随机类具有在给定范围内生成随机int的方法。 例如：
Random r = new Random(); int x = r.nextInt(100);这将产生一个大于等于0且小于100的int数。我想用长数字做完全相同的操作。
long y = magicRandomLongGenerator(100);随机类只有nextLong（），但它不允许设置范围。
Random class has a method to generate random int in a given range. For example:
Random r = new Random(); int x = r.nextInt(100);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 class has only nextLong(), but it doesn't allow to set range.
最满意答案
从Java 7 （或Android API Level 21 = 5.0+）开始，您可以直接使用ThreadLocalRandom.current().nextLong(n) （0≤x<n）和ThreadLocalRandom.current().nextLong(m, n) （对于m≤x<n）。 详见@Alex的答案。
如果您遇到Java 6 （或Android 4.x），则需要使用外部库（例如org.apache.commons.math3.random.RandomDataGenerator.getRandomGenerator().nextLong(0, n-1) ，请参阅@ mawaldne的答案），或实现你自己的nextLong(n) 。
根据http://java.sun.com/j2se/1.5.0/docs/api/java/util/Random.html nextInt被实现为
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; }所以我们可以修改这个来执行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; }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.
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).
According to http://java.sun.com/j2se/1.5.0/docs/api/java/util/Random.html nextInt is implemented as
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; }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; }更多推荐
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