如何实现支持空格分隔符的F#字符串列表分组函数？

需求说明 #

给定字符串列表（示例：["abceadfapaq"; "asdqwedasca"; " "; "asdasqyhgahfgasdsadasda"]）和字符串分隔符（如" "），需返回按分隔符拆分的子列表组成的列表（示例：[["abceadfapaq"; "asdqwedasca"]; ["asdasqyhgahfgasdsadasda"]]）。

现有实现及问题 #

失效函数 #

该函数处理非空格分隔符时正常，但处理空格分隔符时输出不符合预期：

let groupWithNoReplace (inputLines: string list) (separator: string) =
        let complete = 
            seq {
                for line in inputLines do
                    yield! line.Split(' ')
            } |> List.ofSeq
        let folder (a) (cur, acc) = 
            match a with
            | _ when a <> separator -> a::cur, acc
            | _ -> [], cur::acc 
        let result = List.foldBack folder (complete) ([], [])
        (fst result)::(snd result)

// 测试案例及结果
let groupWithNoReplace0 = 
    groupWithNoReplace ["abceadfapaq"; "asdqwedasca"; " "; "asdasqyhgahfgasdsadasda"] " "

val groupWithNoReplace0: string list list =
  [["abceadfapaq"; "asdqwedasca"; ""; ""; "asdasqyhgahfgasdsadasda"]]

let groupWithNoReplace00 = 
    groupWithNoReplace ["abceadfapaq"; "asdqwedasca"; "="; "asdasqyhgahfgasdsadasda"] "="

val groupWithNoReplace00: string list list =
  [["abceadfapaq"; "asdqwedasca"]; ["asdasqyhgahfgasdsadasda"]]

问题核心：当输入行是空格字符串" "时，用Split(' ')会拆分成两个空字符串""，导致最终结果混入无效空元素。

替换实现的有效函数 #

通过将空格替换为特殊符号"§"实现功能，但要求输入不能包含该符号，存在明显局限性：

let groupWithNoReplace (inputLines: string list) (separator: string) =
        let validlines = inputLines |> List.map(fun e -> if e = " " then "§" else e)
        let validsplitter =
            match separator = " " with
            | true -> "§"
            | false -> separator
        let complete = 
            seq {
                for line in validlines do
                    yield! line.Split(' ')
            } |> List.ofSeq
        let folder (a) (cur, acc) = 
            match a with
            | _ when a <> validsplitter -> a::cur, acc
            | _ -> [], cur::acc
                
        let result = List.foldBack folder (complete) ([], [])
        (fst result)::(snd result)

// 测试案例及结果
let groupWithNoReplace1 = 
    groupWithNoReplace 
["abceadfapaq"; "asdqwedasca"; " "; "asdasqyhgahfgasdsadasda"] " "

val groupWithNoReplace1: string list list =
  [["abceadfapaq"; "asdqwedasca"]; ["asdasqyhgahfgasdsadasda"]]

let groupWithNoReplace11 = 
    groupWithNoReplace ["abceadfapaq"; "asdqwedasca"; "="; "asdasqyhgahfgasdsadasda"] "="

val groupWithNoReplace11: string list list =
  [["abceadfapaq"; "asdqwedasca"]; ["asdasqyhgahfgasdsadasda"]]

寻求最优方案 #

需要一种无需依赖输入内容限制的实现，避免使用可能与输入冲突的特殊符号，同时准确完成按指定分隔符分组的需求。

内容的提问来源于stack exchange，提问作者blfuentes