how to get random number in java in range

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Giulia
16 Mar 2018
1package com.mkyong.example.test;
2
3import java.util.Random;
4
5public class TestRandom {
6
7    public static void main(String[] args) {
8
9        for (int i = 0; i < 10; i++) {
10            System.out.println(getRandomNumberInRange(5, 10));
11        }
12
13    }
14
15    private static int getRandomNumberInRange(int min, int max) {
16
17        if (min >= max) {
18            throw new IllegalArgumentException("max must be greater than min");
19        }
20
21        Random r = new Random();
22        return r.nextInt((max - min) + 1) + min;
23    }
24
25}
Jessica
21 May 2017
1
2    private static int getRandomNumberInRange(int min, int max) 
3    {
4
5        if (min >= max) {
6            throw new IllegalArgumentException("max must be greater than min");
7        }
8
9        Random r = new Random();
10        return r.nextInt((max - min) + 1) + min;
11    }
12
13
14    //Random().nextInt(int bound) = Random integer from 0 (inclusive) to bound (exclusive)
15    
16    //1. nextInt(range) = nextInt(max - min)
17    new Random().nextInt(5);  // [0...4] [min = 0, max = 4]
18    new Random().nextInt(6);  // [0...5]
19    new Random().nextInt(7);  // [0...6]
20    new Random().nextInt(8);  // [0...7]
21    new Random().nextInt(9);  // [0...8]
22    new Random().nextInt(10); // [0...9]			
23    new Random().nextInt(11); // [0...10]
24    
25    //2. To include the last value (max value) = (range + 1)
26    new Random().nextInt(5 + 1)  // [0...5] [min = 0, max = 5]
27    new Random().nextInt(6 + 1)  // [0...6]
28    new Random().nextInt(7 + 1)  // [0...7]
29    new Random().nextInt(8 + 1)  // [0...8]
30    new Random().nextInt(9 + 1)  // [0...9]
31    new Random().nextInt(10 + 1) // [0...10]			
32    new Random().nextInt(11 + 1) // [0...11]
33    
34    //3. To define a start value (min value) in a range,
35    //   For example, the range should start from 10 = (range + 1) + min
36    new Random().nextInt(5 + 1)  + 10 // [0...5]  + 10 = [10...15]
37    new Random().nextInt(6 + 1)  + 10 // [0...6]  + 10 = [10...16]
38    new Random().nextInt(7 + 1)  + 10 // [0...7]  + 10 = [10...17]
39    new Random().nextInt(8 + 1)  + 10 // [0...8]  + 10 = [10...18]
40    new Random().nextInt(9 + 1)  + 10 // [0...9]  + 10 = [10...19]
41    new Random().nextInt(10 + 1) + 10 // [0...10] + 10 = [10...20]
42    new Random().nextInt(11 + 1) + 10 // [0...11] + 10 = [10...21]
43    
44    // Range = (max - min)
45    // So, the final formula is ((max - min) + 1) + min
46    
47    //4. Test [10...30]
48    // min = 10 , max = 30, range = (max - min)
49    new Random().nextInt((max - min) + 1) + min
50    new Random().nextInt((30 - 10) + 1) + 10
51    new Random().nextInt((20) + 1) + 10
52    new Random().nextInt(21) + 10    //[0...20] + 10 = [10...30]
53    
54    //5. Test [15...99]
55    // min = 15 , max = 99, range = (max - min)
56    new Random().nextInt((max - min) + 1) + min
57    new Random().nextInt((99 - 15) + 1) + 15
58    new Random().nextInt((84) + 1) + 15
59    new Random().nextInt(85) + 15    //[0...84] + 15 = [15...99]
60    
61    //Done, understand?
Paolo
15 May 2018
1import java.util.concurrent.ThreadLocalRandom;
2
3// nextInt is normally exclusive of the top value,
4// so add 1 to make it inclusive
5int randomNum = ThreadLocalRandom.current().nextInt(min, max + 1);
6
Lucas
24 May 2020
1Min + (int)(Math.random() * ((Max - Min) + 1))
2
Sofia
26 Oct 2020
1int rand = ThreadLocalRandom.current().nextInt(x,y);
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