如何在特定范围内生成随机int值?

以下方法存在与整数溢出相关的错误:

randomNum = minimum + (int)(Math.random() * maximum);
// Bug: `randomNum` can be bigger than `maximum`.
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum =  minimum + i;
// Bug: `randomNum` can be smaller than `minimum`.

当前回答

你可以这样做:

import java.awt.*;
import java.io.*;
import java.util.*;
import java.math.*;

public class Test {

    public static void main(String[] args) {
        int first, second;

        Scanner myScanner = new Scanner(System.in);

        System.out.println("Enter first integer: ");
        int numOne;
        numOne = myScanner.nextInt();
        System.out.println("You have keyed in " + numOne);

        System.out.println("Enter second integer: ");
        int numTwo;
        numTwo = myScanner.nextInt();
        System.out.println("You have keyed in " + numTwo);

        Random generator = new Random();
        int num = (int)(Math.random()*numTwo);
        System.out.println("Random number: " + ((num>numOne)?num:numOne+num));
    }
}

其他回答

使用Java 8 IntStream和Collections.shuffle的不同方法

import java.util.stream.IntStream;
import java.util.ArrayList;
import java.util.Collections;

public class Main {

    public static void main(String[] args) {

        IntStream range = IntStream.rangeClosed(5,10);
        ArrayList<Integer> ls =  new ArrayList<Integer>();

        //populate the ArrayList
        range.forEach(i -> ls.add(new Integer(i)) );

        //perform a random shuffle  using the Collections Fisher-Yates shuffle
        Collections.shuffle(ls);
        System.out.println(ls);
    }
}

Scala中的等价项

import scala.util.Random

object RandomRange extends App{
  val x =  Random.shuffle(5 to 10)
    println(x)
}

您可以使用Random类生成随机数,然后使用.nextInt(maxNumber)生成随机数。maxNumber是生成随机数时所需的最大值。请记住,Random类给你的是0到maxNumber-1的数字。

Random r = new Random();
int i = r.nextInt();

另一种方法是使用Math.Random()类,许多学校的课程都要求您使用该类,因为它更高效,而且您不必声明新的Random对象。要使用Math.random()获取随机数,请键入:

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

请原谅我过于挑剔,但大多数人建议的解决方案,即min+rng.nextInt(max-min+1),似乎很危险,因为:

rng.nextInt(n)无法达到整数.MAX_VALUE。当min为负值时,(max-min)可能会导致溢出。

万无一失的解决方案将为[Integer.min_VALUE,Integer.max_VALUE]内的任何min<=max返回正确的结果。请考虑以下简单的实现:

int nextIntInRange(int min, int max, Random rng) {
   if (min > max) {
      throw new IllegalArgumentException("Cannot draw random int from invalid range [" + min + ", " + max + "].");
   }
   int diff = max - min;
   if (diff >= 0 && diff != Integer.MAX_VALUE) {
      return (min + rng.nextInt(diff + 1));
   }
   int i;
   do {
      i = rng.nextInt();
   } while (i < min || i > max);
   return i;
}

尽管效率低下,但请注意while循环中成功的概率始终为50%或更高。

我发现这个例子生成随机数:


此示例生成特定范围内的随机整数。

import java.util.Random;

/** Generate random integers in a certain range. */
public final class RandomRange {

  public static final void main(String... aArgs){
    log("Generating random integers in the range 1..10.");

    int START = 1;
    int END = 10;
    Random random = new Random();
    for (int idx = 1; idx <= 10; ++idx){
      showRandomInteger(START, END, random);
    }

    log("Done.");
  }

  private static void showRandomInteger(int aStart, int aEnd, Random aRandom){
    if ( aStart > aEnd ) {
      throw new IllegalArgumentException("Start cannot exceed End.");
    }
    //get the range, casting to long to avoid overflow problems
    long range = (long)aEnd - (long)aStart + 1;
    // compute a fraction of the range, 0 <= frac < range
    long fraction = (long)(range * aRandom.nextDouble());
    int randomNumber =  (int)(fraction + aStart);    
    log("Generated : " + randomNumber);
  }

  private static void log(String aMessage){
    System.out.println(aMessage);
  }
} 

此类的示例运行:

Generating random integers in the range 1..10.
Generated : 9
Generated : 3
Generated : 3
Generated : 9
Generated : 4
Generated : 1
Generated : 3
Generated : 9
Generated : 10
Generated : 10
Done.

我已经创建了一个方法来获取给定范围内的唯一整数。

/*
      * minNum is the minimum possible random number
      * maxNum is the maximum possible random number
      * numbersNeeded is the quantity of random number required
      * the give method provides you with unique random number between min & max range
*/
public static Set<Integer> getUniqueRandomNumbers( int minNum , int maxNum ,int numbersNeeded ){

    if(minNum >= maxNum)
        throw new IllegalArgumentException("maxNum must be greater than minNum");

    if(! (numbersNeeded > (maxNum - minNum + 1) ))
        throw new IllegalArgumentException("numberNeeded must be greater then difference b/w (max- min +1)");

    Random rng = new Random(); // Ideally just create one instance globally

    // Note: use LinkedHashSet to maintain insertion order
    Set<Integer> generated = new LinkedHashSet<Integer>();
    while (generated.size() < numbersNeeded)
    {
        Integer next = rng.nextInt((maxNum - minNum) + 1) + minNum;

        // As we're adding to a set, this will automatically do a containment check
        generated.add(next);
    }
    return generated;
}