arrays - 在swift中查找字符串数组的所有组合

Question

我有一个字符串数组，我想找到它的元素的所有可能组合

For Example :

Array = [A,B,C,D]

should produce result as :

[A,AB,AC,AD,ABC,ABD,ACD,ABCD,B,BC,BD,BCD,C,CD,D]

这是我的逻辑:

  var array = ["A", "B", "C","D"]
  var list = [String]()
  for i in 0..<array.count{
    let c = array[i]
    list.append(c)
    var d = c
    for count in 1..<array.count{

        if i+count < array.count{
            for j in i+count..<array.count{
                var a = d
                a.appendContentsOf(array[j])
                print("a : \(a)")
                list.append(a)
            }

        }
        d = c
        d.appendContentsOf(array[count])
        print("d : \(d)")
    }
}
print(list.description)

Its Output is :

["A", "AB", "AC", "AD", "ABC", "ABD", "ACD", "B", "BC", "BD", "BBD", "C", "CD", "D"]

此输出缺少 ABCD 并将 BCD 错误地打印为 BBD

任何人请通过增强我的代码或为此提出您自己的逻辑来帮助我。

Answer 1

@yannick 的回答非常接近。

通过计算集合的幂集，您可以获得所有可能的子集(包括原始集和空集)。

获得幂集后，您只需将子集连接成一个字符串即可获得所需的结果。

这是完整的解决方案(连同更新的代码和大量注释):

extension Array {
    var powerset: [[Element]] {
        guard count > 0 else {
            return [[]]
        }

        // tail contains the whole array BUT the first element
        let tail = Array(self[1..<endIndex])

        // head contains only the first element
        let head = self[0]

        // computing the tail's powerset
        let withoutHead = tail.powerset

        // mergin the head with the tail's powerset
        let withHead = withoutHead.map { $0 + [head] }

        // returning the tail's powerset and the just computed withHead array
        return withHead + withoutHead
    }
}

let myArray = ["A", "B", "C", "D"]
print(myArray.powerset) // -> [["D", "C", "B", "A"], ["C", "B", "A"], ["D", "B", "A"], ["B", "A"], ["D", "C", "A"], ["C", "A"], ["D", "A"], ["A"], ["D", "C", "B"], ["C", "B"], ["D", "B"], ["B"], ["D", "C"], ["C"], ["D"], []]

// joining the subsets
let myResult = myArray.powerset.map { $0.sort().joinWithSeparator("") }
print(myResult) // -> ["A", "AB", "ABC", "ABCD", "ABD", "AC", "ACD", "AD", "B", "BC", "BCD", "BD", "C", "CD", "D", ""]

附言

请注意，此解决方案使用递归方法，而您的解决方案使用迭代方法。

PPS

如果您不想在您的解决方案中使用空字符串 ""，您可以将其过滤掉:

let myResult = myArray.powerset.map({ $0.sort().joinWithSeparator("") }).filter({ $0 != "" })

print(myResult) // -> ["A", "AB", "ABC", "ABCD", "ABD", "AC", "ACD", "AD", "B", "BC", "BCD", "BD", "C", "CD", "D"]