通过从另一个结构订购来迅速排序结构

Question

我有2个结构(游戏，otgame(，其中包含一个名为[COD]的NSString属性。我尝试将第二个结构与[COD]相同的第一个结构相同，然后将剩余添加到队列中。

game = [cod("001234"),cod("001111"),cod("002222"),cod("005555")]
otGame = [cod("002222"),cod("005555"),cod("001111")]
struct game {
    var cod: NSString
    var des: Float?
}
struct otGame {
    var cod: NSString
    var sor: Float?
}

我希望otgame的输出为

[cod("001111"),cod("002222"),cod("005555"),cod("001234")]

Answer 1

我认为您应该做这样的事情：

struct Game {
    var cod: String
    var des: Float?
}
struct OtGame {
    var cod: String
    var sor: Float?
}
let games = [Game(cod: "001234", des: nil),
             Game(cod: "001111", des: nil),
             Game(cod:"002222", des: nil),
             Game(cod:"005555", des: nil)]
var otGames = [OtGame(cod: "002222", sor: nil),
               OtGame(cod: "005555", sor: nil),
               OtGame(cod: "001111", sor: nil)]
//Sort otGames based on indexes of games
otGames.sort { (o1, o2) -> Bool in
    if let index1 = games.firstIndex(where: {$0.cod == o1.cod}), let index2 = games.firstIndex(where: {$0.cod == o2.cod}) {
        return index1 < index2
    }
    return false
}
// Add missing games
for game in games {
    if !otGames.contains(where: {$0.cod == game.cod}) {
        otGames.append(OtGame(cod: game.cod, sor: game.des))
    }
}

请注意，这种性能非常不好，我建议您找到不同的数据结构或解决方案以防止此计算。