List 返回仅在一个列表中出现的元素列表-SML

如果L1=[1,2,3,4,5]和L2[4,5,6,7,8]，我想返回[1,2,3,5,7,8]，这些元素只出现在一个列表中。我已经编写了一个函数，返回两个列表中出现的项目列表

fun exists x nil = false | exists x (h::t) = (x = h) orelse (exists x t);

fun listAnd _ [] = []
  | listAnd [] _ = []
  | listAnd (x::xs) ys = if exists x ys  then x::(listAnd xs ys) 
else listAnd xs ys

我要找的名单应该是L1@L2-列表L1和L2。我还发现了删除元素和删除重复项的函数。我多次尝试稍微更改remDup函数，以便它不会留下任何多次出现的项目的痕迹。无法让它工作。我不知道如何使用和组合所有这些功能，使其工作

fun delete A nil = nil
| delete A (B::R) = if (A=B) then (delete A R) else (B::(delete A R));

fun remDups nil = nil
| remDups (A::R) = (A::(remDups (delete A R)));

如果存在一个diff函数，其中diff xs ys返回xs中的所有元素，但不返回ys中的所有元素，则可以轻松实现listOr函数：

fun listOr xs ys = diff (xs@ys) (listAnd xs ys)

diff函数的编写方式与listAnd类似：

fun diff xs [] = xs
  | diff [] _ = []
  | diff (x::xs) ys = if exists x ys  
                      then diff xs ys 
                      else x::(diff xs ys)