Android面试题之Kotlin常见集合操作技巧

161 阅读1分钟

list

创建和修改

不可变list,listOf

var list = listOf("a","d","f")
println(list.getOrElse(3){"Unknown"})
println(list.getOrNull(3) ?: "Unknown")

创建和修改可变list,mutableList

val mutableList = mutableListOf("a","d","f")
mutableList.removeIf { it.contains("a") }

遍历和解构

forIn,forEach,forEachIndexed

用下划线_来跳过不需要的元素

for (s in list){
    println(s)
}

list.forEach{
    println(it)
}

list.forEachIndexed { index, item ->
    println("$index, $item")
}
//解构
val(a,_,c) = list 

"_"表示跳过这个元素,跳过的元素不会被读取

转换

list.toMutableList()

去重

list.distinct()

快速筛选和排序

可以结合filter、forEach、sortedByDescending、sortedBy来快速对一个list进行筛选和排序

val resultList = mutableListOf<AppInfo>()
val recommendations = listOf(AppA(), AppB(), AppC())
//根据app的安装时间按照倒序排列
recommendations.filter { 
    //filter里跟筛选的条件,符合条件的数据会继续走后面的排序
    it.isAppInstalled() 
}  
.sortedByDescending { 
     //这里是排序的条件,sortedBy是正序
    it.getAppInstalledTime() 
}.forEach { resultList.add(it) }


set

不可重复的,去重里面的元素,函数操作基本同list

map

fun main() {
    val map = mapOf("Jack" to 20, "Jim" to 18)
    mapOf(Pair("Jim",20), Pair("json",30))
    println(map["Jack"])
    println(map.getValue("Jack"))
    println(map.getOrElse("java"){"unknown"})
    println(map.getOrDefault("java", 0))

    map.forEach{
        println("${it.key}, ${it.value}")
    }

    val mutableMap = mutableMapOf("Jack" to 20, "Jim" to 18)
    //添加元素
    mutableMap += "jim" to 30
    mutableMap["jim"] = 31
    mutableMap.getOrPut("role"){19}
    println(mutableMap)
}

to,实际上是一个函数,返回一个Pair结果

去重

val clone = filteredDirectories.clone() as ArrayList<Directory>
        //数据去重。
callback(clone.distinctBy { it.path.getDistinctPath() } as ArrayList<Directory>)

👀关注公众号:Android老皮!!!欢迎大家来找我探讨交流👀