Haskell：频繁的价值观

Question

我正在尝试使用细分树

来解决frequent values

This博客文章采用了类似的方法

我想将列表拆分为以下内容：

-1 -1 1 1 1 1 3 10 10 10变为(0, 2) (2, 6) (6, 7), (7, 10)

我有一个代码：

g s = map (\x->(head x, length x)) . group . sort $ s

但它没有给出正确的输出。

是否可以使用频繁的值？

Answer 1

I'd do it as

f = neighbors . prefixSums . counts
  where
    counts = map length . group . sort
    prefixSums = scanl (+) 0
    neighbors xs = zip xs (tail xs)

This starts by computing the counts of elements, so your (arbitrary permutation of) [-1, -1, 1, 1, 1, 1, 3, 10, 10, 10] becomes [2, 4, 1, 3].

Then, prefixSums computes the sums of prefixes, and so we get [0, 2, 6, 7, 10] for our running example.

To get the end result, we simply take the consecutive neighbors of this list.