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Sebastian Lofaro
Front End Web Development Techdegree Graduate 29,101 PointsCode Challenge
I am unable to solve the code challenge. Does anyone have any ideas?
5 Answers
luke jones
8,915 PointsI done it like so:
func duplicate<T>(item: T, numberOfTimes: Int) -> [T] {
var arrayOfT = [T]()
for index in 0..<numberOfTimes {
arrayOfT.append(item)
}
return arrayOfT
}
duplicate(item: 1, numberOfTimes: 4)
But i feel like there might a slightly cleaner way to do it
James J. McCombie
Python Web Development Techdegree Graduate 21,199 PointsJust posting this as an alternative to the answer already supplied...
func duplicate<T>(item: T, numberOfTimes: Int) -> [T] {
return Array(repeating: item, count: numberOfTimes)
}
Xavier D
Courses Plus Student 5,840 Points...did mine with repeat while...
func duplicate<T>(item:T,numberOfTimes:Int)->[T]
{
var repeatWhile=0;var
returnArray:[T]=[]
repeat{returnArray.append(item);repeatWhile+=1}while repeatWhile<(numberOfTimes)
return returnArray
}
duplicate(item: 1, numberOfTimes: 4)
kols
27,007 PointsWell, this is definitely not the cleanest way to solve this (I prefer James J. McCombie's answer — a more elegant / simple solution 
func duplicate<T>(item: T, numberOfTimes: Int) -> [T] {
var myArray: [T] = []
var counter = numberOfTimes
repeat {
myArray.append(item)
counter -= 1
} while counter > 0
return myArray
}
Joey Liu
3,982 PointsI don't understand why error pop up and said as below: Binary operator '..<' cannot be applied to operands of type 'Int' and 'T'
func duplicate<T>(item: T, numberOfTimes: T) -> [T] { var number:[T] = [] for index in 0..<numberOfTimes { array.append(item) } return array }