Question

生成整数因子的scala方法是什么？这是我的看法1：

def factorize(x: Int): List[Int] = {

  def foo(x: Int, a: Int): List[Int] = {

    if (a > Math.pow(x, 0.5))
      return List(x)

    x % a match {
      case 0 => a :: foo(x / a, a)
      case _ => foo(x, a + 1)
    }
  }

  foo(x, 2)
}

factorize（360）//列表（2,2,2,3,3,5）

根据@ SpiderPig和@ seth-tisue的评论取2

def factorize(x: Int): List[Int] = {
  def foo(x: Int, a: Int): List[Int] = {
    (a*a < x, x % a) match {
        case (true, 0) => a :: foo(x/a, a)
        case (true, _) => foo(x, a+1)
        case (false, _) => List(x)
      }
  }
  foo(x, 2)
}

Answer 1

尾递归解决方案：

def factorize(x: Int): List[Int] = {
  @tailrec
  def foo(x: Int, a: Int = 2, list: List[Int] = Nil): List[Int] = a*a > x match {
    case false if x % a == 0 => foo(x / a, a    , a :: list)
    case false               => foo(x    , a + 1, list)
    case true                => x :: list
  }
  foo(x)
}

Answer 2

＆＃34; Take 2＆＃34;从问题：

def factorize(x: Int): List[Int] = {
  def foo(x: Int, a: Int): List[Int] = x % a match {
     case _ if a * a > x  => List(x)
     case 0 => a :: foo(x / a, a)
     case _ => foo(x, a + 1)
  }
  foo(x, 2)
}

此外，以下方法可能会更快一些（在上一次迭代中没有x % a计算）：

def factorize(x: Int): List[Int] = {
  def foo(x: Int, a: Int): List[Int] = if (a * a > x) List(x) else
    x % a match {
      case 0 => a :: foo(x / a, a)
      case _ => foo(x, a + 1)
    }
  foo(x, 2)
}

scala方式生成一个数字的素因子

2 个答案: