Swift 错误无法转换“[String.Element]”类型的值（又名“Array<character> ') 到预期的参数类型 '[String]'</character>

Question

I am trying to create an array of letters from a given word by using the following Swift code (I have an array of words for allWords, but for simplicity I've just added an example word there for now):

let allWords = ["Leopards"]
var arrayOfLetters = Array(allWords[0])

let everyPossibleArrangementOfLetters = permute(list: arrayOfLetters)

func permute(list: [String], minStringLen: Int = 3) -> Set<String> {
    func permute(fromList: [String], toList: [String], minStringLen: Int, set: inout Set<String>) {
        if toList.count >= minStringLen {
            set.insert(toList.joined(separator: ""))
        }
        if !fromList.isEmpty {
            for (index, item) in fromList.enumerated() {
                var newFrom = fromList
                newFrom.remove(at: index)
                permute(fromList: newFrom, toList: toList + [item], minStringLen: minStringLen, set: &set)
            }
        }
    }

    var set = Set<String>()
    permute(fromList: list, toList:[], minStringLen: minStringLen, set: &set)
    return set
}

I obtained this code from: Calculate all permutations of a string in Swift

But the following error is presented: Cannot convert value of type '[String.Element]' (aka 'Array') to expected argument type '[String]'

I attempted the following, which works, but it takes over 10 seconds per word (depending on number of repeat letters) and I was hoping to find a better solution.

var arrayOfLetters: [String] = []

     for letter in allWords[0] {
     arrayOfLetters.append(String(letter))
     }

     let everyPossibleArrangementOfLetters = permute(list: arrayOfLetters)

I wasn't able to get the following solution to work, although I think is has promise I couldn't get past the productID name of items in the array whereas my array items aren't named... Migration from swift 3 to swift 4 - Cannot convert String to expected String.Element

I'm also creating another array and checking each of those words to ensure their validity, and I run into the same error which I correct in the same way with array.append which is adding a lot more time in that location as well.

var everyPossibleArrangementOfLettersPartDeux: [String] = []
     for word in everyPossibleArrangementOfLetters {
     everyPossibleArrangementOfLettersPartDeux.append(word)
     }

numberOfRealWords = possibleAnagrams(wordArray: everyPossibleArrangementOfLettersPartDeux)

func possibleAnagrams(wordArray: [String]) -> Int {

    func isReal(word: String) -> Bool {
        let checker = UITextChecker()
        let range = NSMakeRange(0, word.utf16.count)
        let misspelledRange = checker.rangeOfMisspelledWord(in: word, range: range, startingAt: 0, wrap: false, language: "en")

        return misspelledRange.location == NSNotFound
    }

    var count = 0

    for word in wordArray {
        if isReal(word: word) {
            count += 1
            //print(word)
        }

        }
        return count
        }

I'm hoping the same replacement for array.append will work in both spots.

Answer 1

The problem is that Array(allWords[0]) produces [Character] and not the [String] that you need.

You can call map on a String (which is a collection of Character s and use String.init on each character to convert it to a String ). The result of the map will be [String] :

var arrayOfLetters = allWords[0].map(String.init)

---

Notes:

1. When I tried this in a Playground, I was getting the mysterious message Fatal error: Only BidirectionalCollections can be advanced by a negative amount . This seems to be a Playground issue, because it works correctly in an app.
2. Just the word "Leopards" produces 109,536 permutations.

---

Another Approach

Another approach to the problem is to realize that permute doesn't have to work on [String] . It could use [Character] instead. Also, since you are always starting with a String , why not pass that string to the outer permute and let it create the [Character] for you.

Finally, since it is logical to think that you might just want anagrams of the original word, make minStringLen an optional with a value of nil and just use word.count if the value is not specified.

func permute(word: String, minStringLen: Int? = nil) -> Set<String> {
    func permute(fromList: [Character], toList: [Character], minStringLen: Int, set: inout Set<String>) {
        if toList.count >= minStringLen {
            set.insert(String(toList))
        }
        if !fromList.isEmpty {
            for (index, item) in fromList.enumerated() {
                var newFrom = fromList
                newFrom.remove(at: index)
                permute(fromList: newFrom, toList: toList + [item], minStringLen: minStringLen, set: &set)
            }
        }
    }

    var set = Set<String>()
    permute(fromList: Array(word), toList:[], minStringLen: minStringLen ?? word.count, set: &set)
    return set
}

Examples:

print(permute(word: "foo", minStringLen: 1))

 ["of", "foo", "f", "fo", "o", "oof", "oo", "ofo"] 

print(permute(word: "foo"))

 ["foo", "oof", "ofo"] 

Answer 2

This line is returning a Character array, not a String one:

var arrayOfLetters = Array(allWords[0])

You can convert this to a String array like so:

var arrayOfLetters = Array(allWords[0]).map{String($0)}

You could alternatively write:

var arrayOfLetters = allWords[0].characters.map{String($0)}

Answer 3

If, it is optional character or string

usedLetters.append(currentWord.randomElement().map(String.init)!)

Here usedLetters is Array[String] currentWord is Optional string