Exercism - Perfect Numbers(with.Kotlin)

[문제]

Determine if a number is perfect, abundant, or deficient based on Nicomachus' (60 - 120 CE) classification scheme for natural numbers.

The Greek mathematician Nicomachus devised a classification scheme for natural numbers, identifying each as belonging uniquely to the categories of perfect, abundant, or deficient based on their aliquot sum. The aliquot sum is defined as the sum of the factors of a number not including the number itself. For example, the aliquot sum of 15 is (1 + 3 + 5) = 9

• Perfect: aliquot sum = number
  • 6 is a perfect number because (1 + 2 + 3) = 6
  • 28 is a perfect number because (1 + 2 + 4 + 7 + 14) = 28
• Abundant: aliquot sum > number
  • 12 is an abundant number because (1 + 2 + 3 + 4 + 6) = 16
  • 24 is an abundant number because (1 + 2 + 3 + 4 + 6 + 8 + 12) = 36
• Deficient: aliquot sum < number
  • 8 is a deficient number because (1 + 2 + 4) = 7
  • Prime numbers are deficient

Implement a way to determine whether a given number is perfect. Depending on your language track, you may also need to implement a way to determine whether a given number is abundant or deficient.

 

[Solution1]

enum class Classification {
    DEFICIENT, PERFECT, ABUNDANT
}

fun classify(naturalNumber: Int): Classification {

    if(naturalNumber < 1) throw RuntimeException()

    val listNum = mutableListOf<Int>()

    for (i in 1 until naturalNumber){
        if(naturalNumber % i == 0){
            listNum.add(i)
        }
    }
    
    return when{
        naturalNumber == listNum.sum() -> Classification.PERFECT
        naturalNumber > listNum.sum() -> Classification.DEFICIENT
        naturalNumber < listNum.sum() -> Classification.ABUNDANT
        else -> throw RuntimeException()
    }

}

for 문을 사용해서 처리

[Solution2]

enum class Classification {
    DEFICIENT, PERFECT, ABUNDANT
}

fun classify(naturalNumber: Int): Classification {

    if(naturalNumber < 1) throw RuntimeException()

    val aliquotSum = (1 until naturalNumber).filter { naturalNumber % it == 0 }.sum()

    return when{
        naturalNumber == aliquotSum -> Classification.PERFECT
        naturalNumber > aliquotSum -> Classification.DEFICIENT
        naturalNumber < aliquotSum -> Classification.ABUNDANT
        else -> throw RuntimeException()
    }

}

배열로 나타낸 다음 filter을 사용하여 구분된 값들을 sum()으로 모아서 처리하도록 변경