Arrays 在单个回路中更新多个蓄能器时遇到问题

这是一个函数，它接受输入“change”，并返回要返回给客户的25美分、10美分、5美分和1美分的明细

我的方法是：假设零钱是0.62美元

检查是零钱>四分之一。如果是，减去0.25美元。每次减去四分之一，四分之一累加器加1。重复此操作，直到无法再从剩余的更改值中减去四分之一。（这将导致两个季度）

此步骤后剩余的更改为$0.12

移到一角硬币上-每次从零钱中减去一角硬币时，在一角硬币累加器中加1

对镍币和便士重复上述步骤

虽然我可以轻松地遍历coinTray数组并减去值，但如何指示一旦数组移动到下一个项目，就切换到下一个累加器并开始向其添加

我在这里解决了这个问题，没有使用循环，但是有更有效的方法吗

func breakdown(change: Double) -> String {

    var coinTray: [Double] = [0.25, 0.10, 0.05, 0.01]
    var totalChange = change

    var quarters = 0
    var dimes = 0
    var nickels = 0
    var pennies = 0

    while totalChange >= coinTray[0] {
        totalChange -= coinTray[0]
        quarters += 1
    }

    while totalChange >= coinTray[1] {
        totalChange -= coinTray[1]
        dimes += 1
    }

    while totalChange >= coinTray[2] {
        totalChange -= coinTray[2]
        nickels += 1
    }

    while totalChange >= coinTray[3] {
        totalChange -= coinTray[3]
        pennies += 1
    }

return "You should have \(quarters) quarter(s), \(dimes) dime(s), \(nickels) nickel(s), and \(pennies) penny/ies."

}

---

我相信这在没有循环的情况下也能做同样的事情

func breakDown(change: Double) -> String {
    var coinTray: [Double] = [0.25, 0.10, 0.05, 0.01]
    var totalChange = change

    var quarters = 0
    var dimes    = 0
    var nickels  = 0
    var pennies  = 0

    quarters    = Int(totalChange/coinTray[0])
    totalChange = totalChange.truncatingRemainder(dividingBy: coinTray[0])
    dimes       = Int(totalChange/coinTray[1])
    totalChange = totalChange.truncatingRemainder(dividingBy: coinTray[1])
    nickels     = Int(totalChange/coinTray[2])
    totalChange = totalChange.truncatingRemainder(dividingBy: coinTray[2])
    pennies     = Int(totalChange * 100)

    return "You should have \(quarters) quarter(s), \(dimes) dime(s), \(nickels) nickel(s), and \(pennies) penny/ies."
}

编辑：似乎OP想要这个答案

var coinTray: [Double] = [0.25, 0.10, 0.05, 0.01]

var quarters = 0
var dimes   = 0
var nickels = 0
var pennies = 0

var coins = [quarters, dimes, nickels, pennies]

func breakDown(change: Double) -> String {

    var totalChange = change
    for i in 0..<coinTray.count {
        coins[i] = Int(totalChange/coinTray[i])
        totalChange = totalChange.truncatingRemainder(dividingBy: coinTray[i])
    }
    return "You should have \(quarters) quarter(s), \(dimes) dime(s), \(nickels) nickel(s), and \(pennies) penny/ies."
}

var coinTray:[Double]=[0.25,0.10,0.05,0.01]
var四分之一=0
变量dimes=0
var镍币=0
var pennies=0
var硬币=[四分之一硬币、一角硬币、五分镍币、便士]
func细分（更改：双）->字符串{
var totalChange=变更
对于0中的i.“我在这里解决了它，没有使用循环”，呃……您似乎在使用4。@RakeshaShastri是的，有4个while循环。但我想知道是否有办法将这些循环清理成一个循环，因为除了累加器之外，所有的循环都在重复。不使用任何循环如何？就像下面发布的答案一样？：）我已经编辑了答案。我想我明白您在寻找什么。但是这对于动态数量的变量是不实际的，在这种情况下，您必须使用元组数组，这对于这样一个简单的事情来说是复杂的。