How to reverse array in Swift without using ".reverse()"? - arrays

Question

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"]

Answer 1

Swift 3:
var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversedNames : [String] = Array(names.reversed())
print(reversedNames) // ["Asus", "Lenovo", "Sony", "Microsoft", "Apple"]

Answer 2

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])
}

Answer 3

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 })

Answer 4

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"]

Answer 5

There's also stride to generate a reversed index:
let names = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversed = [String]()
for index in (names.count - 1).stride(to: -1, by: -1) {
reversed.append(names[index])
}
It also works well with map:
let reversed = (names.count - 1).stride(to: -1, by: -1).map { names[$0] }
Note: stride starts its index at 1, not at 0, contrary to other Swift sequences.
However, to anyone reading this in the future: use .reverse() instead to actually reverse an array, it's the intended way.

Answer 6

var names:[String] = [ "A", "B", "C", "D", "E","F","G"]
var c = names.count - 1
var i = 0
while i < c {
swap(&names[i++],&names[c--])
}
Cristik
while i < c {
swap(&names[i],&names[c]
i += 1
c -= 1
}

Answer 7

Here you go:
var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversedNames = [String]()
for var arrayIndex = names.count - 1 ; arrayIndex >= 0 ; arrayIndex-- {
reversedNames.append(names[arrayIndex])
}

Answer 8

Ignoring checks for emptiness..
var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversedNames = [String]()
for name in names {
reversedNames.insert((name), at:0)
}
print(reversedNames)

Answer 9

Here is the most simpler way.
let names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversedNames = [String]()
for name in names {
reversedNames.insert(name, at: 0)
}

Answer 10

Edited as generic
// Swap the first index with the last index.
// [1, 2, 3, 4, 5] -> pointer on one = array[0] and two = array.count - 1
// After first swap+loop increase the pointer one and decrease pointer two until
// conditional is not true.
func reverse<T>(with array: [T]) -> [T] {
var array = array
var first = 0
var last = array.count - 1
while first < last {
array.swapAt(first, last)
first += 1
last -= 1
}
return array
}
input-> [1, 2, 3, 4, 5] output ->[5, 4, 3, 2, 1]
input-> ["a", "b", "c", "d"] output->["d", "c", "b", "a"]
// Or a shorter version divide and conquer
func reversed<T>(with arr: [T]) -> [T] {
var arr = arr
(0..<arr.count / 2).forEach { i in
arr.swapAt(i, arr.count - i - 1)
}
return arr
}

Answer 11

The most efficient way is to swap the items at start- and endIndex and move the indices bidirectional to the middle. This is a generic version
extension Array {
mutating func upsideDown() {
if isEmpty { return }
var 👉 = startIndex
var 👈 = index(before: endIndex)
while 👉 < 👈 {
swapAt(👉, 👈)
formIndex(after: &👉)
formIndex(before: &👈)
}
}
}

Answer 12

Like this, maybe:
names = names.enumerate().map() { ($0.index, $0.element) }.sort() { $0.0 > $1.0 }.map() { $0.1 }
Oh, wait.. I have to use for loop, right? Then like this probably:
for (index, name) in names.enumerate().map({($0.index, $0.element)}).sort({$0.0 > $1.0}).map({$0.1}).enumerate() {
names[index] = name
}

Answer 13

Swift 5:
let names: [String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversenames: [String] = []
let count = names.count
for index in 0..<count {
reversenames.insert(names[count-index-1], at: index)
}
print(reversenames)

Answer 14

This will work with any sized array.
import Cocoa
var names:[String] = [ "A", "B", "C", "D", "E","F"]
var c = names.count - 1
for i in 0...(c/2-1) { swap(&names[i],&names[c-i]) }
print(names)

Answer 15

Here is how I did it and there is no warning for Swift 3
let names = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversedNames = [String]()
for name in names.enumerate() {
let newIndex = names.count - 1 - name.index
reversedNames.append(names[newIndex])
}
or just simply
reversedNames = names.reverse()

Answer 16

Here the code for swift 3
let array = ["IOS A", "IOS B", "IOS C"]
for item in array.reversed() {
print("Found \(item)")
}

Answer 17

func reverse(array: inout [String]) {
if array.isEmpty { return }
var f = array.startIndex
var l = array.index(before: array.endIndex)
while f < l {
swap(array: &array, f, l)
array.formIndex(after: &f)
array.formIndex(before: &l)
}
}
private func swap( array: inout [String], _ i: Int, _ j: Int) {
guard i != j else { return }
let tmp = array[i]
array[i] = array[j]
array[j] = tmp
}
Or you can write extension of course

Answer 18

var rArray : [Int] = [2,5,6,8,3,8,9,10]
var ReArray = [Int]()
var a : Int = 1
func reversed (_ array: [Int]) -> [Int] {
for i in array {
ReArray.append(array[array.count-a])
a += 1
}
rArray = ReArray
return rArray
}
reversed(rArray)
print(rArray)

Answer 19

var arr = [1, 2, 3, 4, 5] // Array we want to reverse
var reverse: [Int]! // Array where we store reversed values
reverse = arr
for i in 0...(arr.count - 1) {
reverse[i] = arr.last! // Adding last value of original array at reverse[0]
arr.removeLast() // removing last value & repeating above step.
}
print("Reverse : \(reverse!)")
A more simple way :)

Answer 20

Recently I had an interview and I was asked this question, how to reverse an array without using reversed(). Here is my solution below:
func reverseArray( givenArray:inout [Int]) -> [Int] {
var reversedArray = [Int]()
while givenArray.count > 0 {
reversedArray.append(givenArray.removeLast())
}
return reversedArray
}
var array = [1,2,3,4,5,6,7]
var reversed = reverseArray(givenArray: &array)

Answer 21

First, need to find the middle of array. This method is faster than the linear time O(n) and slower than the constant time O(1) complexity.
func reverse<T>(items: [T]) -> [T] {
var reversed = items
let count = items.count
let middle = count / 2
for i in stride(from: 0, to: middle, by: 1) {
let first = items[i]
let last = items[count - 1 - i]
reversed[i] = last
reversed[count - 1 - i] = first
}
return reversed
}

Answer 22

Do it like this for reversed sorting.
let unsortedArray: [String] = ["X", "B", "M", "P", "Z"]
// reverse sorting
let reversedArray = unsortedArray.sorted() {$0 > $1}
print(reversedArray) // ["Z", "X", "P", "M", "B"]

Answer 23

I like simple codes.
var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
var reversedNames = [""]
for name in names {
reversedNames.insert(name, at: 0)
}
print(reversedNames)

Answer 24

var names = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]
// 'while' loop
var index = 0
let totalIndices = names.count - 1
while index < names.count / 2 {
names.swapAt(index, totalIndices-index)
index += 1
}
// 'for' loop
for index in 0..<names.count/2 {
names.swapAt(index, names.count-index-1)
}
// output: "["Asus", "Lenovo", "Sony", "Microsoft", "Apple"]"

Answer 25

You can use the swift3 document:
let names = ["Chris", "Alex", "Ewa", "Barry", "Daniella"]
let reversedNames = names.sorted(by: >)
// reversedNames is equal to:
// ["Ewa", "Daniella", "Chris", "Barry", "Alex"]