如果两个数组包含相同的元素,无论这些元素出现的顺序如何,我如何检入Swift？

Question

假设有两个阵列......

var array1 = ["a", "b", "c"]
var array2 = ["b", "c", "a"]

我希望这两个数组的比较结果是真的,以下是......

var array1 = ["a", "b", "c"]
var array2 = ["b", "c", "a", "d"]

......是假的 我怎样才能在Swift中实现这一目标？我试图将两个数组转换为集合,但由于某种原因,Set()不断删除数组包含的一些(通常是重复的)对象.

任何帮助,将不胜感激.

Answer 1

斯威夫特3,4

extension Array where Element: Comparable {
    func containsSameElements(as other: [Element]) -> Bool {
        return self.count == other.count && self.sorted() == other.sorted()
    }
}

// usage
let a: [Int] = [1, 2, 3, 3, 3]
let b: [Int] = [1, 3, 3, 3, 2]
let c: [Int] = [1, 2, 2, 3, 3, 3]

print(a.containsSameElements(as: b)) // true
print(a.containsSameElements(as: c)) // false

Answer 2

斯威夫特 5.2 解决方案

var array1 = ["a", "b", "c"]
var array2 = ["b", "c", "a"]

if array1.sorted() == array2.sorted() {
    print("array 1 & array 2 are same")
}

Answer 3

你可以这样做:

  array1.sortInPlace()
  array2.sortInPlace()

  print(array1,array2)

  if array1 == array2 {
    print("equal")
  } else {
  print("not equal") 
  }

如果不想改变原始阵列,我们可以做到

 let sorted1 = array1.sort()
 let sorted2 = array2.sort()

  if sorted1 == sorted2 {
    print("equal")
  }else {
    print("not equal")
  }

Answer 4

使用 Set

let array1 = ["a", "b", "c"]
let array2 = ["b", "c", "a", "c"]

let set1 = Set(array1)
let set2 = Set(array2)

if (set1.count == set2.count && set1 == set2) { //if you compare big sets it is recommended to compare the count of items in the sets beforehand
    //they are identical
}

该Set工具Hashable做任务是贯彻落实哈希函数与工作Set