比较两个数组在 swift 中具有任何一个相同的元素

Question

需要比较两个数组，数组 1 = [1,2,3,4]数组 2 = [1,5,6,7,8]结果应为布尔值。需要检查数组 1 是否包含数组 2 中的任何类似元素。所以有人帮我解决了上述问题。

Answer 1

也许这种方式将满足您的需求：

let array1 = [1,2,3,4], array2 = [1,5,6,7,8]
let set1 = Set<Int>(array1), set2 = Set<Int>(array2)
let containsSimilar = set1.intersection(set2).count > 0

根据评论进行编辑

let array1: [Any] = [1,2,"a"], array2: [Any] = ["a","b","c"]
let set1 = Set<JananniObject>(array1.flatMap({ JananniObject(any: $0) }))
let set2 = Set<JananniObject>(array2.flatMap({ JananniObject(any: $0) }))
let containsSimilar = set1.intersection(set2).count > 0

JananniObject在哪里：

enum JananniObject: Equatable, Hashable {
    case integer(Int)
    case string(String)
    static func ==(lhs: JananniObject, rhs: JananniObject) -> Bool {
        switch (lhs, rhs) {
        case (.integer(let integer0), .integer(let integer1)):
            return integer0 == integer1
        case (.string(let string0), .string(let string1)):
            return string0 == string1
        default:
            return false
        }
    }
    var hashValue: Int {
        switch self {
        case .integer(let integer):
            return integer.hashValue
        case .string(let string):
            return string.hashValue
        }
    }
    init?(any: Any) {
        if let integer = any as? Int {
            self = .integer(integer)
        } else if let string = any as? String {
            self = .string(string)
        } else {
            return nil
        }
    }
}

Answer 2

您可以使用

Set执行此操作：

let one = [1, 2, 3]
let two = [2, 1, 3]
let three = [2, 3, 4]
print(Set(one).isSubset(of: two)) // true
print(Set(one).isSubset(of: three)) // false