How to reverse array in Swift without using ".reverse()"?
I have array and need to reverse it without Array.reverse
method, only with a for
loop.
var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
Swift 3:
var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversedNames : [String] = Array(names.reversed())
print(reversedNames) // ["Asus", "Lenovo", "Sony", "Microsoft", "Apple"]
Here is @Abhinav 's answer translated to Swift 2.2 :
var names: [String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversedNames = [String]()
for arrayIndex in (names.count - 1).stride(through: 0, by: -1) {
reversedNames.append(names[arrayIndex])
}
Using this code shouldn't give you any errors or warnings about the use deprecated of C-style for-loops or the use of --
.
Swift 3 - Current:
let names: [String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversedNames = [String]()
for arrayIndex in stride(from: names.count - 1, through: 0, by: -1) {
reversedNames.append(names[arrayIndex])
}
Alternatively, you could loop through normally and subtract each time:
let names = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
let totalIndices = names.count - 1 // We get this value one time instead of once per iteration.
var reversedNames = [String]()
for arrayIndex in 0...totalIndices {
reversedNames.append(names[totalIndices - arrayIndex])
}
Do you really need a for
loop? If not, you can use reduce
.
I guess that this is the shortest way to achieve it without reversed() method (Swift 3.0.1):
["Apple", "Microsoft", "Sony", "Lenovo", "Asus"].reduce([],{ [$1] + $0 })
Only need to make (names.count/2) passes through the array. No need to declare temporary variable, when doing the swap...it's implicit.
var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
let count = names.count
for i in 0..<count/2 {
(names[i],names[count - i - 1]) = (names[count - i - 1],names[i])
}
// Yields: ["Asus", "Lenovo", "Sony", "Microsoft", "Apple"]