Generate a random integer with a specified number of digits Java

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How do I generate random integers within a specific range in Java? 57 answers

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private long generateRandomNumber(int n) {
    long min = (long) Math.pow(10, n - 1);
    return ThreadLocalRandom.current().nextLong(min, min * 10);
}

nextLong produces random numbers between lower bound inclusive and upper bound exclusive so calling it with parameters (1_000, 10_000) for example results in numbers 1000 to 9999. Old Random did not get those nice new features unfortunately. But there is basically no reason to continue to use it anyways.

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public static int randomInt(int digits) {
    int minimum = (int) Math.pow(10, digits - 1); // minimum value with 2 digits is 10 (10^1)
    int maximum = (int) Math.pow(10, digits) - 1; // maximum value with 2 digits is 99 (10^2 - 1)
    Random random = new Random();
    return minimum + random.nextInt((maximum - minimum) + 1);
}

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You can simply disregard the numbers that are not in the required range. That way your modified pseudo random number generator guarantees that it generates a number in the given range uniformly at random:

public class RandomOfDigits {
    public static void main(String[] args) {
        int nd = Integer.parseInt(args[0]);
        int loIn = (int) Math.pow(10, nd-1);
        int hiEx = (int) Math.pow(10, nd);
        Random r = new Random();
        int x;
        do {
            x = r.nextInt(hiEx);
        } while (x < loIn);
        System.out.println(x);
    }
}

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