需要将列表按照元素的升序（Haskell）按升序排列为列表 [英] Need to partition a list into lists based on breaks in ascending order of elements (Haskell)

问题描述

  [4,5,6,7,1,2,3， 4,5,6,1,2] 
  

我需要一个Haskell函数将这个列表转换成由原始列表的段组成的列表，它们按升序形成一系列列表。所以结果应该如下所示：

  [[4,5,6,7]，[1,2,3， 4,5,6]，[1,2]] 
  

有何建议？

解决方案

  ascend :: Ord a => [a]  - > [[a]] 
 ascend xs = foldr f [] xs 
其中
 f a [] = [[a]] 
 f a xs'@（y：ys）| a<头y =（a：y）：ys 
 |否则= [a]：xs'
  

在ghci

  * Main>上升[4,5,6,7,1,2,3,4,5,6,1,2] 
 [[4,5,6,7]，[1,2,3,4， 5,6]，[1,2]] 
  

Say I have any list like this:

[4,5,6,7,1,2,3,4,5,6,1,2]

I need a Haskell function that will transform this list into a list of lists which are composed of the segments of the original list which form a series in ascending order. So the result should look like this:

[[4,5,6,7],[1,2,3,4,5,6],[1,2]]

Any suggestions?

解决方案

ascend :: Ord a => [a] -> [[a]]
ascend xs = foldr f [] xs
  where
    f a []  = [[a]]
    f a xs'@(y:ys) | a < head y = (a:y):ys
                   | otherwise = [a]:xs'

In ghci

*Main> ascend [4,5,6,7,1,2,3,4,5,6,1,2]
[[4,5,6,7],[1,2,3,4,5,6],[1,2]]

这篇关于需要将列表按照元素的升序（Haskell）按升序排列为列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！