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

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"]

Use the same basic logic you would in any language, E.g. [C# Reverse an array without using reverse method of array](http://stackoverflow.com/questions/6088287/reverse-an-array-without-using-reverse-method-of-array) — Alex K., Nov 01 '15 at 14:57
@AlexK. The point is that there's already an efficient method in the Swift stdlib to do that. — Eric Aya, Nov 01 '15 at 14:58
@EricD. Yea but the OP knows that; this is likely a comp sci assigment — Alex K., Nov 01 '15 at 14:59
I'd like an example of reverse implemented in a functional way, without mutating anything, e.g. similar to the traditional Lisp recursive reverse function. — Jean-Denis Muys, Aug 24 '16 at 14:13
I’m voting to close this question because [the official algorithm](https://github.com/apple/swift/blob/main/stdlib/public/core/Reverse.swift) is open source. `reverse` is not special. Just look up whatever you need in the repo. — , Nov 27 '22 at 12:02

score 29 · Answer 1 · answered Aug 04 '17 at 13:02

29

Swift 3:

var names:[String] = ["Apple", "Microsoft", "Sony", "Lenovo", "Asus"]

var reversedNames : [String] = Array(names.reversed())

print(reversedNames)  // ["Asus", "Lenovo", "Sony", "Microsoft", "Apple"]

answered Aug 04 '17 at 13:02

SpaceX

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ZGski · Accepted Answer · 2019-05-20T14:49:02.493

22

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])
}

edited May 20 '19 at 14:49

answered May 05 '16 at 17:56

ZGski

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No warnings but bad code. You can avoid `arradIndex` and `_` with `for arrayIndex in (names.count - 1).stride(through: 0, by: -1)`. – Sulthan May 05 '16 at 18:09
Thanks @Sulthan! I've edited my answer reflect your much better solution. – ZGski May 05 '16 at 18:19
2

by the way, you could also mix it up with a `map` into `let reversedNames = (names.count - 1).stride(through: 0, by: -1).map { names[$0] }`. That solution is actually in [Eric's answer](http://stackoverflow.com/a/33463720/669586). – Sulthan May 05 '16 at 18:27

score 18 · Answer 3 · edited Nov 15 '19 at 03:30

18

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 })

edited Nov 15 '19 at 03:30

Ronan Boiteau

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answered Dec 20 '16 at 17:40

brunobasas

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score 6 · Answer 4 · answered Nov 08 '15 at 01:51

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"]

Eric Aya · Answer 5 · 2015-11-01T15:32:10.013

4

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.

edited Nov 01 '15 at 15:32

answered Nov 01 '15 at 15:16

Eric Aya

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I would use `stride(through: 0, by: -1)`. – Sulthan May 05 '16 at 18:30
To anyone reading this in the present (May '16, Xcode 7.3, Swift 2.2): .reverse() creates an ReverseRandomAccessCollection> which is not the same as an array; you cannot, for instance, add elements to it. – green_knight Jun 01 '16 at 20:27
@green_knight Yes and no. Yes, it does by default. No, *it's not a problem* because you can either force the type to array like `let result:[String] = names.reverse()` or use the array initializer like `let result = Array(names.reverse())`. – Eric Aya Jun 01 '16 at 20:31
Thanks. For me it _was_ a problem, because the documentation lags behind the implementation (it states that reverse() will return a reversed array, which it no longer does, though it has in the past), and casting to an array failed. That pretty much leaves one stranded. – green_knight Jun 02 '16 at 06:58
1

@milos-mandic Hey there. Could you please make up your mind and stop changing the accepted answer? Thanks! ;) – Eric Aya Jul 20 '16 at 13:26
1

@EricD honestly, I think your answer should be accepted over mine. – ZGski Jul 20 '16 at 14:39

user3441734 · Answer 6 · 2021-07-07T06:19:44.263

4

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
  
}

edited Jul 07 '21 at 06:19

answered Nov 01 '15 at 15:58

user3441734

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This no longer works, as the prefix and postfix incrementing/decrementing operators are no longer part of the Swift programming language. – Cristik Jul 04 '21 at 04:57
The new code snipped doesn't behave like the original one, there the post-increment and post-decrement operations were taking place after the swap, while in the new code they happen before the swap. And I think this make the code miss swapping the first element with the last. – Cristik Jul 05 '21 at 14:26

score 3 · Answer 7 · answered Nov 01 '15 at 15:22

3

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])
}

answered Nov 01 '15 at 15:22

Abhinav

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I got a warning that 'C-style for statement is depreciated and will be removed in future' :( – senty Apr 22 '16 at 02:07
Is this still valid? @Abhinav – Lukesivi May 05 '16 at 11:48
@senty I've taken the liberty of translating and updating the accepted answer to Swift 2.2 as to clear any warnings within Xcode. [My Answer](http://stackoverflow.com/a/37057200/4577897). – ZGski May 05 '16 at 18:02
1

This no longer works, Swift has removed support for the postfix decrementing operators. – Cristik Jul 04 '21 at 04:59

score 3 · Answer 8 · edited Jun 23 '21 at 15:29

3

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)

edited Jun 23 '21 at 15:29

achinthaishane

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answered May 23 '16 at 21:14

Pilkie

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vadian · Answer 9 · 2022-09-25T07:05:44.203

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: &)
        }
    }
}

score 2 · Answer 10 · answered Nov 21 '18 at 20:48

2

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)
}

answered Nov 21 '18 at 20:48

Muhammad Khan

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Solution already given in https://stackoverflow.com/a/37400672 by Pilkie – Eric Aya Nov 22 '18 at 10:56

Israel Manzo · Answer 11 · 2021-07-05T01:34:13.373

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
     }

func reversed(with arr: [T]) -> [T] { var arr = arr (0.. – GeoSD Apr 13 '22 at 14:17 — GeoSD, Apr 13 '22 at 14:17

score 2 · Answer 12 · edited Oct 12 '22 at 22:15

2

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)

edited Oct 12 '22 at 22:15

Tomerikoo

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answered Jan 14 '21 at 09:06

Saikiran Komirishetty

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0x416e746f6e · Answer 13 · 2015-11-01T15:19:43.033

1

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
}

edited Nov 01 '15 at 15:19

answered Nov 01 '15 at 15:11

0x416e746f6e

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Simon O'Doherty · Answer 14 · 2015-11-01T15:35:41.007

0

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)

edited Nov 01 '15 at 15:35

answered Nov 01 '15 at 15:26

Simon O'Doherty

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If I remember correctly, swap() cannot exchange an element with itself in Swift 2. That would happen here for an odd-sized array. Or did they fix that in Swift 2.1? – Martin R Nov 01 '15 at 15:32
@MartinR fixed. Although it will have issues with 1 item array. Although you could skip the whole loop if names.count = 1. – Simon O'Doherty Nov 01 '15 at 15:34
meh.. not a great answer. – Simon O'Doherty Nov 01 '15 at 15:37

score 0 · Answer 15 · answered May 06 '16 at 08:46

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()

score 0 · Answer 16 · answered Apr 13 '17 at 12:16

0

Here the code for swift 3

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

answered Apr 13 '17 at 12:16

Inder_iOS

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The question is very specific about NOT using reverse/reversed. It's even mentioned in the comments under the question. When OP asked this, *everybody* suggested using reverse/reversed but OP needed something else. – Eric Aya Apr 13 '17 at 14:45
@ Eric Aya: I came here for this answer. This is such a classic question so I think many different approaches should be allowed. – Sentry.co Apr 24 '17 at 11:12

score 0 · Answer 17 · answered Aug 04 '17 at 13:48

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

score 0 · Answer 18 · edited Aug 09 '17 at 05:56

0

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)

edited Aug 09 '17 at 05:56

Anh Pham

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answered Aug 09 '17 at 05:18

md.rayhan_bhuiyan

1

score 0 · Answer 19 · edited Nov 21 '19 at 16:15

0

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 :)

edited Nov 21 '19 at 16:15

Sapphire_Brick

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answered Aug 06 '19 at 12:47

Mayank Neema

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score 0 · Answer 20 · answered Nov 29 '21 at 08:22

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)

Kuvonchbek Yakubov · Answer 21 · 2022-01-22T18:39:54.510

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
}

Pushker pandey · Answer 22 · 2023-08-11T03:25:24.547

There are so many answers, Here in swift 5 by using generic method to reverse any type of array: modified from @tomeriko 's solution

func reverseArray<Element:Equatable>(input : [Element])-> [Element]{
   var count = input.count
   var result_ = input
   for index in 0..<input.count{
      result_.insert(input[count-index-1], at: index)
      result_.removeLast()
    }
   return result_
 }
 print(reverseArray(input: [1,2,3,4,5,6]))
 print(reverseArray(input: ["1","2","3","4"]))

output:

[6, 5, 4, 3, 2, 1]

["4", "3", "2", "1"]

Divyanshu Kumar · Answer 23 · 2018-11-06T05:39:24.590

-1

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"]

edited Nov 06 '18 at 05:39

answered Nov 06 '18 at 05:32

Divyanshu Kumar

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This doesn't answer the question because it reverse sorts the array, it doesn't reverse the original order as OP is asking. – Eric Aya Nov 06 '18 at 10:28
@Moritz you are right, but there are many newbies out there who will find this in search of reverse sorting, for their sake i have written this solution, and i have clearly mentioned "for reversed sorting". – Divyanshu Kumar Nov 12 '18 at 09:46

score -1 · Answer 24 · answered Nov 17 '19 at 06:26

-1

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)

answered Nov 17 '19 at 06:26

Dre Adesina

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Obyadur · Answer 25 · 2022-09-25T03:19:02.440

-1

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"]"

edited Sep 25 '22 at 03:19

answered Sep 25 '22 at 03:09

Obyadur

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As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – abdo Salm Sep 28 '22 at 17:44

score -2 · Answer 26 · edited Sep 28 '17 at 23:14

-2

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"]

edited Sep 28 '17 at 23:14

bfontaine

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answered Sep 28 '17 at 20:16

Paresh Hirpara

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This doesn't answer the question, it sorts the array, it doesn't reverse it. – Eric Aya Sep 29 '17 at 09:10
2

Please post answers relevant to the question.This answers sorts an array not reverses it. – Abhijeet Mallick Feb 16 '20 at 13:27

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

26 Answers26

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