随机化一个 BigInteger

I'm looking to randomize a BigInteger. The intent is to pick a number from 1 to 8180385048. Though, from what I noticed, the BigInteger(BitLen, Random) does it from n to X²-1, I'd want some unpredictable number. I tried to make a method that would do it, but I keep running into bugs and have finally given in to asking on here. :P Does anyone have any suggestions on how to do this?

从文档来看Random.nextInt(int n) http://download-llnw.oracle.com/javase/6/docs/api/java/util/Random.html#nextInt%28int%29显然需要解决同样的问题，他们似乎得出的结论是，你不能比“超出范围时重新采样”做得更好，但惩罚预计可以忽略不计。

来自文档：

The algorithm is slightly tricky. It rejects values that would result in an uneven distribution (due to the fact that 2³¹ is not divisible by n). The probability of a value being rejected depends on n. The worst case is n=2³⁰+1, for which the probability of a reject is 1/2, and the expected number of iterations before the loop terminates is 2.

我建议你简单地使用随机化构造函数 http://download.oracle.com/javase/6/docs/api/java/math/BigInteger.html#BigInteger%28int,%20java.util.Random%29您提到并迭代，直到达到范围内的值，例如如下所示：

public static BigInteger rndBigInt(BigInteger max) {
    Random rnd = new Random();
    do {
        BigInteger i = new BigInteger(max.bitLength(), rnd);
        if (i.compareTo(max) <= 0)
            return i;
    } while (true);
}

public static void main(String... args) {
    System.out.println(rndBigInt(new BigInteger("8180385048")));
}

对于您的特定情况（最大值 = 8180385048），必须重申的概率（即使一次）约为 4.8%，所以不用担心:-)