Haskell生成子集

Question

我有一个函数'subsets'，它生成给定集合的所有子集：

subsets :: [Int] -> [[Int]]
subsets []  = [[]]
subsets (x:xs) = subsets xs ++ map (x:) (subsets xs)

如何在另一个函数中组合map，foldl和filter以返回包含总和为0的元素的所有子集？

**示例：**

set    = [1,-1,5,2,-2,3]
result = [[1,-1],[2,-2],[-1,-2,3]]

Answer 1

你已经有了子集。所以我们需要一个函数

filterSubs :: [[Int]] -> [[Int]]
filterSubs = --remove all subsets which don't sum to 0

接下来我们需要一个谓词

sumZero :: [Int] -> Bool
sumZero xs = sum xs == 0

现在，使用此filter可以轻松构建filterSubs。我将把这个留给你，弄清楚它是如何工作的。然后我们的解决方案是微不足道的

zeroSubs = filterSubs . subsets

Answer 2

这个想法是你想要获取所有子集，然后只保持那些sum 0

subsetsSum0 ls = filter (\ss -> (sum ss) == 0) (subsets ls)

让我们让这段代码免费

subsetsSum0 = filter ((==0) . sum) . subsets