1.集合的可变与不可变

kotlin存在三种常用的集合
1.List是一个有序的，可以通过索引访问元素的集合，可以存在重复元素；
2.Set是唯一元素的集合，一般来说无固定顺序；
3.Map是一种键值对集合，键值是唯一的，值可以重复。

1.1不可变集合

fun main() {
    val stringList = listOf("one","two","one")
    // stringList.add("four") 错误，不可变集合不能添加元素了
    println(stringList)

    val stringSet = setOf<String>("one","two","one")
    //两个one，只输出一个
    println(stringSet)

    val stringMap = mapOf<Int,String>(Pair(1,"w"),Pair(2,"q"),Pair(1,"r"))
     println(stringMap)
}

输出如下：
[one, two, one]
[one, two]
{1=r, 2=q}
[2, 3, 4, 3]

1.2可变集合

    val stringList = mutableListOf("one","two","one")
    println(stringList)

    val stringSet = mutableSetOf<String>("one","two","one")
    println(stringSet)

    val stringMap = mutableMapOf<Int,String>(Pair(1,"w"),Pair(2,"q"),Pair(1,"r"))
     println(stringMap)
   //输出如1.1中的一样

可见每个集合都有对应的可变与不可集合

1.3集合排序

kotlin封装了一些基本的集合排序方法，如下

    fun sort() {
        val numberList = mutableListOf(1, 2, 34, 5)
        //随机打乱数据顺序
        numberList.shuffle()
        println(numberList)
        //从小到大排序
        numberList.sort()
        println(numberList)
        //从大到小排序
        numberList.sortDescending()
        println(numberList)
    }

输出结果如下：
[2, 5, 34, 1]
[1, 2, 5, 34]
[34, 5, 2, 1]

    fun stuSort(){
        val stuList : MutableList<Stu> = mutableListOf()
        stuList.add(Stu("xiaoLi",10))
        stuList.add(Stu("xiaoHua",21))
        stuList.add(Stu("xiaoMen",11))
        stuList.add(Stu("xiaoK",13))
        //按照age升序排列
        stuList.sortBy { it.age }
        println(stuList)
        //优先name排序，再按照age升序排列
        stuList.sortWith(compareBy({it.name},{it.age}))
        println(stuList)
        
    }
输出结果如下
[xiaoLi-10, xiaoMen-11, xiaoK-13, xiaoHua-21]
[xiaoHua-21, xiaoK-13, xiaoLi-10, xiaoMen-11]

1.4集合中的Set与Map

    fun setAndMap(){
        //自动过滤重复元素
        val hello = mutableSetOf("h","e","l","l","o")
        hello.remove("o")
        println(hello)
        //集合的加减操作
        hello+= setOf("w","o","r","l","d")
        hello.add("p")
        println(hello)

        val numbersMap = mapOf("key1" to 1,"key2" to 2,"key3" to 3,"key4" to 4)
        println("All keys : ${numbersMap.keys}")
        println("All values : ${numbersMap.values}")
        if ("key2" in numbersMap) println("value by key \"key2\":${numbersMap["key2"]}")
        if (1 in numbersMap.values) println("1 is in the map")
        if (numbersMap.containsValue(1)) println("1 is in the map")
    }
输出结果如下：
[h, e, l]
[h, e, l, w, o, r, d]
All keys : [key1, key2, key3, key4]
All values : [1, 2, 3, 4]
value by key "key2":2
1 is in the map
1 is in the map