我怎样才能交错两个数组？

Question

如果我有两个数组，例如

let one = [1,3,5]
let two = [2,4,6]

我想要 merge/interleave 以下模式中的数组 [one[0]、two[0]、one[1]、two[1] 等....]

//prints [1,2,3,4,5,6]
let comibned = mergeFunction(one, two)
print(combined)

什么是实现组合功能的好方法？

func mergeFunction(one: [T], _ two: [T]) -> [T] {
    var mergedArray = [T]()
    //What goes here
    return mergedArray
}

Answer 1

如果你只是想交错两个数组，你可以这样做：

let maxIndex = max(one.count, two.count)
var mergedArray = Array<T>()
for index in 0..<maxIndex {
    if index < one.count { mergedArray.append(one[index]) }
    if index < two.count { mergedArray.append(two[index]) }
}

return mergedArray

Answer 2

如果两个数组的长度相同，那么这是一个可能的解决方案：

let one = [1,3,5]
let two = [2,4,6]

let merged = zip(one, two).flatMap { [[=10=], ] }

print(merged) // [1, 2, 3, 4, 5, 6]

这里zip()并行枚举数组，returns一个序列 一对（2 元素元组），每个数组中有一个元素。 flatMap() 从每对创建一个 2 元素数组并将结果连接起来。

如果数组可以有不同的长度，那么你追加 结果的较长数组的额外元素：

func mergeFunction<T>(one: [T], _ two: [T]) -> [T] {
    let commonLength = min(one.count, two.count)
    return zip(one, two).flatMap { [[=11=], ] } 
           + one.suffixFrom(commonLength)
           + two.suffixFrom(commonLength)
}

---

Swift3 的更新：

func mergeFunction<T>(_ one: [T], _ two: [T]) -> [T] {
    let commonLength = min(one.count, two.count)
    return zip(one, two).flatMap { [[=12=], ] } 
           + one.suffix(from: commonLength)
           + two.suffix(from: commonLength)
}

Answer 3

使用Swift5，您可以使用以下Playground示例代码之一来解决您的问题。

---

#1。使用 zip(_:_:) function and Collection's flatMap(_:) 方法

let one = [1, 3, 5, 7]
let two = [2, 4, 6]

let array = zip(one, two).flatMap({ [[=10=], ] })
print(array) // print: [1, 2, 3, 4, 5, 6]

苹果states:

If the two sequences passed to zip(_:_:) are different lengths, the resulting sequence is the same length as the shorter sequence.

---

#2。使用符合 Sequence and IteratorProtocol 协议

的对象

struct InterleavedSequence<T>: Sequence, IteratorProtocol {

    private let firstArray: [T]
    private let secondArray: [T]
    private let thresholdIndex: Int
    private var index = 0
    private var toggle = false

    init(firstArray: [T], secondArray: [T]) {
        self.firstArray = firstArray
        self.secondArray = secondArray
        self.thresholdIndex = Swift.min(firstArray.endIndex, secondArray.endIndex)
    }

    mutating func next() -> T? {
        guard index < thresholdIndex else { return nil }
        defer {
            if toggle {
                index += 1
            }
            toggle.toggle()
        }
        return !toggle ? firstArray[index] : secondArray[index]
    }

}

let one = [1, 3, 5, 7]
let two = [2, 4, 6]

let sequence = InterleavedSequence(firstArray: one, secondArray: two)
let array = Array(sequence)
print(array) // print: [1, 2, 3, 4, 5, 6]

Answer 4

  /// Alternates between the elements of two sequences.
  /// - Parameter keepSuffix:
  /// When `true`, and the sequences have different lengths,
  /// the suffix of `interleaved`  will be the suffix of the longer sequence.
  func interleaved<Sequence: Swift.Sequence>(
    with sequence: Sequence,
    keepingLongerSuffix keepSuffix: Bool = false
  ) -> AnySequence<Element>
  where Sequence.Element == Element {
    keepSuffix
    ? .init { () -> AnyIterator<Element> in
      var iterators = (
        AnyIterator( self.makeIterator() ),
        AnyIterator( sequence.makeIterator() )
      )
      return .init {
        defer { iterators = (iterators.1, iterators.0) }
        return iterators.0.next() ?? iterators.1.next()
      }
    }
    : .init(
      zip(self, sequence).lazy.flatMap { [[=10=], ] }
    )
  }

let oddsTo7 = stride(from: 1, to: 7, by: 2)
let evensThrough10 = stride(from: 2, through: 10, by: 2)
let oneThrough6 = Array(1...6)

XCTAssertEqual(
  Array( oddsTo7.interleaved(with: evensThrough10) ),
  oneThrough6
)

XCTAssertEqual(
  Array(
    oddsTo7.interleaved(with: evensThrough10, keepingLongerSuffix: true)
  ),
  oneThrough6 + [8, 10]
)