Java:0 <= x <中的随机长数n 范围

Java: random long number in 0 lt;= x lt; n range(Java:0 = x 中的随机长数n 范围)

问题描述

Random 类具有在给定范围内生成随机 int 的方法.例如:

Random class has a method to generate random int in a given range. For example:

Random r = new Random(); 
int x = r.nextInt(100);

这将生成一个大于或等于 0 且小于 100 的 int 数.我想对 long 数做同样的事情.

This would generate an int number more or equal to 0 and less than 100. I'd like to do exactly the same with long number.

long y = magicRandomLongGenerator(100);

Random 类只有 nextLong()，但它不允许设置范围.

Random class has only nextLong(), but it doesn't allow to set range.

推荐答案

从Java 7(或Android API Level 21 = 5.0+)开始你可以直接使用ThreadLocalRandom.current().nextLong(n)(对于 0 ≤ x ThreadLocalRandom.current().nextLong(m, n)(对于 m ≤ x

Starting from Java 7 (or Android API Level 21 = 5.0+) you could directly use ThreadLocalRandom.current().nextLong(n) (for 0 ≤ x < n) and ThreadLocalRandom.current().nextLong(m, n) (for m ≤ x < n). See @Alex's answer for detail.

如果您使用 Java 6(或 Android 4.x)，您需要使用外部库(例如 org.apache.commons.math3.random.RandomDataGenerator.getRandomGenerator().nextLong(0, n-1)，请参阅 @mawaldne 的答案)，或实施您的自己的 nextLong(n).

If you are stuck with Java 6 (or Android 4.x) you need to use an external library (e.g. org.apache.commons.math3.random.RandomDataGenerator.getRandomGenerator().nextLong(0, n-1), see @mawaldne's answer), or implement your own nextLong(n).

根据 https://docs.oracle.com/javase/1.5.0/docs/api/java/util/Random.html nextInt 实现为

 public int nextInt(int n) {
     if (n<=0)
                throw new IllegalArgumentException("n must be positive");

     if ((n & -n) == n)  // i.e., n is a power of 2
         return (int)((n * (long)next(31)) >> 31);

     int bits, val;
     do {
         bits = next(31);
         val = bits % n;
     } while(bits - val + (n-1) < 0);
     return val;
 }

所以我们可以修改它来执行 nextLong:

So we may modify this to perform nextLong:

long nextLong(Random rng, long n) {
   // error checking and 2^x checking removed for simplicity.
   long bits, val;
   do {
      bits = (rng.nextLong() << 1) >>> 1;
      val = bits % n;
   } while (bits-val+(n-1) < 0L);
   return val;
}

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