How do I generate a random integer between min and

问题:

What method returns a random int between a min and max? Or does no such method exist?

What I'm looking for is something like this:

NAMEOFMETHOD (min, max) 

(where min and max are ints), that returns something like this:

8

(randomly)

If such a method does exist could you please link to the relevant documentation with your answer.

Thanks.

---

UPDATE

Attempting to implement the full solution and I get the following error message:

class TestR
{
    public static void main (String[]arg) 
    {   
        Random random = new Random() ;
        int randomNumber = random.nextInt(5) + 2;
        System.out.println (randomNumber) ; 
    } 
} 

I'm still getting the same errors from the compiler:

TestR.java:5: cannot find symbol
symbol  : class Random
location: class TestR
        Random random = new Random() ;
        ^
TestR.java:5: cannot find symbol
symbol  : class Random
location: class TestR
        Random random = new Random() ;
                            ^
TestR.java:6: operator + cannot be applied to Random.nextInt,int
        int randomNumber = random.nextInt(5) + 2;
                                         ^
TestR.java:6: incompatible types
found   : <nulltype>
required: int
        int randomNumber = random.nextInt(5) + 2;
                                             ^
4 errors

What's going wrong here?

回答1:

Construct a Random object at application startup:

Random random = new Random();

Then use Random.nextInt(int):

int randomNumber = random.nextInt(max + 1 - min) + min;

Note that the both lower and upper limits are inclusive.

回答2:

You can use Random.nextInt(n). This returns a random int in [0,n). Just using max-min+1 in place of n and adding min to the answer will give a value in the desired range.

回答3:

public static int random_int(int Min, int Max)
{
     return (int) (Math.random()*(Max-Min))+Min;
}

random_int(5, 9); // For example

回答4:

As the solutions above do not consider the possible overflow of doing max-min when min is negative, here another solution (similar to the one of kerouac)

public static int getRandom(int min, int max) {
    if (min > max) {
        throw new IllegalArgumentException("Min " + min + " greater than max " + max);
    }      
    return (int) ( (long) min + Math.random() * ((long)max - min + 1));
}

this works even if you call it with:

getRandom(Integer.MIN_VALUE, Integer.MAX_VALUE) 

回答5:

With Java 7 or above you could use

ThreadLocalRandom.current().nextInt(int origin, int bound)

Javadoc: ThreadLocalRandom.nextInt

回答6:

Using the Random class is the way to go as suggested in the accepted answer, but here is a less straight-forward correct way of doing it if you didn't want to create a new Random object :

min + (int) (Math.random() * (max - min + 1));

回答7:

This generates a random integer of size psize

public static Integer getRandom(Integer pSize) {

    if(pSize<=0) {
        return null;
    }
    Double min_d = Math.pow(10, pSize.doubleValue()-1D);
    Double max_d = (Math.pow(10, (pSize).doubleValue()))-1D;
    int min = min_d.intValue();
    int max = max_d.intValue();
    return RAND.nextInt(max-min) + min;
}

回答8:

import java.util.Random;