集合(Set)

• 定義：Set<元素類型>，無法使用類型推斷，可省略類型

let num : Set = [1, 2, 3, 1, 4]      //{1,2,3,4}

• 用數(shù)組字面量創(chuàng)建集合

let citys : Set = ["Beijing", "Shanghai", "Guangzhou", "Shenzhen"]

• ①元素計數(shù)：count，是否為空：isEmpty

citys.count        //4
citys.isEmpty      //false

• ②插入：insert

citys.insert("Changsha")

• ③移除：remove

citys.remove("Changsha")

• ④是否包含某元素：contains

citys.contains("Beijing")

• ⑤轉(zhuǎn)換成數(shù)組：sorted

let cityArray = citys.sorted()      //["Beijing", "Guangzhou", "Shanghai", "Shenzhen"]

集合間的運算：交差并補

• 1??交集 intersection

var x :Set = [1, 2, 3, 4]
let y :Set = [3, 4, 5, 6]
x.intersection(y)      // {3, 4}

• 2??差集 subtract

x.subtract(y)      //結(jié)果賦值給x：{1, 2}

• 3??并集 union

x.union(y)      //結(jié)果賦值給x：{1,2,3,4,5,6}

• 4??補集 symmetricDifference

x.symmetricDifference(y)      //結(jié)果賦值給x：{1,2,5,6}

• 子集：isSubset(可以相等)，嚴格子集isStrictSubset

let a :Set = [1, 2, 3]
let b :Set = [3, 2, 1]
let c :Set = [1, 2, 3, 4]
a.isSubset(of: b)              // true
a.isStrictSubset(of: b)        // false
a.isStrictSubset(of: c)        // true

• 父集：isSupersetOf(可以相等)，嚴格父集isStrictSuperSetOf
• 無交集：isDisjoint

let j : Set = ["游戲", "動漫"]
let k: Set = ["吃", "睡"]
j.isDisjoint(with: k)      //true