如何生成所有可能的组合？

Question

我目前正在尝试从 Strings 的 Array 中创建所有可能组合的 Set，每个元素只包含一个字母。

Array 本身可以包含相同的字母两次或更多次，它们只应在出现时使用。

Set稍后应包含从最少 2 个字母到给定 Array 长度的所有组合。

我在 Whosebug 上进行了搜索，但只找到了忽略事实的排列函数，即每个字母只应在出现时使用。

这是我的第一个 Swift 2 项目，所以请原谅我的新手:)

我想要的

var array = ["A", "B", "C","D"]
var combinations: Set<String>

... <MAGIC> ...

print(combinations)
// "AB", "ABC", "ABD", "ABCD", "ABDC", "AC", "ACB", "ACD", "ACBD", "ACDB", and so on ...

我目前的做法

func permuation(arr: Array<String>) {

    for (index, elementA) in arr.enumerate() {
        //1..2..3..4
        var tmpString = elementA
        var tmpArray = arr
        tmpArray.removeAtIndex(index)

        for (index2, elementB) in tmpArray.enumerate() {
            // 12..13..14
            var tmpString2 = tmpString + elementB
            var tmpArray2 = tmpArray

            //[3,4]
            tmpArray2.removeAtIndex(index2)

            results.append(tmpString2)
        }
    }

}
permuation(array)
print(results)
// "["AB", "AC", "AD", "BA", "BC", "BD", "CA", "CB", "CD", "DA", "DB", "DC"]"

我知道，这在很多方面都是非常错误的，但我受困于这段代码，不知道如何添加递归功能。

Answer 1

试试这个。

一般算法是让一个 fromList 包含您尚未使用的字母，一个 toList 是您目前已建立的字符串。这使用递归来构建所有可能的字符串，并在长度为 2 或更大时将它们添加到集合中：

func permute(fromList: [String], toList: [String] = [String](), var set: Set<String> = Set<String>()) -> Set<String> {
    if toList.count >= 2 {
        set.insert(toList.joinWithSeparator(""))
    }
    if !fromList.isEmpty {
        for (index, item) in fromList.enumerate() {
            var newFrom = fromList
            newFrom.removeAtIndex(index)
            set = permute(newFrom, toList: toList + [item], set: set)
        }
    }
    return set
}

permute(["A", "B", "C"])
// {"BA", "AC", "ABC", "AB", "BCA", "CB", "BC", "CAB", "ACB", "CA", "CBA", "BAC"}

permute(["A", "A", "B"])
// {"BA", "BAA", "AAB", "AB", "ABA", "AA"}

更快的回答：

正如@MartinR 在他的 post 中指出的那样，由于所有的集合创建和复制，上面的解决方案有点慢。我最初是使用一个 inout 变量来编写它的，但将其更改为功能更强大的接口以使其更易于调用。

这是我最初的（更快的）实现，另外我将它嵌入了一个 permute 中，它只需要一个 [String] 和 returns 一个 Set<String>。它创建 set 和 toList 数组，然后调用 permute 的内部版本来完成真正的工作：

func permute(list: [String], minStringLen: Int = 2) -> Set<String> {
    func permute(fromList: [String], toList: [String], minStringLen: Int, inout set: Set<String>) {
        if toList.count >= minStringLen {
            set.insert(toList.joinWithSeparator(""))
        }
        if !fromList.isEmpty {
            for (index, item) in fromList.enumerate() {
                var newFrom = fromList
                newFrom.removeAtIndex(index)
                permute(newFrom, toList: toList + [item], minStringLen: minStringLen, set: &set)
            }
        }
    }

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

permute(["A", "B", "C"])
// {"BA", "AC", "ABC", "AB", "BCA", "CB", "BC", "CAB", "ACB", "CA", "CBA", "BAC"}

permute(["A", "A", "B"])
// {"BA", "BAA", "AAB", "AB", "ABA", "AA"}

permute(["A", "A", "B"], minStringLen: 1)
// {"BA", "A", "BAA", "AB", "AA", "B", "AAB", "ABA"}

permute(["A", "A", "B"], minStringLen: 3)
// {"ABA", "BAA", "AAB"}

编辑： 我添加了一个 minStringLen 参数（默认值为 2）而不是对该值进行硬编码。

有关性能比较，请参阅@MartinR 的回答。

Swift 3 和 Swift 4:

func permute(list: [String], minStringLen: Int = 2) -> 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
}

print(permute(list: ["A", "B", "C"]))
// ["ABC", "CA", "BAC", "ACB", "BA", "CAB", "BC", "CB", "BCA", "CBA", "AB", "AC"]

print(permute(list: ["A", "A", "B"]))
// ["AA", "AAB", "ABA", "AB", "BA", "BAA"]

print(permute(list: ["A", "A", "B"], minStringLen: 1))
// ["AAB", "ABA", "B", "BA", "A", "BAA", "AA", "AB"]

print(permute(list: ["A", "A", "B"], minStringLen: 3))
// ["AAB", "ABA", "BAA"]

Answer 2

这与@vacawama 的回答非常相似，但希望有所不同足以值得一个单独的答案:)

这里构建了一个所有组合的数组（解释内嵌评论):

func combinations(array : [String]) -> [String] {

    // Recursion terminates here:
    if array.count == 0 { return [] }

    // Concatenate all combinations that can be built with element #i at the
    // first place, where i runs through all array indices:
    return array.indices.flatMap { i -> [String] in

        // Pick element #i and remove it from the array:
        var arrayMinusOne = array
        let elem = arrayMinusOne.removeAtIndex(i)

        // Prepend element to all combinations of the smaller array:
        return [elem] + combinations(arrayMinusOne).map { elem + [=10=] }
    }
}

然后就可以过滤至少两个字母的字符串，并且将其转换为 Set:

let c = Set(combinations(["A", "B", "C"]).filter { [=11=].characters.count >= 2 })
print(c)
// ["BA", "AC", "ABC", "AB", "BCA", "CB", "BC", "CAB", "ACB", "CA", "CBA", "BAC"]

我做了一个简单的性能对比（在Release模式下编译在 Macbook Pro 上):

let array = ["A", "B", "C", "D", "E", "F", "G"]

let t1 = NSDate()
let c1 = Set(combinations(array).filter { [=12=].characters.count >= 2 })
let t2 = NSDate()
let c2 = permute(array)
let t3 = NSDate()

print(c1 == c2) // true
print(t2.timeIntervalSinceDate(t1))
print(t3.timeIntervalSinceDate(t2))

结果取决于输入数组的大小，但是@vacawama 的更新方法是最快的：

# of array   This      vacawama's   vacawama's
elements:    method:   1st method:  2nd method:

  2          0.00016   0.00005      0.00001
  3          0.00043   0.00013      0.00004
  4          0.00093   0.00062      0.00014
  5          0.00335   0.00838      0.00071
  6          0.01756   0.24399      0.00437
  7          0.13625   11.90969     0.03692

Answer 3

在您的输出示例中，您也不清楚您真正想要的是什么：

它们的所有组合和排列：

["AB", "BA", "AC", "CA", "AD", "DA", ..., "ABCD", "ABDC", "ACBD", "ACDB", ...]

所有组合:

["AB", "AC", "AD", "BC", "BD", "CD", "ABC", "ABD", ...]

我可以推荐@oisdk 很棒的 Swift 库：SwiftSequence for both of them, it has lots of useful functions. In the Combinations section he even shows an example of its usage with the Power Set，这几乎正是您在案例 1 中寻找的。导入他的库文件后，您可以创建powerSet 在 CollectionType 上的功能如下：

extension CollectionType {
    func powerSet() -> LazySequence<FlattenSequence<LazyMapSequence<Self, ComboSeq<Self.Generator.Element>>>>{
        var i = 0
        return lazy.flatMap{ _ in self.lazyCombos(++i) }
    }
}

此方法延迟求值，这意味着它仅在真正需要时才求值。现在您提到您只想拥有至少 2 个元素的组合。使用 filter 方法可以轻松完成此操作：

let combinations = ["A", "B", "C", "D"].powerSet().filter{ [=13=].count >= 2 }
    // As an array: [["A", "B"], ["A", "C"], ["A", "D"], ["B", "C"], ["B", "D"], ["C", "D"], ["A", "B", "C"], ["A", "B", "D"], ["A", "C", "D"], ["B", "C", "D"], ["A", "B", "C", "D"]]

对于情况 2，如果您还需要这些排列，您可以这样做：

let combPerms = combinations.flatMap{ [=14=].permutations() }
    // As an array: [["A", "B"], ["B", "A"], ["A", "C"], ["C", "A"], ["A", "D"], ["D", "A"], ["B", "C"], ["C", "B"], ["B", "D"], ["D", "B"], ["C", "D"], ["D", "C"], ["A", "B", "C"], ["A", "C", "B"], ["B", "A", "C"], ["B", "C", "A"], ["C", "A", "B"], ["C", "B", "A"], ["A", "B", "D"], ["A", "D", "B"], ["B", "A", "D"], ["B", "D", "A"], ["D", "A", "B"], ["D", "B", "A"], ["A", "C", "D"], ["A", "D", "C"], ["C", "A", "D"], ["C", "D", "A"], ["D", "A", "C"], ["D", "C", "A"], ["B", "C", "D"], ["B", "D", "C"], ["C", "B", "D"], ["C", "D", "B"], ["D", "B", "C"], ["D", "C", "B"], ["A", "B", "C", "D"], ["A", "B", "D", "C"], ["A", "C", "B", "D"], ["A", "C", "D", "B"], ["A", "D", "B", "C"], ["A", "D", "C", "B"], ["B", "A", "C", "D"], ["B", "A", "D", "C"], ["B", "C", "A", "D"], ["B", "C", "D", "A"], ["B", "D", "A", "C"], ["B", "D", "C", "A"], ["C", "A", "B", "D"], ["C", "A", "D", "B"], ["C", "B", "A", "D"], ["C", "B", "D", "A"], ["C", "D", "A", "B"], ["C", "D", "B", "A"], ["D", "A", "B", "C"], ["D", "A", "C", "B"], ["D", "B", "A", "C"], ["D", "B", "C", "A"], ["D", "C", "A", "B"], ["D", "C", "B", "A"]]

您可以将它们转换为 String 的 Set 或 Array:

let array = Array(combPerms)
let set = Set(combPerms)

但我强烈建议使用惰性版本 ;) 是的，要删除重复项，您可以只使用 Set(["A", "B", "C", "D"]) 而不是 ["A", "B", "C", "D"]

你也可以像这样一次性完成案例2：

let x = ["A", "B", "C", "D"]

let result = Set(
    x.indices
        .flatMap{ x.lazyCombos([=16=] + 1) }
        .filter{ [=16=].count >= 2 }
        .flatMap{ [=16=].permutations() }
        .map{ [=16=].joinWithSeparator("") })

Answer 4

这里有一个 Swift 3 函数，速度要快一些。它基于 Array 类型的扩展，可用于具有任何元素类型的数组。

public func allCombinations(_ array:[String], minLength:Int=2) -> [String]
{
   var result:[String] = []
   for n in minLength...array.count
   {
      result = result + array.combinations(of:n).map{ [=10=].joined(separator:"") }
   }
   return result
}

extension Array
{
    public func combinations(of group:Int) -> [[Element]]
    {
       if group > count  { return [] }

       if group == count { return [self] }

       var result:[[Element]] = []

       var comboIndexes = (0..<group).map{[=10=]}

       let fullCombo   = group - 1
       let indexLimit  = count - fullCombo

       var carry = fullCombo

       while carry >= 0
       {
          if carry == fullCombo
          { result.append(comboIndexes.map{self[[=10=]]}) }

          comboIndexes[carry] += 1

          if comboIndexes[carry] == carry + indexLimit 
          { carry -= 1 ; continue }

          while carry < fullCombo
          {
             carry += 1
             comboIndexes[carry] = comboIndexes[carry-1] + 1 
          }       
       }

       return result
   }
}

在我的测试中，它运行在 7 个字母上比 vacawama 的第二个版本快大约 40 倍。

[编辑] 后来我意识到这个函数会产生组合（按照 OP 中的要求），而 vacawama 的函数会产生排列。我测试了一个等效的排列算法，它只比 vacawama 的算法快 55%。

extension Array
{
   public func permutations(of group:Int? = nil) -> [[Element]]
   {
      let group       = group ?? count
      var result      : [[Element]] = []
      var permutation : [Element]   = []

      func permute(from baseIndex:Int)
      {
         if baseIndex == permutation.count - 1
         { 
           result.append(permutation)
           return 
         }

         permute(from:baseIndex+1)

         for index in baseIndex+1..<permutation.count
         {
            swap(&permutation[baseIndex],&permutation[index]) 
            permute(from:baseIndex+1)
         }
         let baseElement = permutation[baseIndex]
         permutation.remove(at:baseIndex)
         permutation.append(baseElement)
      }

      var comboIndexes = (0..<group).map{[=11=]}

      let fullCombo   = group - 1
      let indexLimit  = count - fullCombo

      var carry = fullCombo

      while carry >= 0
      {
         if carry == fullCombo
         { 
           permutation = comboIndexes.map{self[[=11=]]}
           permute(from:0)
         }

         comboIndexes[carry] += 1

         if comboIndexes[carry] == carry + indexLimit 
         { carry -= 1 ; continue }

         while carry < fullCombo
         {
            carry += 1
            comboIndexes[carry] = comboIndexes[carry-1] + 1 
         }       
      }

      return result
   }
}

如何生成所有可能的组合？

How to generate all possible combinations?

arrays

recursion

set

ios

swift