我正在做一项额外的作业,其中 SML 中的快速排序的分区函数只能使用 foldr
完成,并且不能使用任何库函数。我已经很好地进行了分区,并且掌握了快速排序的基础知识,但是遇到了问题,似乎将错误的列表合并在一起/以错误的顺序合并。
(*comp is a placeholder, I will be given a comparator function and list as input*)
fun comp a b = a > b;
(*Both partition functions are working as intended.*)
fun getGreater (a::b) c = foldr (fn (a, lst) => if (comp a c) then a::lst else lst) [] (a::b);
fun getLesserOrEqual (a::b) c = foldr (fn (a, lst) => if not (comp a c) then a::lst else lst) [] (a::b);
fun merge (a::b) (c::d) = foldr (op::) (c::d) (a::b);
fun tripleMerge (a::b) c (d::e) = merge (a::b) (c::(d::e));
(*Removes head, uses head as pivot. Recursive calls on results on getLesserOrEqual and getGreater*)
fun sort [] = [] | sort (a::b) = if ([a] = (a::b)) then [a] else
tripleMerge (sort(getLesserOrEqual b a)) a (sort(getGreater b a));
作为示例,以下是我正在运行的一些测试。当我遵循纸上逻辑时,我没有得到与错误项目相同的答案:
sort [2, 6, 3, 6, 4, 100, 0];
val it = [0,2,3,6,4,6,100] : int list (*Incorrect*)
sort [1, 2, 3, 4, 5];
val it = [1,2,3,4,5] : int list
sort [5, 4, 3, 2, 1];
val it = [5,4,3,2,1] : int list (*incorrect*)
sort [1, 100, 10, 1000, 0];
val it = [0,1,10,100,1000] : int list
sort [1, 2, 1, 2, 1, 2, 5, 1];
val it = [1,1,1,1,2,2,2,5] : int list
我的错误对任何人来说都是显而易见的吗?
最佳答案
以下是一些反馈:
您可以引用
op >
(a, b)
(即,将一个运算符变成一个需要一对的函数。据我所知,您的实际比较函数是作为输入给出的。一个令人困惑的小点可能是,在标准库中,名为
compare
的函数返回 order 类型,即LESS
、EQUAL
或GREATER
,而不是 bool。但没关系。正如 John 所说,当函数针对空列表定义良好时,没有真正理由使用模式
a::b
。在函数名称前加上
get
有点多余,因为所有(纯)函数都将某物作为其主要目的。您可以通过制作更高阶的变体来组合两个过滤函数:
fun flip f x y = f y x fun filter p xs = foldr (fn (x, ys) => if p x then x::ys else ys) [] xs fun greaterThan x xs = filter (flip comp x) xs fun lessThanOrEqual x xs = filter (comp x) xs
似乎
sort
有两种基本情况,一种使用模式处理,另一种使用 if-then-else 处理。为什么?fun sort [] = [] | sort [x] = [x] | sort (x::xs) = sort (lessThanOrEqual x xs) @ x :: sort (greaterThan x xs)
或者,如果出于某种原因,您想要最后列出的基本案例:
fun sort (x::(xs as (_::_))) = sort (lessThanOrEqual x xs) @ x :: sort (greaterThan x xs) | sort xs = xs
尽管this isn't Quicksort ,所以效率还差得远,你可以做的一件事是改进它,就是只通过
xs
一次,而不是两次。由于我们只被允许foldr
,所以我们必须生成一对列表:fun partition p xs = foldr (fn (x, (ys, zs)) => if p x then (x::ys, zs) else (ys, x::zs)) ([], []) xs fun sort [] = [] | sort [x] = [x] | sort (x::xs) = let val (lesser, greater) = partition (comp x) xs in sort lesser @ x :: sort greater end
关于使用foldr在SML中快速排序,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42074945/