Swift Anagram checker

时间:2015-08-25 16:54:45

标签: ios swift swift2 anagram

我正在尝试为swift构建一个anagram检查器。这是我的代码。如果您不知道anagram检查器检查两个字符串是否包含相同的字符,但顺序无关紧要。

func checkForAnagram(#firstString: String, #secondString: String) -> Bool {
    var firstStringArray: [Character] = []
    var secondStringArray: [Character] = []
    /* if case matters delete the next four lines
    and make sure your variables are not constants */
    var first = firstString
    var second = secondString
    first = first.lowercaseString
    second = second.lowercaseString
    for charactersOne in first {
        firstStringArray += [charactersOne]
    }
    for charactersTwo in second {
        secondStringArray += [charactersTwo]
    }
    if firstStringArray.count != secondStringArray.count {
        return false
    } else {
        for elements in firstStringArray {
            if secondStringArray.contains(elements){
                return true
            } else {
                return false
            }
        }

}
}


var a = "Hello"
var b = "oellh"
var c = "World"
checkForAnagram(firstString: a, secondString: b)

我收到错误信息。

'[Character]' does not have a member 'contains'

11 个答案:

答案 0 :(得分:16)

你应该试试

System.out.println(map.containsValue("mark"));

答案 1 :(得分:16)

接受的答案是紧凑而优雅,但与其他解决方案相比效率非常低。

我现在将提出并讨论一些anagram检查器变体的实现。为了测量性能,我将使用不同的变体来从50,000多个单词的数组中找到给定单词的字谜。

// Variant 1: Sorting of Character
// Measured time: 30.46 s
func anagramCheck1(a: String, b: String) -> Bool {
    return a.characters.sorted() == b.characters.sorted()
}

这实际上是用Swift 3语法编写的接受答案的解决方案。它非常慢,因为与NSString不同,Swift的String基于Character,它可以正确处理Unicode字符。

更高效的解决方案利用NSCountedSet类,它允许我们将字符串表示为一组字符,每个字符都有自己的计数。如果两个字符串映射到相同的NSCountedSet,则它们是字符串。 注意:检查字符串长度作为前提条件会使实现始终更有效。

// Variant 2: NSCountedSet of Character
// Measured time: 4.81 s
func anagramCheck2(a: String, b: String) -> Bool {
    guard a.characters.count == b.characters.count else { return false }
    let aSet = NSCountedSet()
    let bSet = NSCountedSet()
    for c in a.characters {
        aSet.add(c)
    }
    for c in b.characters {
        bSet.add(c)
    }
    return aSet == bSet
}

更好,但不是很好。在这里,其中一个“罪魁祸首”是使用原生的Swift Character类型(来自Swift的String)。回到旧的Objective-C类型(NSString和unichar)可以提高效率。

// Variant 3: NSCountedSet of unichar
// Measured time: 1.31 s
func anagramCheck3(a: String, b: String) -> Bool {
    let aString = a as NSString
    let bString = b as NSString
    let length = aString.length
    guard length == bString.length else { return false }
    let aSet = NSCountedSet()
    let bSet = NSCountedSet()
    for i in 0..<length {
        aSet.add(aString.character(at: i))
        bSet.add(bString.character(at: i))
    }
    return aSet == bSet
}

使用NSCountedSet很好,但在我们比较两个NSCountedSet对象之前,我们完全填充它们。一个有用的替代方法是仅为两个字符串中的一个完全填充NSCountedSet,然后,当我们为另一个字符串填充NSCountedSet时,如果另一个字符串包含在第一个字符串的NSCountedSet中找不到的字符,则我们会提前失败字符串。

// Variant 4: NSCountedSet of unichar and early exit
// Measured time: 1.07 s
func anagramCheck4(a: String, b: String) -> Bool {
    let aString = a as NSString
    let bString = b as NSString
    let length = aString.length
    guard length == bString.length else { return false }
    let aSet = NSCountedSet()
    let bSet = NSCountedSet()
    for i in 0..<length {
        aSet.add(aString.character(at: i))
    }
    for i in 0..<length {
        let c = bString.character(at: i)
        if bSet.count(for: c) >= aSet.count(for: c) {
            return false
        }
        bSet.add(c)
    }
    return true
}

这是我们将获得的最佳时机(使用Swift)。但是,为了完整起见,让我再讨论一下这种变体。

下一个替代方案利用类型[unichar:Int]的Swift Dictionary来存储每个字符的重复次数,而不是NSCountedSet。它比前两个版本略慢,但我们可以稍后重复使用以获得更快的实现。

// Variant 5: counting repetitions with [unichar:Int]
// Measured time: 1.36
func anagramCheck5(a: String, b: String) -> Bool {
    let aString = a as NSString
    let bString = b as NSString
    let length = aString.length
    guard length == bString.length else { return false }
    var aDic = [unichar:Int]()
    var bDic = [unichar:Int]()
    for i in 0..<length {
        let c = aString.character(at: i)
        aDic[c] = (aDic[c] ?? 0) + 1
    }
    for i in 0..<length {
        let c = bString.character(at: i)
        let count = (bDic[c] ?? 0) + 1
        if count > aDic[c] ?? 0 {
            return false
        }
        bDic[c] = count
    }
    return true
}

请注意,与Variant 3相对应的使用NSCountedSet的vanilla Objective-C实现比以前的所有版本都要快得多。

// Variant 6: Objective-C and NSCountedSet
// Measured time: 0.65 s
- (BOOL)anagramChecker:(NSString *)a with:(NSString *)b {
    if (a.length != b.length) {
        return NO;
    }
    NSCountedSet *aSet = [[NSCountedSet alloc] init];
    NSCountedSet *bSet = [[NSCountedSet alloc] init];
    for (int i = 0; i < a.length; i++) {
        [aSet addObject:@([a characterAtIndex:i])];
        [bSet addObject:@([b characterAtIndex:i])];
    }
    return [aSet isEqual:bSet];
}

我们可以改进之前尝试的另一种方法是观察,如果我们需要找到给定单词的字谜,我们不妨将该单词视为固定,因此我们可以构建相应的结构(NSCountedSet,字典,...)只有一次这个词。

// Finding all the anagrams of word in words
// Variant 7: counting repetitions with [unichar:Int]
// Measured time: 0.58 s
func anagrams(word: String, from words: [String]) -> [String] {
    let anagrammedWord = word as NSString
    let length = anagrammedWord.length
    var aDic = [unichar:Int]()
    for i in 0..<length {
        let c = anagrammedWord.character(at: i)
        aDic[c] = (aDic[c] ?? 0) + 1
    }
    let foundWords = words.filter {
        let string = $0 as NSString
        guard length == string.length else { return false }
        var bDic = [unichar:Int]()
        for i in 0..<length {
            let c = string.character(at: i)
            let count = (bDic[c] ?? 0) + 1
            if count > aDic[c] ?? 0 {
                return false
            }
            bDic[c] = count
        }
        return true
    }
    return foundWords
}

现在,在上一个版本中,我们使用了[unichar:Int]字典。这证明比使用NSCountedSet的unichar更有效,无论是提前退出(0.60秒)还是没有(0.87秒)。

答案 2 :(得分:1)

// This answer also would work
// Convert your parameters on Array, then sorted them and compare them
func ana(str1: String, str2: String)->Bool{

    let a = Array(str1)
    let b = Array(str2)

    if a.sorted() == b.sorted() {
        return true
    }
    return false
}

答案 3 :(得分:1)

不要忘记空格

     func isAnagram(_ stringOne: String, stringTwo: String) -> Bool {
         return stringOne.lowercased().sorted().filter { $0 != " "} stringTwo.lowercased().sorted().filter { $0 != " "}

}

答案 4 :(得分:0)

// Make sure name your variables correctly so you won't confuse
// Mutate the constants parameter, lowercase to handle capital letters and the sorted them to compare both. Finally check is there are equal return true or false.

func anagram(str1: String, srt2: String)->Bool{

    let string1 = str1.lowercased().sorted()
    let string2 = srt2.lowercased().sorted()

    if string1 == string2 {
        return true
    }
    return false
}

答案 5 :(得分:0)

我们可以使用字典来构造一个新的数据结构容器。然后按字符串的键/字符比较该值。

func anagram(str1: String, str2 : String) -> Bool {

    var dict1 = [Character: Int](), dict2 = [Character: Int]()

    str1.forEach { (char) in
        if dict1[char] != nil {
            dict1[char]! += 1
        } else {
            dict1[char] = 1
        }
    }

    str2.forEach { (char) in
        if dict2[char] != nil {
            dict2[char]! += 1
        } else {
            dict2[char] = 1
        }
    }

    return dict1 == dict2 ? true : false

}

// input -> "anna", "aann"
// The count will look like: 
// ["a": 2, "n": 2] & ["a": 2, "n": 2] 
// then return true

答案 6 :(得分:0)

func checkAnagrams(str1: String, str2: String) -> Bool {
        guard str1.count == str2.count else { return false }
        var dictionary = Dictionary<Character, Int>()
        for index in 0..<str1.count {
            let value1 = str1[str1.index(str1.startIndex, offsetBy: index)]
            let value2 = str2[str2.index(str2.startIndex, offsetBy: index)]
            dictionary[value1] = (dictionary[value1] ?? 0) + 1
            dictionary[value2] = (dictionary[value2] ?? 0) - 1
        }
        return !dictionary.contains(where: {(_, value) in
            return value != 0
        })
}

时间复杂度-O(n)

答案 7 :(得分:0)

    func isAnagram(word1: String, word2: String) -> Bool {
        let set1 = Set(word1)
        let set2 = Set(word2)
        return set1 == set2
    }

or

    func isAnagram(word1: String,word2: String) -> Bool {
        return word1.lowercased().sorted() == word2.lowercased().sorted()
    }

答案 8 :(得分:0)

  

Swift 4.1函数将为您提供3个关于Anagram的问题答案:-

     

1。输入字符串(a,b)是Anagram吗? //布尔

     

2。如果不是Anagram,那么计数的数量需要更改字符串(a,b)中的字符以使它们成为anagram? // Int

     

3。如果不是Anagram,则需要更改字符串(a,b)中的字符列表以使它们成为anagram? // [字符]

步骤1:-复制并粘贴以下内容到所需的班级:-

  //MARK:-  Anagram checker
    func anagramChecker(a:String,b:String) -> (Bool,Int,[Character]) {
        var aCharacters = Array(a)
        var bCharacters = Array(b)
        var count = 0
        var isAnagram = true
        var replacementRequiredWords:[Character] = [Character]()
        if aCharacters.count == bCharacters.count {
            let listA = aCharacters.filter { !bCharacters.contains($0) }
            for i in 0 ..< listA.count {
                if !replacementRequiredWords.contains(listA[i]) {
                    count = count + 1
                    replacementRequiredWords.append(listA[i])
                    isAnagram = false
                }
            }
            let listB = bCharacters.filter { !aCharacters.contains($0) }
            for i in 0 ..< listB.count {
                if !replacementRequiredWords.contains(listB[i]) {
                    count = count + 1
                    replacementRequiredWords.append(listB[i])
                     isAnagram = false
                }
            }
        }else{
            //cant be an anagram
            count = -1
        }
         return (isAnagram,count,replacementRequiredWords)
    }

第2步:-输入两个要测试的输入字符串

// Input Strings
var a = "aeb"
var b = "abs"

步骤3:-打印结果:-

print("isAnagram : \(isAnagram(a: a, b: b).0)")

print("number of count require to change strings in anagram  : \(isAnagram(a: a, b: b).1)")//-1 will come in case of cant be a Anagram

print("list of  Characters needs to be change : \(isAnagram(a: a, b: b).2)")
  

上述练习的结果:-

isAnagram : false
number of count require to change strings in anagram  : 2
list of  Characters needs to be change : ["e", "s"]
  

希望这个10分钟的练习将对我的Swift有所帮助   家族,轻松解决Anagram相关问题。 :)

答案 9 :(得分:0)

使用Swift中的inout方法检查两个字符串是否为字谜

func checkAnagramString(str1: inout String, str2: inout String)-> Bool{

  var result:Bool = false
  str1 = str1.lowercased().trimmingCharacters(in: .whitespace)
  str2 = str2.lowercased().trimmingCharacters(in: .whitespaces)

  if (str1.count != str2.count) {
     return result
  }

  for c in str1 {
     if str2.contains(c){
         result = true
    }
    else{
        result = false
        return result
    }
  }
   return result
}

调用函数以检查字符串是否为字谜

 var str1 = "tommarvoloriddle"
 var str2 = "iamlordvoldemort"

 print(checkAnagramString(str1: &str1, str2: &str2)) //Output = true.

答案 10 :(得分:0)

我刚刚意识到在 Swift 5.X

中执行Anagram函数的另一种简便方法
func checkForAnagram(firstString firstString: String, secondString: String) -> Bool {
    return !firstString.isEmpty && firstString.sorted() == secondString.sorted()
}