如何使用 BigInteger 类在 Java 中生成一个随机的 n 位整数？
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How do I generate a random n digit integer in Java using the BigInteger class?
提问by makaveli2178
I am unsure about how to generate a random n digit integer in Java using the BigInteger class.
我不确定如何使用 BigInteger 类在 Java 中生成一个随机的 n 位整数。
采纳答案by Eyal Schneider
private static Random rnd = new Random();
public static String getRandomNumber(int digCount) {
StringBuilder sb = new StringBuilder(digCount);
for(int i=0; i < digCount; i++)
sb.append((char)('0' + rnd.nextInt(10)));
return sb.toString();
}
And then you can use it:
然后你可以使用它：
new BigInteger(getRandomNumber(10000))
回答by Carl
According to the docs, there is a constructor to do what you want in java 6: BigInteger(int, java.util.Random)
根据文档，有一个构造函数可以在 java 6 中执行您想要的操作：BigInteger(int, java.util.Random)
To that, you need only add a randomly selected 5000th digit-i.e. Use the rng constructor to 4999 digits, the add the last in via a separate random process. Actually, since you want to just sample performance for large values, you could generate the bits, and tack a one bit on the big end, rather than slave to decimal notation.
为此，您只需要添加一个随机选择的第 5000 位数字，即使用 rng 构造函数添加到 4999 位数字，通过单独的随机过程添加最后一位。实际上，由于您只想对大值的性能进行采样，您可以生成位，并在大端添加一位，而不是从属十进制表示法。
回答by Jon Skeet
The simplest way would probably to be to fill a char[] array with 5000 random digits, convert that to a string, and then call the BigInteger(String)constructor.
最简单的方法可能是用 5000 个随机数字填充 char[] 数组，将其转换为字符串，然后调用BigInteger(String)构造函数。
If any of those steps gives you problems, please give more details.
如果这些步骤中的任何一个给您带来问题，请提供更多详细信息。
Alternatively, you coulddo something like this:
或者，您可以执行以下操作：
Random rng = new Random(); // But use one instance throughout your app
BigInteger current = BigInteger.ZERO;
for (int i = 0; i < 5000; i++) {
BigInteger nextDigit = BigInteger.valueOf(rng.nextInt(10));
current = current.multiply(BigInteger.TEN).add(nextDigit);
}
I suspect that would be rather less efficient though.
不过，我怀疑这会效率较低。
You could reduce the number of steps required by generating nine random digits at a time, with rng.nextInt(1000000000).
您可以通过一次生成九个随机数字来减少所需的步骤数，使用rng.nextInt(1000000000).
回答by Sean Patrick Floyd
Here are two versions, one takes a Random as parameter (in case you want to re-use it):
这里有两个版本，一个以 Random 作为参数（以防你想重用它）：
public static BigInteger getRandomNumber(final int digCount){
return getRandomNumber(digCount, new Random());
}
public static BigInteger getRandomNumber(final int digCount, Random rnd){
final char[] ch = new char[digCount];
for(int i = 0; i < digCount; i++){
ch[i] =
(char) ('0' + (i == 0 ? rnd.nextInt(9) + 1 : rnd.nextInt(10)));
}
return new BigInteger(new String(ch));
}
The resulting BigInteger will always have the specified length.
生成的 BigInteger 将始终具有指定的长度。
回答by venu88
If n is between 1 to 12 then following method helps
如果 n 介于 1 到 12 之间，则以下方法会有所帮助
private String getRandom(int length) {
if (length < 1 && length > 12) {
throw new IllegalArgumentException("Random number generator length should be between 1 to 12");
}
long nextLong = Math.abs(random.nextLong());
return String.valueOf(nextLong).substring(0, length);
}
One more thing to note is that it is not well tested code.
还要注意的一件事是，它不是经过良好测试的代码。
回答by Dheeraj Joshi
Take a string with 5000 digits in it then convert it into BigInteger.
取一个包含 5000 位数字的字符串，然后将其转换为 BigInteger。
