Kotlin對list進行排序的一種方法
阿新 • • 發佈:2018-12-30
今天看了java排序的方法,自己用kotlin寫了一個小的list的排序,程式碼如下。
1.首先定義一個內部類,用於比較list中兩個數的大小
inner class User : Comparable<User> { var name: String? = null var order: Int? = null override fun compareTo(other: User): Int { return this.order!!.compareTo(other.order!!) } }
2.對list進行排序
val list = ArrayList<User>() private fun dataOrder() { val user1:User? = User() val user2:User? = User() val user3:User? = User() user1?.name = "a" user1?.order = 3 user2?.name = "b" user2?.order = 2 user3?.name = "c" user3?.order = 5 //此處add user2再add user1 user1?.let { list.add(it) } user2?.let { list.add(it) } user3?.let { list.add(it) } Collections.sort(list) for (u in list) { Log.e("sds","paixusuanfa>>>>>>" + u.name) } }