Swift语言基础笔记（三）

接上一篇：Swift语言基础笔记（二）

介绍了字符、字符串、元组、可选型，这篇介绍Swift语言的容器类Collections，数组、字典、集合，他们有各自的特点，数组是有序的；集合是无序的，且具有唯一性，提供集合操作，特殊的计算、快速查找；字典是以键值对的形式存在，我们在实际使用过程中要选择合适的数据结构；对一个数据结构的学习就是学习它的增、删、改、查。

数组

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

import UIKit
//声明一个数组
var numbers = [2, 3, 4, 5, 6]
//默认是String
var vowels = ["T", "E", "D", "K"]

var numbers1: [Int] = [2, 3, 4, 5, 6]
//默认是String
var vowels1: [String] = ["T", "E", "D", "K"]

var numbers2: Array<Int> = [2, 3, 4, 5, 6]
//默认是String
var vowels2: Array<String> = ["T", "E", "D", "K"]


//声明一个空的数组
var emptyArray1: [Int] = []
var emptyArray2: Array<Int> = []
var emptyArray3 = [Int]()
var emptyArray4 = Array<Int>()

//使用构造方法声明数组
var allZeros = [Int](count: 6, repeatedValue: 0)
var allZeros1 = Array<Int>(count: 6, repeatedValue: 0)
//不建议用这种方式声明数组
var array = []

//数组的基本用法 查
vowels.count

numbers.isEmpty

emptyArray1.isEmpty

vowels[3]

//得到的是一个可选型，因为数组有可能为空，得到的就是nil
vowels.first
vowels.last

emptyArray1.first

//自动解包
if let first = emptyArray1.first{
    print("emptyArray1 first")
}

if let first = vowels.first{
    print("vowels first = \(first)")
}
//强制解包
vowels.first!
//也可以通过下标得到最后一个元素
vowels[vowels.count-1]

//得到数组中的最小值
numbers.minElement()
vowels.maxElement()
//得到数组的一个范围值
numbers[1..<3]
numbers[2..<numbers.count]
//是否包含
vowels.contains("T")
vowels.contains("C")
//得到字符的索引值
vowels.indexOf("E")
vowels.indexOf("L")

//遍历数组
for index in 0..<numbers.count{
    print("numbers index \(index) is \(numbers[index])")
}
//推荐使用
for number in numbers{
    print("number is \(number)")
}
//需要使用元组下标可以使用以下方法
for (index, vowel) in vowels.enumerate(){
    print("numbers index \(index) is \(vowel)")
}

//判断数组是否相等
var nums = [2, 3, 4, 5, 6];
numbers == nums

//数组的更多操作 增 删 改
var courses = ["第一课", "第二课", "第三课"]
courses.count
//添加一个元素
courses.append("第四课")
courses.count
print(courses)
//另一个添加元素的方法,=号的另一边要是同类型才能相加
courses += ["第五课"]
courses.count
print(courses)
//插入指定位置
courses.insert("新添加", atIndex: 2)
print(courses)

//删除
courses.removeFirst()
print(courses)
courses.removeLast()
print(courses)
courses.removeAtIndex(2)
print(courses)
courses.removeRange(0..<2)
print(courses)
courses.removeAll();
print(courses)

courses.append("第一课")
courses.append("第二课")
courses.append("第三课")
courses.append("第四课")
courses.append("第五课")
print(courses)
//改
courses[0] = "lesson1"
print(courses)
//修改1〜3的下标，如果没有就删除对应下标
courses[1...3] = ["lesson2", "lesson3"]
print(courses)
courses[0..<3] = ["lensson"]
print(courses)

//二维数组
// 数组的定义以下三种方式都可以定义二维数组
var board = [[276, 73, 4, 0], [456, 45, 49, 0], [409, 45, 4, 0]]
var board1: [Array<Int>] = [[276, 73, 4, 0], [456, 45, 49, 0], [409, 45, 4, 0], [675, 73, 4, 0]]
var board2: Array<Array<Int>> = [[276, 73, 4, 0], [456, 45, 49, 0], [409, 45, 4, 0], [675, 73, 4, 0]]
board[0]
board[0][0]

board.count
board[0].count

//NSArray NSArray与String的区别一个是引用，一个是值
var array2 = [2, 3, 4, 5, 6] as NSArray
var array3: NSArray = [4, "Hello", 8]

字典的学习

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

import UIKit
//字典的介绍，字典是无序的数据结构
var dict = ["swift":"面向对象", "C语言":"面向过程", "java":"面向对象"]
if let value = dict["swift"]{
    print("swift is \(value)")
}
dict.count
dict.isEmpty
dict.first
//得到所有的key
Array(dict.keys)
//得到所有的value
Array(dict.values)
for key in dict.keys{
    print(key)
}

for value in dict.values{
    print(value)
}

for (index, value) in dict.enumerate(){
    print("index is \(index) value is \(value)")
}

for (key, value) in dict{
    print("key is \(key) value is \(value)")
}



//声明空的字典
var emptyDic1: [String: Int] = [:]
var emptyDic2: Dictionary<Int, String> = [:]
var emptyDic3 = [String: String]()
var emptyDic4 = Dictionary<Int, Int>()

let dict1 = [1:"A", 2:"B", 3:"C"]
let dict2 = [1:"A", 2:"B", 3:"C", 4:"D"]
dict1 == dict2

//字典的增，删，改
var user = ["name":"dzt", "pwd":"111", "age":"30"]
//修改
user["pwd"] = "000"
print(user)
user.updateValue("49", forKey: "age")
print(user)
let oldPwd = user.updateValue("000", forKey: "pwd")
//判断修改前与修改后的密码是否一致
if let oldPwd = oldPwd, newPwd = user["pwd"] where oldPwd == newPwd{
    print("注意，修改后的密码与修改前密码一样")
}

//增，
user["msg"] = "hello"
print(user)
user.updateValue("insert", forKey: "title")
print(user)

//删除
user["title"] = nil
print(user)
if let msg = user.removeValueForKey("msg"){
   print("msg is \(msg) delete success")
}
user.removeAll()
print(user)

集合的学习

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

import UIKit
//集合的使用，集合是无序的，可以保证数据是唯一的
var lang: Set<String> = ["java", "swift"]
//声明一个空集合
var emptySet1: Set<Int> = []
var emptySet2 = Set<Int>()

//可以有多种声明方法
var vowels = Set(["A", "E", "I", "I"])

var langs: Set = ["Swift", "java", "C#"];
lang.count
lang.isEmpty
emptySet1.isEmpty

//随机取出的数
let e = lang.first

lang.contains("swift")
for l in lang{
    print(l)
}

lang == langs

//集合的增删改
var skillsA: Set<String> = ["java", "swift"]
var skillsB: Set<String> = ["javascript", "oc", "html"]
var skillsC: Set<String> = []

//增
skillsC.insert("oc")
skillsC.insert("css")
skillsC.insert("swift")
skillsC.insert("oc")
print(skillsC)

//删除
skillsC.remove("oc")
print(skillsC)
if let skill = skillsC.remove("css"){
    print("\(skill) is delete")
}

skillsC.removeAll()

//计算并集
skillsA.union(skillsB)
print(skillsA)
//skillsA.unionInPlace(skillsB)
//print(skillsA)

//交集
skillsA.intersect(skillsB)

//集合减法
skillsA.subtract(skillsB)

//集合异或
skillsA.exclusiveOr(skillsB)

skillsA.union(["java", "c#"])
print(skillsA)
//判断字集
var skillsD: Set = ["swift"]
skillsD.isSubsetOf(skillsA)
skillsD.isStrictSubsetOf(skillsA)
//判断超集
skillsA.isSupersetOf(skillsD)
skillsA.isStrictSupersetOf(skillsD)

//判断是否相离，没有公共部分
skillsA.isDisjointWith(skillsB)
skillsA.isDisjointWith(skillsC)

以上代码都是在Xcode的Playground上运行的。