//: Playground - noun: a place where people can play

import UIKit

// # 集合的可变性

// # 数组

// 数组内部储存类类型,相当于存储了一堆指针
class Human {
    var name: String
    init(name: String) {
        self.name = name
    }
}
let personOne = Human(name: "Tom")
var twoPeople = [personOne, personOne]
twoPeople[0].name = "Pussy"
twoPeople[1].name
// 数组整体被复制给其他数组,新数组还是和老数组共享数据
var theTwo = twoPeople
theTwo[0].name = "Jerry"
twoPeople[0].name

// trick: 如果内容已经提供了类型信息，比如说作为函数的实际参数或者已经分类了的变量或常量，可以通过空数组字面量来创建一个空数组
func a(b: [Int]) {
    print(b.count)
}
a(b: [])

// 使用默认值创建同值的数组,用+来连接两个数组,类型推断出数组的类型
var threeNames = Array(repeating: "China", count: 3)
var threeOtherNames = Array(repeating: "Russia", count: 3)
var sixNames = threeNames + threeOtherNames
sixNames[sixNames.startIndex]
sixNames[0]
// trick: 可以使用下标脚本语法来一次改变一个范围的值，就算替换与范围长度不同的值的合集也行。
sixNames[0...2] = ["Russia"]
sixNames.count
// trick: 移除最后一个元素
sixNames.removeLast()
// trick: 获取元素同时获取整数索引
for (index, value) in sixNames.enumerated() {
    print("Name \(index + 1), \(value)")
}
// 顺便复习一下元组
for thing in sixNames.enumerated() {
    print("Name \(thing.0 + 1), \(thing.1)")
}

// # 合集
// 同一类型不重复无序.必须遵循Hashable协议.
var i = 0
i.hashValue
"".hashValue
"0".hashValue
// 合集没有简写形式
var letters = Set<Character>()
letters.insert("a") // 有返回值
letters = []
// 合集必须显式地声明Set类型
var favoriteGenres: Set<String> = ["Rock", "Classical", "Hip hop"]
// 可以写成var favoriteGenres: Set = ["Rock", "Classical", "Hip hop"]
favoriteGenres.remove("H") // 这是Array和String所没有的,返回的是一个可选
favoriteGenres.remove(at: favoriteGenres.startIndex)
// for-in遍历无序,若要以特定顺序(用小于符号整理)遍历:
for genre in favoriteGenres.sorted() {
    print("\(genre)")
}

// # 执行合集操作
let oddDigits: Set = [1, 3, 5, 7, 9]
let evenDigits: Set = [0, 2, 4, 6, 8]
let singleDigitPrimeNumbers: Set = [2, 3, 5, 7]
// 并集
oddDigits.union(evenDigits).sorted()
// 交集
oddDigits.intersection(evenDigits).sorted()
// 减
oddDigits.subtracting(singleDigitPrimeNumbers).sorted()
// 并集减交集
oddDigits.symmetricDifference(singleDigitPrimeNumbers).sorted()
let houseAnimals: Set = ["🐶", "🐱"]
let farmAnimals: Set = ["🐮", "🐔", "🐑", "🐶", "🐱"]
let cityAnimals: Set = ["🐦", "🐭"]
houseAnimals.isSubset(of: farmAnimals)
houseAnimals.isStrictSubset(of: farmAnimals)
farmAnimals.isSuperset(of: houseAnimals)
farmAnimals.isStrictSuperset(of: houseAnimals)
farmAnimals.isDisjoint(with: cityAnimals) // 无交集

// # 字典
// 无序,键类型相同,值类型相同,键与值相互关联.Key类型必须遵循Hashable协议
var namesOfIntegers = [Int: String]()
namesOfIntegers[16] = "sixteen"
namesOfIntegers = [:] // 重新设为空值
// .count, isEmpty都可以使用,遵循collection protocol
// updateValue(_:forKey:)有则改之，无则加之,返回一个字典值类型的可选项值,是原来的值
var airports = [String: String]()
airports.updateValue("Dubin Airport", forKey: "DUB")
// 下标语法，也满足有则改之，无则加之,被赋值是返回一个值的可选项类型
airports["LRT"] = "London Heathrow"
var a = airports["LRT"]
// 移除一个键值对
airports["APL"] = "Apple International"
airports.count
airports["APL"] = nil
airports.count // 可见其count的是非nil的键值对
let removedValue = airports.removeValue(forKey: "DUB") // optional
// 遍历字典时,返回元组
for (airportCode, airportName) in airports {
    print("\(airportCode): \(airportName)")
}
for airportCode in airports.keys {
    print("Airport code: \(airportCode)")
}
for airportName in airports.values {
    print("Airport name: \(airportName)")
}
// trick: 用[String]的构造器来构造[String],毕竟,xxx.keys不是[String]
let airportCodes = [String](airports.keys)
let airportNames = [String](airports.values)