SML：看说功能

Question

我在递归编写 look and say 函数时遇到问题。它应该获取一个整数列表并评估为一个整数列表 "reads as spoken." 例如，

look_and_say([1, 2, 2]) = "one one two twos" = [1, 1, 2, 2]

和

look_and_say([2, 2, 2]) = "three twos" = [3, 2]

我在弄清楚如何在整个递归调用中向列表添加元素（并跟踪该列表）时遇到了一些困难。

这是我写的一个辅助函数，应该有用：

fun helper(current : int, count : int, remainingList : int list) : int list =
    if (current = hd remainingList) then
        helper(current, count + 1, tl remainingList)
    else
        (* add count number of current to list *)
        helper(hd remainingList, 1, tl remainingList);

下面是我的主要功能的粗略概述：

fun look_and_say(x::y : int list) : int list =
    if x = nil then
    (* print *)
    else
        helper(x, 1, y);

想法？

Answer 1

您的想法似乎是正确的，尽管看起来您的助手永远不会终止。这是一种无需助手即可实现的方法。

fun look_and_say [] = []
  | look_and_say (x::xs) =
      case look_and_say xs of
        [] => [1,x]
      | a::b::L => if x=b then (a+1)::b::L 
                   else 1::x::a::b::L

下面是使用您的助手实现它的方法。

fun helper(current, count, remainingList) =
  if remainingList = [] then
    [count, current]
  else if current = hd remainingList then
    helper(current, count + 1, tl remainingList)
  else count::current::look_and_say(remainingList)
and look_and_say [] = []
  | look_and_say (x::y) = helper(x, 1, y)

SML：看说功能

SML: Look and Say Function

functional-programming

sml