问题描述
我最近开始使用HackerRank,并且正在尝试“匹配销售”。在开发Kotlin的函数编程功能方面,我已经找到了满足的解决方案。但是,我没有得到预期的答案...
问题摘要:
给出一个数组: ->查找并返回对总数。
即:
输入 -> [10,20,20,10,10,30,50,10,20]
对数 -> 3
这是我的代码和一些解释它的注释:
fun sockMerchant(n: Int,pile: Array<Int>): Int{
var count = 0
mutableMapOf<Int,Int>().withDefault { 0 }.apply {
// the [attempted] logic behind this piece of code here is
// that as we iterate through the list,the 'getorPut()'
// function will return either the value for the given [key]
// or throw an exception if there is no such key
// in the map. However,since our map was created by
// [withDefault],the function resorts to its `defaultValue` <-> 0
// instead of throwing an exception.
for (e in values) {
// this simplifies our code and inserts a zero [+1] where needed.
// if (key exists)
// // return its associated value and MOD it:
// case: even -> increment counter
// else -> do nothing
// else if (key dne)
// // insert default value <-> [0] + 1
// ....
// ....
// ....
if (getorPut(e,{ getValue(e) + 1 } ) % 2 == 0) count++
}
}
return count
}
fun main(args: Array<String>) {
val scan = Scanner(System.`in`)
val n = scan.nextLine().trim().toInt()
val ar = scan.nextLine().split(" ").map{ it.trim().toInt() }.toTypedArray()
val result = sockMerchant(n,ar)
println(result)
}
解决方法
我可以通过将数字分组在一起,获取结果列表,然后求和每个对包含几对来做到这一点:
fun sockMerchant(n: Int,pile: Array<Int>): Int =
pile.groupBy { it }.values.sumBy { it.size / 2 }
完成pile.groupBy { it }
之后,我们将具有以下结构:
{10=[10,10,10],20=[20,20,20],30=[30],50=[50]}
我们取这些值,然后将每个值的大小相加并除以2。这将把半对向下舍入为0,将完整对舍入为1。
注意:在这种情况下,我还不太清楚n
的目的是什么。
我对其进行了一些修改,使其更易于测试,但这是修复方法:
import java.util.*
fun sockMerchant(n: Int,pile: Array<Int>): Int{
var count = 0
mutableMapOf<Int,Int>().withDefault { 0 }.apply {
// the [attempted] logic behind this piece of code here is
// that as we iterate through the list,the 'getOrPut()'
// function will return either the value for the given [key]
// or throw an exception if there is no such key
// in the map. However,since our map was created by
// [withDefault],the function resorts to its `defaultValue` <-> 0
// instead of throwing an exception.
for (e in pile) {
// this simplifies our code and inserts a zero [+1] where needed.
// if (key exists)
// // return its associated value and MOD it:
// case: even -> increment counter
// else -> do nothing
// else if (key dne)
// // insert default value <-> [0] + 1
// ....
// ....
// ....
println(e)
put(e,getValue(e) + 1)
if (getValue(e) % 2 == 0) count++
println(entries)
}
}
return count
}
val n = 5
val ar = "10 10 10 10 20 20 30 40".split(" ").map{ it.trim().toInt() }.toTypedArray()
val result = sockMerchant(n,ar)
println(result)
输出:
10
[10=1]
10
[10=2]
10
[10=3]
10
[10=4]
20
[10=4,20=1]
20
[10=4,20=2]
30
[10=4,20=2,30=1]
40
[10=4,30=1,40=1]
3
Pair.kts:3:18: warning: parameter 'n' is never used
fun sockMerchant(n: Int,pile: Array<Int>): Int{
^
Process finished with exit code 0
说明:
- 您遍历了“值”,该值一开始是空的,因此您从未对代码做任何事情
- 即使在
pile
上循环时,您的增量逻辑也未超过1,因此条件从未满足且计数也从未增加。
但是背后的主要理由是正确的。