Feo*_*akt 2 f# linked-list list
如何获得子列表
[1]; [1; 2]; [1; 2; 3]; ...; [1; 2; 3; ...; n]
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从清单
[1; 2; 3; ...; n]
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以最惯用的方式?我所能做的就是:
List.scan (fun acc elem -> elem::acc) [] [1;2;3;4;5]
> val it : int list list =
[[]; [1]; [2; 1]; [3; 2; 1]; [4; 3; 2; 1]; [5; 4; 3; 2; 1]]
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谢谢。
您的实现很好。这是我的选择:
let source = [1..10]
let counts = [0..source.Length] // Or start at 1 if you don't want to start with an empty list
counts |> List.map (fun count -> source |> List.take count)
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