Ocaml 中讨论的递归函数
我定义了一个类型和一个函数：
type element = ...
let merge (x0: element) (x1: element): element * bool = ...
merge 返回的第二部分表示 x0 和 x1 是否可合并。如果是这样，则返回的第一部分是合并的结果，否则第一部分可以被忽略。
然后我想实现一个功能restruct：element list -> element list 只要列表中的任何 2 个元素是可合并的（我们用它们合并的结果替换这 2 个元素），它就会保持合并，合并的顺序并不重要。
我想这一定是一个递归函数，现阶段对我来说有点复杂，有人可以帮忙吗？
非常感谢
I have defined a type and a function:
type element = ...
let merge (x0: element) (x1: element): element * bool = ...
The second part of the return of merge represents if x0 and x1 are merge-able. If so, the first part of the return is the result of merging, otherwise the first part could be ignored.
Then I would like to realize a function restructure: element list -> element list which keeps merging as long as any 2 elements of the list are merge-able (we replace the 2 elements by the result of their merging of couse) the order of merging is not important.
I guess it must be a recursive function, and a little bit complicated for me at this stage, could anyone help?
Thank you very much
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好的，我可以发布一个解决方案，但这对您来说既不有趣也没有什么启发:)。相反，让我尝试给出一些提示，如果您仍然遇到困难，我会尽力提供更多帮助。
merge函数应该返回option(element)，这是返回函数可选结果的经典方法。restruct，我建议您从一个函数merge_into(res : element, elts : element list)开始，它将尝试将elts合并到<代码>资源。希望这应该比重组更容易。merge_into实现重组？如果您仍然遇到问题，请随时告诉我...编码愉快！ :)
Ok, I could post a solution but that's neither fun nor very instructive for you :). Instead let me try to give a few hints and if you're still stucked I'll try to help more.
mergefunction should returnoption(element), which is a classical way to return optional results of functions.restructureI suggest you start with a functionmerge_into(res : element, elts : element list)which will try to mergeeltsintores. That should, hopefully, be easier thanrestructure.restructureusingmerge_into?Don't hesitate to let me know if you still have problems... happy coding! :)