Kotlin list，set，map集合练习代码

val list = listOf(1 ,2 ,3, 4, 5) //创建不可变list val mutableList = mutableListOf("s","b") //创建可变MutbaleList val set = setOf(1, 2, 3, 4, 5)//创建不可变Set val mutableSet = mutableSetOf("haha","wotainanle")//创建可变MutableSet val map = mapOf(1 to "a",2 to "b", 3 to "c")//创建不可变Map val mutableMap = mutableMapOf(1 to "l", 2 to "bw", 3 to "nb")//创建可变mutalemap //空集合 val emptyList: List<Int> = listOf() val emptySet: Set<Int> = setOf() val emptyMap: Map<Int, String> = mapOf() fun main(args: Array<String>) { //遍历list集合 list.forEach{ println(it) } //遍历set集合 set.forEach{ println(it) } //遍历mutablemap集合 mutableMap.forEach{ println("K=${it.key},V = ${it.value}") } //遍历list的索引+值 list.forEachIndexed { index, value -> println("list index = ${index} , value =${value}") } //遍历set的索引和值 set.forEachIndexed { index, value -> println("set index = ${index} , value = ${value}") } //遍历map集合 map.entries.forEach({ println("key = " + it.key + " value = " + it.value)}) }

输出

1 2 3 4 5 1 2 3 4 5 K=1,V = l K=2,V = bw K=3,V = nb list index = 0 , value =1 list index = 1 , value =2 list index = 2 , value =3 list index = 3 , value =4 list index = 4 , value =5 set index = 0 , value = 1 set index = 1 , value = 2 set index = 2 , value = 3 set index = 3 , value = 4 set index = 4 , value = 5 key = 1 value = a key = 2 value = b key = 3 value = c

映射函数，过滤函数，排序函数在集合中运用

val list3 = listOf(5 ,4 ,3 ,2 ,1 ,1) val list2 = listOf(1, 2, 3, 4, 5) val set2 = setOf(1, 2, 3, 4, 5) val map2 = mapOf(1 to "a", 2 to "b", 3 to "c") //创建数据类Students data class Students(var id:Int,var name: String,var age:Int,var score:Int){ override fun toString(): String { return "Students(id=$id ,name='$name',age=$age,score=$score)" } } //添加students对象进集合 var studentList = listOf( Students(1,"lbw",18,55), Students(2,"lbf",88,55), Students(3,"lbfnb",28,55) ) fun main(args: Array<String>) { //映射函数运用 println(list2.map { it*it }) println(set2.map { it + 1 }) println(map2.map { it.key + 1 }) //过滤函数运用 println(studentList.filter { it.age > 18 }) println(studentList.filter { it.id == 3 }) println(list2.filterIndexed { index,it -> index > 2 && it > 4}) //排序函数运用 println(list2.reversed()) println(set2.reversed()) println(list3.sorted()) println(list3.distinct()) }

输出结果：

[1, 4, 9, 16, 25] [2, 3, 4, 5, 6] [2, 3, 4] [Students(id=2 ,name='lbf',age=88,score=55), Students(id=3 ,name='lbfnb',age=28,score=55)] [Students(id=3 ,name='lbfnb',age=28,score=55)] [5] [5, 4, 3, 2, 1] [5, 4, 3, 2, 1] [1, 1, 2, 3, 4, 5] [5, 4, 3, 2, 1]