将整数向上舍入为3的倍数-Roundupanintegertomultipleof3

11  

I got this working in a Playground:

我让它在游乐场工作:

import Foundation

func roundToThree(value: Int) -> Int{
    var fractiOnNum= Double(value) / 3.0
    let roundedNum = Int(ceil(fractionNum))
    return roundedNum * 3
}

roundToThree(2)
roundToThree(7)

11  

My 2 ct:

我的2克拉:

func roundUp(value: Int, divisor: Int) -> Int {
    let rem = value % divisor
    return rem == 0 ? value : value + divisor - rem
}

0  

I implemented it this way:

我这样实现了它:

func roundUp(value: Int, divisor: Int) -> Int {
    let modulo = value % divisor
    var multiplier = value / divisor
    if modulo > 0 {
        ++multiplier
    }
    return divisor * multiplier
}

The function calculates:

该函数计算:

• modulo: the modulo of the division of the value by the divisor

模数:除数除以值的模数

• multiplier: the integer division of the value by the divisor

乘数:乘以除数的整数除法

• if modulo is greater than zero, the multiplier is incremented

如果modulo大于零,则乘数递增

• the resulting value is the divisor multiplied by the multiplier

结果值是除数乘以乘数

Using integers only, this algorithm should be faster than any other based on floating point calculations.

仅使用整数,此算法应该比基于浮点计算的任何其他算法更快。

I made the divisor a parameter, you can optionally decide to remove it and replace it in teh function body with the 3 literal, or assign it a default value:

我将除数设为参数,您可以选择将其删除并将其替换为函数体中的3字面值,或者为其指定默认值:

func roundUp(value: Int, divisor: Int = 3) -> Int 

0  

A swift3 Float extension with a configurable multiplier (adapted from the accepted answer from @erdekhayser)

带有可配置乘法器的swift3 Float扩展(改编自@erdekhayser接受的答案)

extension Float {
    func roundUpTo(multiplier: Int) -> Int{
        let fractiOnNum= self / Float(multiplier)
        return Int(ceil(fractionNum)) * multiplier
    }
}

0  

This is like Martin R's answer but works by rounding to the nearest multiple and not just by rounding up

这就像Martin R的答案,但是通过四舍五入到最接近的倍数而不仅仅是四舍五入

Swift 3/4

func roundUp(value: Double, divisor: Double) -> Double {
    let rem = value.truncatingRemainder(dividingBy: divisor)
    if(rem >= divisor / 2)
    {
        return rem == 0 ? value : value + divisor - rem
    }
    else
    {
        return rem == 0 ? value : value - rem
    }
}