任意型リストのsplit(その2 by id:htid46)

splitBy2.fsx

let splitBy2 sep xs =
    let a, b = List.foldBack (fun x (a, b) ->
        if x = sep then (b :: a, []) else (a, x :: b)) xs ([], [])
    b :: a
let test1 = splitBy2 0 [1;2;3;0;4;5;0;6] = [[1;2;3];[4;5];[6]] // -> true
let test2 = splitBy2 0 [] = [[]] // -> true
let test3 = splitBy2 0 [1;0;2;0] = [[1];[2];[]] // -> true