我的解决方案，它似乎可以在O(n)时间，因为哈希映射访问是O(1)，过滤器是O(n)。它还使用by闭包来选择按顺序区分元素的属性。

extension Sequence {

    func distinct<T: Hashable>(by: (Element) -> T) -> [Element] {
        var seen: [T: Bool] = [:]
        return self.filter { seen.updateValue(true, forKey: by($0)) == nil }
    }
}

这里有一个解决方案

不使用传统NS类型 对于O(n)来说是相当快的 简洁 保持元素顺序

extension Array where Element: Hashable {

    var uniqueValues: [Element] {
        var allowed = Set(self)
        return compactMap { allowed.remove($0) }
    }
}

让我提出一个类似于斯科特·加德纳的答案，但使用了更简洁的reduce语法。 此解决方案从自定义对象数组中删除重复项(保持初始顺序)

// Custom Struct. Can be also class. 
// Need to be `equitable` in order to use `contains` method below
struct CustomStruct : Equatable {
      let name: String
      let lastName : String
    }

// conform to Equatable protocol. feel free to change the logic of "equality"
func ==(lhs: CustomStruct, rhs: CustomStruct) -> Bool {
  return (lhs.name == rhs.name && lhs.lastName == rhs.lastName)
}

let categories = [CustomStruct(name: "name1", lastName: "lastName1"),
                  CustomStruct(name: "name2", lastName: "lastName1"),
                  CustomStruct(name: "name1", lastName: "lastName1")]
print(categories.count) // prints 3

// remove duplicates (and keep initial order of elements)
let uniq1 : [CustomStruct] = categories.reduce([]) { $0.contains($1) ? $0 : $0 + [$1] }
print(uniq1.count) // prints 2 - third element has removed

如果你想知道这个约简魔法是如何工作的，这里是完全相同的，只是使用了更扩展的约简语法

let uniq2 : [CustomStruct] = categories.reduce([]) { (result, category) in
  var newResult = result
  if (newResult.contains(category)) {}
  else {
    newResult.append(category)
  }
  return newResult
}
uniq2.count // prints 2 - third element has removed

你可以简单地复制粘贴这段代码到Swift Playground中。

您可以直接使用set集合删除重复，然后将其转换回数组

var myArray = [1, 4, 2, 2, 6, 24, 15, 2, 60, 15, 6]
var mySet = Set<Int>(myArray)

myArray = Array(mySet) // [2, 4, 60, 6, 15, 24, 1]

然后你可以按你想要的顺序排列你的数组

myArray.sort{$0 < $1} // [1, 2, 4, 6, 15, 24, 60]

我相信提供一个uniq()和uniqInPlace()函数通过删除数组的值来改变数组是很好的。这类似于Swift提供的sort()和sortInPlace()函数。此外，由于它是一个数组，它应该保持它的原始元素顺序。

extension Array where Element: Equatable {

    public func uniq() -> [Element] {
        var arrayCopy = self
        arrayCopy.uniqInPlace()
        return arrayCopy
    }

    mutating public func uniqInPlace() {
        var seen = [Element]()
        var index = 0
        for element in self {
            if seen.contains(element) {
                removeAtIndex(index)
            } else {
                seen.append(element)
                index++
            }
        }
    }
}

你只能在变量数组(即var)上使用uniqInPlace()，因为你不能改变常量数组(即let)。

一些用法示例:

var numbers = [1, 6, 2, 2, 4, 1, 5]
numbers.uniqInPlace() // array is now [1, 6, 2, 4, 5]

let strings = ["Y", "Z", "A", "Y", "B", "Y", "Z"]
let uniqStrings = strings.uniq() // uniqStrings is now ["Y", "Z", "A", "B"]

Xcode 10.1 - Swift 4.2简单而强大的解决方案

func removeDuplicates(_ nums: inout [Int]) -> Int {
    nums = Set(nums).sorted()
    return nums.count
}

例子

var arr = [1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9]
removeDuplicates(&arr)

print(arr) // [1,2,3,4,5,6,7,8,9]