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leojdiaz wants to merge 2 commits into from
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samuel review #6

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3a4a184
my solution
leojdiaz Sep 6, 2021
577f91e
add readme
leojdiaz Sep 13, 2021
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11 changes: 10 additions & 1 deletion README.md
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JavaCodeReview
This class determine if a number is perfection or not.

Definition of a Perfect Number:
In mathematics a perfect number is defined as an integer which is the sum of its proper
positive divisors; that is, the sum of the positive divisors not including the number itself.
Some examples of perfect numbers are:

6 = 1 + 2 + 3
28 = 1 + 2 + 4 + 7 + 14
496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248
42 changes: 42 additions & 0 deletions src/main/java/payapal/Perfection.java
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package main.java.payapal;

public class Perfection {

private static Perfection perf;

public synchronized static Perfection getPerf() {
if (perf == null) {
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@JanioSamuel JanioSamuel Sep 13, 2021

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Please, change for a ternary if.
return perf != null ? perf : new Perfection();

perf = new Perfection();
}
return perf;
}


public static boolean isPerfect(long candidate) {
boolean retVal;
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@JanioSamuel JanioSamuel Sep 13, 2021

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Please, change the name to returnValue.

long[] divisors = GetDivisors(candidate);
int sum = 0;
for (int d = 0; d < 1000; d++)
{
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@JanioSamuel JanioSamuel Sep 13, 2021

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Please, use the braces in the same line as the code.

sum = sum + divisors[d];
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@JanioSamuel JanioSamuel Sep 13, 2021

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Please, change sum to long or use divisors[d].intValue() if possible.
Also, please use sum +=

}
if (sum == candidate)
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@JanioSamuel JanioSamuel Sep 13, 2021

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Please, change candidate to Int.
candidate.intValue()

retVal = true;
return retVal;
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@JanioSamuel JanioSamuel Sep 13, 2021

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Maybe, try to use return sum == candidate;, with this way, you could remove the retVal variable.

}


private static long[] GetDivisors(long candidate) {
long[] divisors = new long[];
int d = 0;
for (long i = 0; i < candidate; i++) {
long foo = candidate / i;
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@JanioSamuel JanioSamuel Sep 13, 2021

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Note: try to avoid long divide operations, because you will lose precision. Try to use Bigdecimal.

if (foo * i == candidate) {
divisors[d] = i;
d = d + 1;
}
}
return divisors;
}

}