Kotlin 从列表中获取特定过滤器的唯一列表值

我有一段代码:

/**
 * You can edit, run, and share this code.
 * play.kotlinlang.org
 */

class Person(val firstName: String, val lastName: String, var age: Int, var isEmployed: Boolean, var companyName: String)
{
    override fun toString(): String = firstName + " " + lastName + ", age:" + 
    age + ", isEmployed: " + isEmployed + ", companyName:" + companyName + "\n"
}


fun main() {
    
    var alist = emptyList<Person>()
    alist += (Person("Alan", "Walker", 23, true, "Google"))
    alist += (Person("Bee", "dog", 24, true, "Google"))
    alist += (Person("John", "Cena", 25, true, "Google"))
    alist += (Person("See", "S", 26, true, "Stackoverflow"))
    alist += (Person("Soya", "A", 27, true, "Stackoverflow"))
    alist += (Person("Zander", "Cage", 28, true, "Stackoverflow"))
    
    
    println(alist)
    
    var uniqueGoogle = alist.filter{ it.companyName.equals("Google") }.first()
    var newList = alist.filter{ !it.companyName.equals("Google") }
    newList = uniqueGoogle + newList
    
    println(newList)
}

我需要从上面的名单上的人的名单是唯一的公司名称只为谷歌.我真的不擅长使用列表聚合，这里需要一些帮助.

println(newList)的预期输出:

[Alan Walker, age:23, isEmployed: true, companyName:Google
, See S, age:23, isEmployed: true, companyName:Stackoverflow
, Soya A, age:23, isEmployed: true, companyName:Stackoverflow
, Zander Cage, age:23, isEmployed: true, companyName:Stackoverflow
]

因为从谷歌我只想要第一个元素.我的解决方案做到了这一点，但它对我来说似乎不是很有效或很好，它给了我一些错误，不知道为什么……