在scala中,我需要编写一个方法来查找给定数字中每个数字的乘积。我有下面的片段。
def productDigits(number: Int): Int = {
def helper(current: Int, accumulator: Int): Int = {
current match {
case current if current < 10 => accumulator * current
case _ => helper(current / 10, accumulator * (current % 10))
}
}
helper(number, 1)
}有更好的方法吗?
发布于 2020-01-08 19:36:13
下面是OP递归解决方案的jmh基准与Luis‘one-liner的比较。
执行
sbt "jmh:run -i 10 -wi 10 -f 2 -t 1 bench.So59652263"
哪里
@State(Scope.Benchmark)
@BenchmarkMode(Array(Mode.Throughput))
class So59652263 {
def _dexter2305(number: Int): Int = {
def helper(current: Int, accumulator: Int): Int = {
current match {
case current if current < 10 => accumulator * current
case _ => helper(current / 10, accumulator * (current % 10))
}
}
helper(number, 1)
}
def _luis(number: Int): Int = number.toString.map(_.asDigit).product
val num: Int = (math.random * 100000000).toInt
@Benchmark def dexter2305: Int = _dexter2305(num)
@Benchmark def luis: Int = _luis(num)
}给出
[info] So59652263.dexter2305 thrpt 20 89093066.408 ± 1825286.801 ops/s
[info] So59652263.luis thrpt 20 11585098.230 ± 272966.526 ops/s在我们所看到的递归解决方案中,似乎比一行程序的吞吐量高出7倍。
当number是String而不是Int时进行基准测试
@State(Scope.Benchmark)
@BenchmarkMode(Array(Mode.Throughput))
class So59652263 {
def _dexter2305(number: String): Int = {
def helper(current: Int, accumulator: Int): Int = {
current match {
case current if current < 10 => accumulator * current
case _ => helper(current / 10, accumulator * (current % 10))
}
}
helper(number.toInt, 1)
}
def _luis(number: String): Int = number.map(_.asDigit).product
val num: String = (math.random * 100000000).toInt.toString
@Benchmark def dexter2305: Int = _dexter2305(num)
@Benchmark def luis: Int = _luis(num)
}给出
[info] Benchmark Mode Cnt Score Error Units
[info] So59652263.dexter2305 thrpt 20 36237007.844 ± 1631349.975 ops/s
[info] So59652263.luis thrpt 20 13975984.042 ± 1890083.941 ops/s但是,递归解决方案的吞吐量仍然比number为Int时小2.5倍。
https://stackoverflow.com/questions/59652263
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