如何在Swift中将两个数组合并为一个？

Question

我有两个数组：

var array1: [String] = ["l1", "m2", "r3"]
var array2: [String] = ["l4", "m5", "r6"]

我如何将它们组合成一个看起来像的数组

var combinations: [[String]] = [
["l1", "m2", "r3"], ["l1", "m5", "r3"], ["l1", "m5", "r6"], ["l1", "m2", "r6"],
["l4", "m2", "r3"], ["l4", "m5", "r3"], ["l4", "m5", "r6"], ["l4", "m2", "r6"]
]

编辑1 ：如果您有两个以上的数组

let arrays: [[String]] = [array1, array2, moreArrays]
var combinations: [[String]] = []

let arraysLen: Int = arrays.count
let arrLen: Int = arrays[0].count
let numCombinations: Int = Int(pow(Double(arraysLen), Double(arrLen)))

for i in (0 ..< numCombinations) {
  var combination: [String] = []
  for j in (0 ..< arrLen) {
    let arrIndex: Int = i / Int(pow(Double(arraysLen), Double(j))) % arraysLen
    let elIndex: Int = arrLen - j - 1
    combination.insert(arrays[arrIndex][elIndex], at: 0)
  }
  combinations.append(combination)
}

Answer 1

如果要获得两个数组的可能组合，可以使用一个简单的技巧：

var array1: [String] = ["l1", "m2", "r3"]
var array2: [String] = ["l4", "m5", "r6"]

guard array1.count == array2.count else {
    fatalError("Array must be of the same lenght")
}

let numCombinations = 1 << array1.count
let combinations: [[String]] = (0..<numCombinations).map { (index: Int) in
    return Array((0..<array1.count).map { (elementIndex: Int) in
        // every bit decides which array to choose
        let chooseArray1 = (index & (1 << elementIndex)) == 0
        return chooseArray1 ? array1[elementIndex] : array2[elementIndex]
    })
}
print(combinations) // [["l1", "m2", "r3"], ["l4", "m2", "r3"], ["l1", "m5", "r3"], ["l4", "m5", "r3"], ["l1", "m2", "r6"], ["l4", "m2", "r6"], ["l1", "m5", "r6"], ["l4", "m5", "r6"]]

Answer 2

如果两个数组计数相同，则在每次迭代中都从原始数组迭代两个数组并交换对象，并将其追加到结果中。

var combinations: [[String]] = []
if array1.count == array2.count {
    combinations.append(array1)
    combinations.append(array2)
    for i in 0..<array1.count {
        var arr1 = array1
        var arr2 = array2
        (arr1[i], arr2[i]) = (arr2[i], arr1[i])
        combinations.append(arr1)
        combinations.append(arr2)
    }
    print(combinations)
}

[[“ l1”，“ m2”，“ r3”]，[“ l4”，“ m5”，“ r6”]，[“ l4”，“ m2”，“ r3”]，[“ l1”，“ m5 “，” r6“]，
[“ l1”，“ m5”，“ r3”]，[“ l4”，“ m2”，“ r6”]，[“ l1”，“ m2”，“ r6”]，[“ l4”，“ m5” ，“ r3”]]