Kotlin中的List

Kotlin中的List

在Kotlin中有两种类型的列表:

List：只读列表,列表创建后不能被修改。

MutableList: 可以在创建后被修改，这意味着你可以添加、删除或更新它的元素。

Sample

fun main() {
    val numbers: List<Int> = listOf(1, 2, 3, 4, 5, 6)
    val numbers2 = listOf(1, 2, 3, 4, 5, 6)
    println("List:$numbers")
    println("Size:${numbers.size}")
    println("First:${numbers.first()}")
    println("First:${numbers.last()}")
    println("Contains 4? ${numbers.contains(4)}")
    println("Contains 7? ${numbers.contains(7)}")
}

Run and Out

List:[1, 2, 3, 4, 5, 6]
Size:6
First:1
First:6
Contains 4? true
Contains 7? false

List只读列表

在只读列表中添加或更改元素是不被允许的

创建List

val numbers: List<Int> = listOf(1, 2, 3, 4, 5, 6)
numbers.add(7) //不被允许
numbers[0] = 7 //不被允许

常见方法

reversed

返回一个新列表，其中的元素按相反的顺序排列

sorted

返回一个新列表，其中的元素按升序排序

Sample

val colors = listOf("green", "orange", "blue")

println("Reversed list: ${colors.reversed()}")
//按字母顺序
println("Sorted list: ${colors.sorted()}")
println("List: $colors")
val oddNumbers  = listOf(5, 3, 7,1)
println("List: $oddNumbers")
println("Sorted list: ${oddNumbers.sorted()}")

Run and Out

Reversed list: [blue, orange, green]
Sorted list: [blue, green, orange]
List: [green, orange, blue]
List: [5, 3, 7, 1]
Sorted list: [1, 3, 5, 7]

Mutable List

可变列表是创建后可以修改的列表。您可以添加、删除或更改

创建Mutable List

val  entrees = mutableListOf()

这行code导致错误,listOf 也有这个问题

error info：Not enough information to infer type variable T

原因是：当创建一个可变列表或列表时，Kotlin尝试从传递的参数推断列表包含的元素类型

例如，如果您编写listOf(“noodles”)，Kotlin推断您想要创建一个字符串列表

当初始化一个没有元素的空列表时，Kotlin无法推断元素的类型，所以您必须显式地声明类型。为此，可以在mutabllistof或listOf后的尖括号中添加类型。

正确代码：

2种方式都可以

 val entrees = mutableListOf<String>()
 val entrees:MutableList<String> = mutableListOf()

Sample

        val entrees = mutableListOf<String>()
//        val entrees:MutableList = mutableListOf()
        println("Entrees:$entrees")
        println("Add noodles:${entrees.add("noodles")}")
        println("Entrees:$entrees")
        val more = listOf("ravioli", "lasagna", "fettuccine")
        println("Add list: ${entrees.addAll(more)}")
        println("Entrees: $entrees")
        println("Remove spaghetti: ${entrees.remove("ravioli")}")
        println("Entrees: $entrees")
        println("Remove first element: ${entrees.removeAt(0)}")
        println("Entrees: $entrees")
        entrees.clear()
        println("Entrees: $entrees")
        println("Empty? ${entrees.isEmpty()}")
        //IndexOutOfBoundsException: Index: 0, Size: 0
        println("Empty? ${entrees.removeAt(0)}")

Run and Out

Entrees:[]
Add noodles:true
Entrees:[noodles]
Add list: true
Entrees: [noodles, ravioli, lasagna, fettuccine]
Remove spaghetti: true
Entrees: [noodles, lasagna, fettuccine]
Remove first element: noodles
Entrees: [lasagna, fettuccine]
Entrees: []
Empty? true

java.lang.IndexOutOfBoundsException: Index: 0, Size: 0

循环

Sample

val names = listOf("Jessica", "Henry", "Alicia", "Jose")
        var index = 0

        while (index < names.size) {
            print(names[index])
            index++
        }
        println()
        for (name in names) {
            print(name)
        }
        println()
        for (item in 'b'..'g') {
            print(item)
        }
        println()
        for (item in 1..5) {
            print( "$item")
        }
        println()
        for (item in 5 downTo  1) {
            print(item)
        }
        println()
        for (item in 3..6 step   3) {
            print(item)
        }

输出：

JessicaHenryAliciaJose
JessicaHenryAliciaJose
bcdefg
12345
54321
36

Use Case

看下面几个类：

假如创建一个Vegetables class

val vegetables = Vegetables("Cabbage", "Sprouts", "Onion")

写法1：

class Vegetables(val topping1: String,
                 val topping2: String, 
                 val topping3: String){}

写法2：

class Vegetables(val toppings: List<String>) {}

不过要传递List类型的参数

Vegetables(listOf("cabbage", "sprouts", "onion"))

写法3：

优雅的

class Vegetables(vararg val toppings: String) {
}

在Kotlin中，vararg修饰符允许您将数目可变的相同类型的参数传递给函数或构造函数。通过这种方式，您可以将不同的蔬菜作为单独的字符串而不是列表提供

Sample

class Vegetables(vararg val toppings:String){

    override fun toString(): String {
        return toppings.joinToString()
    }
}
val  vegetables = Vegetables("cabbage","onion");
println(vegetables.toString())

Run and Out

cabbage, onion

joinToString用[separator]分隔的所有元素创建一个字符串,

注意:要指定除逗号以外的其他分隔符，请将所需的分隔符字符串作为参数传递给joinToString()方法。示例:joinToString(" ")以空格分隔每个元素，具体看文档

Class 和List 示例

open class Item(val name: String, val price: Int) {
}

class Noodles() : Item("Noodles", 10) {
    override fun toString(): String {
        return name
    }
}

class Vegetables(vararg val toppings: String) : Item("Vegetables", 5) {
    override fun toString(): String {
        return if (toppings.isEmpty()) {
            "$name Chef's Choice"
        } else {
            name + " " + toppings.joinToString()
        }
    }
}

class Order(val orderNumber: Int){
    private val itemList = mutableListOf<Item>()

   fun addItem(newItem:Item):Order{
       itemList.add(newItem)
       return this
   }
    fun addAll(newItems: List<Item>):Order{
        itemList.addAll(newItems)
        return this
    }
    fun print(){
        println("Order #${orderNumber}")
        var total = 0
        for (item in itemList){
            println("$item:$${item.price}")
            total += item.price
        }
        println("Total:$$total")
    }
}

Run

val ordersList = mutableListOf<Order>()
val order1 = Order(1)
order1.addItem(Noodles())
ordersList.add(order1)

val order2 = Order(2)
order2.addItem(Noodles())
order2.addItem(Vegetables())
ordersList.add(order2)

val order3 = Order(3)
val items = listOf(Noodles(), Vegetables("Carrots","Beans","Celery"))
order3.addAll(items)
ordersList.add(order3)

ordersList.add(Order(4)
    .addItem(Noodles())
    .addItem(Vegetables("Spinach")))

 for (order in ordersList){
     order.print()
     println()
 }

Out

Order #1
Noodles:$10
Total:$10

Order #2
Noodles:$10
Vegetables Chef’s Choice:$5
Total:$15

Order #3
Noodles:$10
Vegetables Carrots, Beans, Celery:$5
Total:$15

Order #4
Noodles:$10
Vegetables Spinach:$5
Total:$15