下面的代码是f＃中mergesort的代码，我必须重写它，因此它使用模式匹配。
let rec msort xs =
let sz = List.length xs
if sz < 2 then xs
else let n = sz / 2
let ys = xs.[0..n-1]
let zs = xs.[n..sz-1]
in merge (msort ys) (msort zs)
到目前为止，我已经到了这里：
let rec msort (vs: int list) =
let sz = List.length xs
match xs with
| List.length < 2 -> vs
| _ ->
let n = sz / 2
let ys = xs.[0..n-1]
let zs = xs.[n..sz-1]
in merge (msort ys) (msort zs)
我似乎无法找出更好的方法。有谁可以帮助我吗？
我可能会这样：
let rec msort (values: int list) =
let n = values.Length / 2
if n = 0
then values
else let rec merge (xs: int list) (ys: int list) =
match (xs, ys) with
| ([], ys) -> ys
| (xs, []) -> xs
| (x :: xs1, y :: ys1) ->
if x < y
then x :: merge xs1 ys
else y :: merge xs ys1
let (first, second) = values |> List.splitAt n
merge (msort first) (msort second)
模式匹配在初始逻辑（列表的长度，长度为0和1的早期退出）上并不是很有用，但是我认为在拆分后匹配子列表的大小写时，它使可读性更高。即使这样，在该msort部分中只有一个if / then，如果您确实想要：
let rec msort (values: int list) =
match values.Length / 2 with
| 0 -> values
| n ->
let rec merge (xs: int list) (ys: int list) =
match (xs, ys) with
| ([], ys) -> ys
| (xs, []) -> xs
| (x :: xs1, y :: ys1) ->
if x < y
then x :: merge xs1 ys
else y :: merge xs ys1
let (first, second) = values |> List.splitAt n
merge (msort first) (msort second)
这仅将if/thenforx<y留在merge实现中，而我不会更改。
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