字典
字典是一种存储多个相同类型的值的容器。每个值（value）都关联唯一的键（key），键作为字典中的这个值数据的标识符。和数组中的数据项不同，字典中的数据项并没有具体顺序。
1.初始化
var games: [String:String] = [“Diablo3”:”2014:8:12”,
“Dragon Age”:”2014:10:07”]
var games = [“Diablo3”:”2014:8:12”,“DragonAge”:”2014:10:07”]
games[“LittleBigPlanet3”] = “2014:11:29"
games[“LittleBigPlanet3”] = “2014:11:30"
var nameOfIntegers = [Int:String]()
namesOfIntegers[16] = “sixteen"
namesOfIntegers = [:]
2.修改已有键值：
if let oldValue = games.updateValue(“2014:8:14”,forKey:”Diablo3”){
println(“Diablo3的旧值：\(oldValue)”)
}
3.获取键值为可选类型：
if let releaseDate = games[“Diablo3”]{
println(“该游戏的发布日期是\(releaseDate)”)
}else {
println(“该游戏的发布日期不在games字典里”)
}
games[“LittleBigPlanet3”] = nil 移除键值
games.removeValueForKey(“Diablo3”) 和updateValue一样
4.字典遍历
let airports = [“TYO”:”Tokyo”,”LHR”:”London”]
for (airportCode,airportName) in airports{
println(“\(airportCode): \(airportName)”)
}
for airportCode in airports.keys{}
for airportNameinairports.values{}
5.示例代码：
var airports:Dictionary<String,String> = ["TKO":"Tokyo","CHA":"China"]
println("the airports Dictionary has \(airports.count) airport")
airports["LON"] = "London"
airports["LON"] = "London weather"
airports["CHA"] = nil
if let oldValue = airports.updateValue("Dublin",forKey: "CHA"){
println("the old value is \(oldValue)")
}else{
println("there is no airport named CHA")
}
for (airportNumber,airportName) in airports{
println("airportNumber:\(airportNumber) airportName:"+airportName)
}
for key in airports.keys{
}
for value in airports.values{
}
let airportCode = Array(airports.keys)
var nameOfIntergers = Dictionary<String,Int>()
nameOfIntergers = [:]
