CODEWITHSUNDEEP
swiftstring
user9470831
user9470831

Reputation:

Reversing Words Functionally in Swift

I can reverse every word in a string functionally without using a loop, but when I try to reverse EVERY OTHER WORD. I run into problems. I can do it with a loop but not functionally. What am I not seeing here?

Functionally (every word):

import UIKit

let input = "This is a sample sentence"

func reverseWords(input: String) -> String {
    let parts = input.components(separatedBy: " ")
    let reversed = parts.map { String($0.reversed()) }
    return reversed.joined(separator: " ")
}

reverseWords(input: input)

With loop (EVERY OTHER WORD): 

var sampleSentence = "This is a sample sentence"

func reverseWordsInSentence(sentence: String) -> String {
    let allWords = sampleSentence.components(separatedBy:" ")
    var newSentence = ""

    for index in 0...allWords.count - 1 {
        let word = allWords[index]
        if newSentence != "" {
            newSentence += " "
        }

        if index % 2 == 1 {
            let reverseWord = String(word.reversed())
            newSentence += reverseWord
        } else {
            newSentence += word
        }
    }
    return newSentence
}
reverseWordsInSentence(sentence: sampleSentence)

Upvotes: 1

Views: 1268

Answers (2)

vacawama
vacawama

Reputation: 154603

With a slight modification of your reverseWords you can reverse every other word. Use enumerated() to combine a word with its position, and then use that to reverse odd words:

let input = "one two three four five"

func reverseOddWords(input: String) -> String {
    let parts = input.components(separatedBy: " ")
    let reversed = parts.enumerated().map { $0 % 2 == 0 ? String($1.reversed()) : $1 }
    return reversed.joined(separator: " ")
}

print(reverseOddWords(input: input))

eno two eerht four evif

Or you could pattern your function after Swift's sort and pass the filter closure to the reverseWords function:

let input = "one two three four five"

func reverseWords(_ input: String, using filter: ((Int) -> Bool) = { _ in true }) -> String {
    let parts = input.components(separatedBy: " ")
    let reversed = parts.enumerated().map { filter($0) ? String($1.reversed()) : $1 }
    return reversed.joined(separator: " ")
}

// default behavior is to reverse all words
print(reverseWords("one two three four five"))

eno owt eerht ruof evif

print(reverseWords("one two three four five", using: { $0 % 2 == 1 }))

one owt three ruof five

print(reverseWords("one two three four five", using: { [0, 3, 4].contains($0) }))

eno two three ruof evif

let everyThirdWord = { $0 % 3 == 0 }

print(reverseWords("one two three four five", using: everyThirdWord))

eno two three ruof five

Upvotes: 3

Duncan C
Duncan C

Reputation: 131426

Use stride() to generate a sequence of indexes of every other word.

Then use forEach() to select each index in the stride array and use it to mutate the word at that index to reverse it. 

import UIKit

let string = "Now is the time for all good programmers to babble incoherently"

var words = string.components(separatedBy: " ")

stride(from: 0, to: words.count, by: 2)
.forEach { words[$0] = String(words[$0].reversed()) }


let newString = words.joined(separator: " ")

print(newString)

The output string is: 

"woN is eht time rof all doog programmers ot babble yltnerehocni"

Upvotes: 2

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