scala - scala中使用递归的硬币兑换算法

Question

我正在尝试使用递归在 Scala 中编写硬币找零问题。我写的代码如下。

def countChange(money: Int, coins: List[Int]): Int = {
  def ways(change: List[Int], size: Int, capacity: Int): Int = {
    if(capacity == 0) 1
    if((capacity < 0) || (size <= 0)) 0

    //println and readLine to check and control each recursive call.

    println("calling ways(",change, change.length-1, capacity,") + ways(",change,   change.length, capacity - change(change.length - 1),")")
    readLine()
    //

    ways(change, change.length-1, capacity) + ways(change, change.length, capacity - change(change.length - 1))
  }
  ways(coins, coins.length, money)
}

在运行代码时，它不会终止并继续调用第一个递归调用。我哪里错了？

Answer 1

漂亮又简单

def countChange(money: Int, coins: List[Int]): Int = {
  if(money == 0)
    1
  else if(money > 0 && !coins.isEmpty)
    countChange(money - coins.head, coins) + countChange(money, coins.tail)
  else
    0
}

Answer 2

这是我的实现：我已经测试过了，它工作正常

def countChange(money: Int, coins: List[Int]): Int = {

  def count(capacity: Int, changes: List[Int]): Int = {
    if (capacity == 0)
      1
    else if (capacity < 0)
      0
    else if (changes.isEmpty && capacity >= 1)
      0
    else
      count(capacity, changes.tail) 
        + count(capacity - changes.head, changes)
  }

  count(money, coins.sortWith(_.compareTo(_) < 0))
}

Answer 3

只是另一种解决方案

def countChange(amount: Int, coins: List[Int]): Int = coins match {
  case _ if amount == 0 => 1
  case h :: t if amount > 0 => countChange(amount - h, h :: t) + countChange(amount, t)
  case _ => 0
}

Answer 4

简单地声明一个值不会让 Scala 返回它。您要么需要明确的退货，要么必须是最后一项。因此：

if (capacity == 0) return 1

或者

if (capacity == 0) 1
else if (...)
else { ... }

Answer 5

嘿，我只是认为不仅要查看数量还要查看它们的列表会更好，所以放在上面的示例之上，例如：

def moneyChanges(money: Int, coins: List[Int]) : Option[List[Seq[Int]]]= {
  var listOfChange=List[Seq[Int]]()
  def changeMoney(capacity: Int, changes: List[Int], listOfCoins: Option[Seq[Int]]): Int = {
    if (capacity == 0) {
      listOfChange = listOfCoins.get :: listOfChange
      1
    } else if (capacity < 0)
      0
    else if (changes.isEmpty && capacity >= 1)
      0
    else {
      changeMoney(capacity, changes.tail, listOfCoins) +
      changeMoney(capacity - changes.head, changes, 
      Some(changes.head +: listOfCoins.getOrElse(Seq())))
    }
  }

  changeMoney(money, coins.sortWith(_.compareTo(_) < 0), None)
  Some(listOfChange)
}

Answer 6

这是一种DP方法，可以减少递归方法中的大量重新计算

object DP {
  implicit val possibleCoins = List(1, 5, 10, 25, 100)
  import collection.mutable.Map

  def countChange(amount: Int)(implicit possibleCoins: List[Int]) = {
    val min = Map((1 to amount).map (_->Int.MaxValue): _*)
    min(0) = 0
    for {
      i <- 1 to amount
      coin <- possibleCoins
      if coin <= i && min(i - coin) + 1 < min(i)
    } min(i) = min(i-coin) + 1
    min(amount)
  }

  def main(args: Array[String]) = println(countChange(97))
}

看DP从新手到高级算法

Answer 7

Below code is similar to one of the above example except I am using match case instead of if else

def countChange(money: Int, coins: List[Int]): Int = {
    def change(m: Int, coinList: List[Int], count: Int): Int =
      m match {
        case _ if m < 0 => count
        case _ if coinList.isEmpty => {
          m match {
            case 0 => count + 1
            case _ => count
          }
        }
        case _ => change(m, coinList.tail, count) + change(m - coinList.head, coinList, count)
      }
    change(money, coins, 0)
  }

Answer 8

这是我的代码：它没有经过优化，但适用于所有测试用例。

这个想法是从 money 中减去列表中的第一个硬币，直到它变为 0。一旦变为 0，它将返回 1，这意味着一个解决方案是可能的。要添加来自不同递归的所有解决方案，我使用了foldLeft.

（使用 迭代列表foldLeft，所以首先进入 1，然后再次进入递归并迭代 (1, 2) 列表）

                    [4, (1, 2)].  
             /(1 as cn)       \ (2 as cn)
            [3, (1, 2)].                 [2, (2)]
         /(-1)       \(-2)                \
      [2, (1, 2)].     [1, (2)].          [0, (2)]   
       /.(-1)    \(-2) 
     [1, (1, 2)].   [0, (2)]
      /. (-1)  \(-2)
    [0, (1, 2)].  [-1, (2)]

def countChange(money: Int, coins: List[Int]): Int = coins.foldLeft(0)((accum, cn) =>
  (money, cn) match {
    case (money, _) if money < 0 => 0
    case (0, _) => 1
    case (curr_money, curr_coin) =>
      val (before_curr_coin, after_curr_coin) = coins.span(_ != curr_coin)
      accum + countChange(curr_money - curr_coin, after_curr_coin)
  })