-
Notifications
You must be signed in to change notification settings - Fork 19.5k
/
PerfectNumber.java
71 lines (65 loc) · 2.09 KB
/
PerfectNumber.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
package com.thealgorithms.maths;
/**
* In number theory, a perfect number is a positive integer that is equal to the
* sum of its positive divisors, excluding the number itself. For instance, 6
* has divisors 1, 2 and 3 (excluding itself), and 1 + 2 + 3 = 6, so 6 is a
* perfect number.
*
* link:https://en.wikipedia.org/wiki/Perfect_number
*/
public final class PerfectNumber {
private PerfectNumber() {
}
/**
* Check if {@code number} is perfect number or not
*
* @param number the number
* @return {@code true} if {@code number} is perfect number, otherwise false
*/
public static boolean isPerfectNumber(int number) {
if (number <= 0) {
return false;
}
int sum = 0;
/* sum of its positive divisors */
for (int i = 1; i < number; ++i) {
if (number % i == 0) {
sum += i;
}
}
return sum == number;
}
/**
* Check if {@code n} is perfect number or not
*
* @param n the number
* @return {@code true} if {@code number} is perfect number, otherwise false
*/
public static boolean isPerfectNumber2(int n) {
if (n <= 0) {
return false;
}
int sum = 1;
double root = Math.sqrt(n);
/*
* We can get the factors after the root by dividing number by its factors
* before the root.
* Ex- Factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50 and 100.
* Root of 100 is 10. So factors before 10 are 1, 2, 4 and 5.
* Now by dividing 100 by each factor before 10 we get:
* 100/1 = 100, 100/2 = 50, 100/4 = 25 and 100/5 = 20
* So we get 100, 50, 25 and 20 which are factors of 100 after 10
*/
for (int i = 2; i <= root; i++) {
if (n % i == 0) {
sum += i + n / i;
}
}
// if n is a perfect square then its root was added twice in above loop, so subtracting root
// from sum
if (root == (int) root) {
sum -= root;
}
return sum == n;
}
}