用11而不是10初始化上限数。

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);