你如何在 Scala 中记忆案例？

Question

将这段使用记忆的代码转换为正确的 Scala 使用案例和函数式编程的最佳方法是什么？

def uniquePathsMemoization(n:Int, m:Int, row:Int, col:Int, seen:Array[Array[Int]]):Int = {
  if (row == m && col == n) 1
  if (row > m || col > n) 0

  if (seen(row+1)(col) == -1) seen(row+1)(col) = uniquePathsMemoization(n, m, row + 1, col, seen)
  if (seen(row)(col + 1) == -1 ) seen(row)(col) = uniquePathsMemoization(n,m, row, col + 1, seen)

seen(row+1)(col) + seen(row)(col + 1)
}

Answer 1

这是使用 match 和 case

的代码修改版本

def uniquePathsMemoization(n:Int, m:Int, row:Int, col:Int, seen:Array[Array[Int]]):Int = (row,col) match{

  case (row,col) if row == m && col == n =>
    1
  case (row,col) if row > m || col > n =>
    0
  case (row,col) =>
    if (seen(row+1)(col) == -1) seen(row+1)(col) = uniquePathsMemoization(n, m, row + 1, col, seen)
    if (seen(row)(col + 1) == -1 ) seen(row)(col) = uniquePathsMemoization(n,m, row, col + 1, seen)

    seen(row+1)(col) + seen(row)(col + 1)
}

由于存储在 seen 数组中的状态，将此代码转换为纯函数版本并不容易。但是这个状态可以为应用程序的其余部分隐藏，使用函数装饰器：

def uniquePathsMemoizationGenerator( maxRows: Int, maxCols:Int ) : (Int,Int,Int,Int) => Int = {


  def uniquePathsMemoization(n:Int, m:Int, row:Int, col:Int, seen:Array[Array[Int]]):Int = (row,col) match{

    case (row,col) if row == m && col == n =>
      1
    case (row,col) if row > m || col > n =>
      0
    case (row,col) =>
      if (seen(row+1)(col) == -1) seen(row+1)(col) = uniquePathsMemoization(n, m, row + 1, col, seen)
      if (seen(row)(col + 1) == -1 ) seen(row)(col) = uniquePathsMemoization(n,m, row, col + 1, seen)

      seen(row+1)(col) + seen(row)(col + 1)
  }

  val seen = Array.fill(maxRows,maxCols)(-1)
  uniquePathsMemoization(_,_,_,_,seen)
}

val maxRows = ???
val maxCols = ???
val uniquePaths = uniquePathsMemoizationGenerator( maxRows, maxCols )

// Use uniquePaths from this point, instead of uniquePathsMemoization

你如何在 Scala 中记忆案例？

How do you memoization with cases in Scala?

scala

dynamic-programming

memoization