如何在Swift中將兩個數組合並為一個？

Question

我有兩個數組：

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

我如何將它們組合成一個看起來像的數組

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"]
]

編輯1 ：如果您有兩個以上的數組

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

如果要獲得兩個數組的可能組合，可以使用一個簡單的技巧：

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

如果兩個數組計數相同，則在每次迭代中都從原始數組迭代兩個數組並交換對象，並將其追加到結果中。

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”]]