如何在Java中生成特定范围内的随机整数?

如何在Java中生成特定范围内的随机整数?,java,random,integer,java-8,Java,Random,Integer,Java 8,如何在特定范围内生成随机int值 我尝试了以下方法,但这些方法不起作用: 尝试1: randomNum = minimum + (int)(Math.random() * maximum); Random rn = new Random(); int n = maximum - minimum + 1; int i = rn.nextInt() % n; randomNum = minimum + i; RandomUtils random = new RandomUtils(); ran

如何在特定范围内生成随机
int

我尝试了以下方法,但这些方法不起作用:

尝试1:

randomNum = minimum + (int)(Math.random() * maximum);
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum =  minimum + i;
RandomUtils random = new RandomUtils();
random.nextInt(0, 0); // returns 0
random.nextInt(10, 10); // returns 10
random.nextInt(-10, 10); // returns numbers from -10 to 10 (-10, -9....9, 10)
random.nextInt(10, -10); // throws assert
import junit.framework.Assert;
import java.util.Random;

public class RandomUtils extends Random {

    /**
     * @param min generated value. Can't be > then max
     * @param max generated value
     * @return values in closed range [min, max].
     */
    public int nextInt(int min, int max) {
        Assert.assertFalse("min can't be > then max; values:[" + min + ", " + max + "]", min > max);
        if (min == max) {
            return max;
        }

        return nextInt(max - min + 1) + min;
    }
}
错误:
randomNum
可以大于
最大值

尝试2:

randomNum = minimum + (int)(Math.random() * maximum);
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum =  minimum + i;
RandomUtils random = new RandomUtils();
random.nextInt(0, 0); // returns 0
random.nextInt(10, 10); // returns 10
random.nextInt(-10, 10); // returns numbers from -10 to 10 (-10, -9....9, 10)
random.nextInt(10, -10); // throws assert
import junit.framework.Assert;
import java.util.Random;

public class RandomUtils extends Random {

    /**
     * @param min generated value. Can't be > then max
     * @param max generated value
     * @return values in closed range [min, max].
     */
    public int nextInt(int min, int max) {
        Assert.assertFalse("min can't be > then max; values:[" + min + ", " + max + "]", min > max);
        if (min == max) {
            return max;
        }

        return nextInt(max - min + 1) + min;
    }
}
错误:
randomNum
可以小于
最小值

使用:

minimum + rn.nextInt(maxValue - minvalue + 1)
Random ran = new Random();
int x = ran.nextInt(6) + 5;

Java 1.7或更高版本中,执行此操作的标准方法如下:

import java.util.concurrent.ThreadLocalRandom;

// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
int randomNum = ThreadLocalRandom.current().nextInt(min, max + 1);
import java.util.Random;

/**
 * Returns a pseudo-random number between min and max, inclusive.
 * The difference between min and max can be at most
 * <code>Integer.MAX_VALUE - 1</code>.
 *
 * @param min Minimum value
 * @param max Maximum value.  Must be greater than min.
 * @return Integer between min and max, inclusive.
 * @see java.util.Random#nextInt(int)
 */
public static int randInt(int min, int max) {

    // NOTE: This will (intentionally) not run as written so that folks
    // copy-pasting have to think about how to initialize their
    // Random instance.  Initialization of the Random instance is outside
    // the main scope of the question, but some decent options are to have
    // a field that is initialized once and then re-used as needed or to
    // use ThreadLocalRandom (if using at least Java 1.7).
    // 
    // In particular, do NOT do 'Random rand = new Random()' here or you
    // will get not very good / not very random results.
    Random rand;

    // nextInt is normally exclusive of the top value,
    // so add 1 to make it inclusive
    int randomNum = rand.nextInt((max - min) + 1) + min;

    return randomNum;
}
看。这种方法的优点是不需要显式初始化实例,如果使用不当,可能会导致混淆和错误

然而,相反地,无法明确设置种子,因此在测试或保存游戏状态或类似情况下,很难重现有用的结果。在这些情况下,可以使用下面所示的Java 1.7之前的技术

在Java 1.7之前,执行此操作的标准方法如下:

import java.util.concurrent.ThreadLocalRandom;

// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
int randomNum = ThreadLocalRandom.current().nextInt(min, max + 1);
import java.util.Random;

/**
 * Returns a pseudo-random number between min and max, inclusive.
 * The difference between min and max can be at most
 * <code>Integer.MAX_VALUE - 1</code>.
 *
 * @param min Minimum value
 * @param max Maximum value.  Must be greater than min.
 * @return Integer between min and max, inclusive.
 * @see java.util.Random#nextInt(int)
 */
public static int randInt(int min, int max) {

    // NOTE: This will (intentionally) not run as written so that folks
    // copy-pasting have to think about how to initialize their
    // Random instance.  Initialization of the Random instance is outside
    // the main scope of the question, but some decent options are to have
    // a field that is initialized once and then re-used as needed or to
    // use ThreadLocalRandom (if using at least Java 1.7).
    // 
    // In particular, do NOT do 'Random rand = new Random()' here or you
    // will get not very good / not very random results.
    Random rand;

    // nextInt is normally exclusive of the top value,
    // so add 1 to make it inclusive
    int randomNum = rand.nextInt((max - min) + 1) + min;

    return randomNum;
}
看。在实践中,这门课往往比其他课程更可取


特别是,当标准库中有一个简单的API来完成任务时,无需重新发明随机整数生成轮。

我想知道库提供的任何随机数生成方法是否符合要求

 rand.nextInt((max+1) - min) + min;
例如:或


或者从中查看RandomUtils。

您可以将第二个代码示例编辑为:

Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum =  rn.nextInt(range) + minimum;

请注意,这种方法比
nextInt
方法更具偏见,效率更低

实现这一点的一个标准模式是:

Min + (int)(Math.random() * ((Max - Min) + 1))
数学库函数Math.random()在
[0,1)
范围内生成一个双精度值。请注意,此范围不包括1

为了首先获得一个特定的值范围,需要乘以要覆盖的值范围的大小

Math.random() * ( Max - Min )
这将返回范围
[0,Max-Min)
中的值,其中不包括“Max-Min”

例如,如果希望
[5,10)
,则需要覆盖五个整数值,以便使用

Math.random() * 5
这将返回范围
[0,5)
内的值,其中不包括5

现在,您需要将此范围向上移动到目标范围。您可以通过添加最小值来完成此操作

Min + (Math.random() * (Max - Min))
现在,您将获得范围
[Min,Max)
的值。按照我们的示例,这意味着
[5,10)

但是,这仍然不包括
Max
,您将获得一个双精度值。为了获得包含的
Max
值,您需要向范围参数
(Max-Min)
中添加1,然后通过强制转换为整数来截断小数部分。这通过以下方式实现:

Min + (int)(Math.random() * ((Max - Min) + 1))
这是一个随机整数值,范围为
[Min,Max]
,或者按照示例
[5,10]

5 + (int)(Math.random() * ((10 - 5) + 1))

中的
Math.Random
类是基于0的。因此,如果您编写以下内容:

Random rand = new Random();
int x = rand.nextInt(10);
x
将介于
0-9
之间

因此,给定以下
25
项的数组,生成
0
(数组的基部)和
array.length
之间的随机数的代码为:

String[] i = new String[25];
Random rand = new Random();
int index = 0;

index = rand.nextInt( i.length );
由于
i.length
将返回
25
,因此
nextInt(i.length)
将返回一个介于
0-24
范围内的数字。另一个选项与
Math.Random
相同

index = (int) Math.floor(Math.random() * i.length);
要更好地理解,请查看论坛帖子。

使用:

minimum + rn.nextInt(maxValue - minvalue + 1)
Random ran = new Random();
int x = ran.nextInt(6) + 5;

整数
x
现在是一个随机数,其结果可能是
5-10

如果掷骰子,它将是介于1到6(而不是0到6)之间的随机数,因此:


这很好。

请原谅我的挑剔,但是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
    Integer.MAX\u值
  • (最大-最小)
    min
    为负值时,可能会导致溢出
一个简单的解决方案将返回任何
min-max)的正确结果{
抛出新的IllegalArgumentException(“无法从无效范围[“+min+”,“+max+”])中随机抽取int;
}
int diff=最大-最小值;
if(diff>=0&&diff!=Integer.MAX_值){
返回值(最小值+最小值+最小值(差值+1));
}
int i;
做{
i=rng.nextInt();
}而(imax);
返回i;
}

尽管效率低下,但请注意,
while
循环中的成功概率始终为50%或更高。

另一个选项只是使用:

public static Random RANDOM = new Random(System.nanoTime());

public static final float random(final float pMin, final float pMax) {
    return pMin + RANDOM.nextFloat() * (pMax - pMin);
}

下面是一个有用的类,可以在包含/排除边界的任意组合范围内生成随机
int

import java.util.Random;

public class RandomRange extends Random {
    public int nextIncInc(int min, int max) {
        return nextInt(max - min + 1) + min;
    }

    public int nextExcInc(int min, int max) {
        return nextInt(max - min) + 1 + min;
    }

    public int nextExcExc(int min, int max) {
        return nextInt(max - min - 1) + 1 + min;
    }

    public int nextIncExc(int min, int max) {
        return nextInt(max - min) + min;
    }
}
我发现这个例子:


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);
  }
} 

此方法可能便于使用:

此方法将返回一个介于提供的最小值和最大值之间的随机数:

public static int getRandomNumberBetween(int min, int max) {
    Random foo = new Random();
    int randomNumber = foo.nextInt(max - min) + min;
    if (randomNumber == min) {
        // Since the random number is between the min and max values, simply add 1
        return min + 1;
    } else {
        return randomNumber;
    }
}
此方法将从提供的最小值和最大值返回一个随机数(因此生成的数字也可以是最小值或最大值):


ThreadLocalRandom
相当于多线程环境中的类
java.util.Random
。生成随机数是在每个线程的本地执行的。因此,通过减少冲突,我们可以获得更好的性能

int rand = ThreadLocalRandom.current().nextInt(x,y);

x
y
-间隔,例如(1,10)

只需执行以下语句即可:

Randomizer.generate(0,10);//最小值为零,最大值为十
下面是它的源代码

Randomizer.java
公共类随机化器{
公共静态int-generate(int-min,int-max){
返回min+(int)(Math.random()*((max-min)+1));
}
}
它只是简单明了。

只需使用t
Random r = new Random();
int[] fiveRandomNumbers = r.ints(5, 0, 11).toArray();
int randomNumber = r.ints(1, 0, 11).findFirst().getAsInt();
public final class IntRandomNumberGenerator {

    private PrimitiveIterator.OfInt randomIterator;

    /**
     * Initialize a new random number generator that generates
     * random numbers in the range [min, max]
     * @param min - the min value (inclusive)
     * @param max - the max value (inclusive)
     */
    public IntRandomNumberGenerator(int min, int max) {
        randomIterator = new Random().ints(min, max + 1).iterator();
    }

    /**
     * Returns a random number in the range (min, max)
     * @return a random number in the range (min, max)
     */
    public int nextInt() {
        return randomIterator.nextInt();
    }
}
RandomUtils random = new RandomUtils();
random.nextInt(0, 0); // returns 0
random.nextInt(10, 10); // returns 10
random.nextInt(-10, 10); // returns numbers from -10 to 10 (-10, -9....9, 10)
random.nextInt(10, -10); // throws assert
import junit.framework.Assert;
import java.util.Random;

public class RandomUtils extends Random {

    /**
     * @param min generated value. Can't be > then max
     * @param max generated value
     * @return values in closed range [min, max].
     */
    public int nextInt(int min, int max) {
        Assert.assertFalse("min can't be > then max; values:[" + min + ", " + max + "]", min > max);
        if (min == max) {
            return max;
        }

        return nextInt(max - min + 1) + min;
    }
}
Random rand = new Random();
randomNum = minimum + rand.nextInt((maximum - minimum) + 1);
public static int generateRandomInteger(int min, int max) {
    SecureRandom rand = new SecureRandom();
    rand.setSeed(new Date().getTime());
    int randomNum = rand.nextInt((max - min) + 1) + min;
    return randomNum;
}
Random rn = new Random();
int result = rn.nextInt(max - min + 1) + min;
System.out.println(result);
 /**
   * @param min - The minimum.
   * @param max - The maximum.
   * @return A random double between these numbers (inclusive the minimum and maximum).
   */
 public static double getRandom(double min, double max) {
   return (Math.random() * (max + 1 - min)) + min;
 }
Random random = new Random();

int max = 10;
int min = 5;
int totalNumber = 10;

IntStream stream = random.ints(totalNumber, min, max);
stream.forEach(System.out::println);
Random r = new Random();
int lowerBound = 1;
int upperBound = 11;
int result = r.nextInt(upperBound-lowerBound) + lowerBound;
int n = new SplittableRandom().nextInt(0, 1_001);
int[] a = new SplittableRandom().ints(100, 0, 1_001).parallel().toArray();
IntStream stream = new SplittableRandom().ints(100, 0, 1_001);