如何在没有forloop的情况下使用递归嵌套for循环在haskell中 [英] How to use recursion nested for loop in haskell without forloop

问题描述

假设有一些嵌套for循环的动态数字来输出所有组合在这个例子中，有3个嵌套的forloop，来生成所有组合的列表，例如
[1,2,3]，[1,3,5]等。

如果有N个嵌套forloop，如何使用递归做haskell？

伪代码

 对于我从1到5做
 for j from 1 to 10 do 
 if i< j然后
 for k从1到50 do 
 if j < k then 
 list1 :: [i，j，k] 
  

未完成的forloop有编译错误

  forloop :: Integer  - >整数 - > [a] 
 forloop n m 
 | n == 1 = 1 
 |否则=如果n> m 
然后[（forloop n-1 m）] ++ [n] 
 else [] 
  

期望一个递归版本的函数，它可以保存在redis中，比如Action类型的.net框架

解决方案

假设每个嵌套层次都有一个范围结束的列表。考虑你如何生成列表，如果你有一个范围较小的列表的解决方案。

  solve :: [ Int]  - > Int  - > [[Int]] 
 solve [] _ = [[]] 
 solve（r：t）i = [j：s | j<  -  [i + 1..r]，s<  - 解决t j] 
  

assume there are dynamic number of nested for loop to output all combinations

in this example, there are 3 nested forloop, to generate a list of all combination such as [1,2,3], [1,3,5] etc.

if there are N nested forloop, how to use recursion to do in haskell?

pseudo code

for i from 1 to 5 do
     for j from 1 to 10 do
          if i < j then
          for k from 1 to 50 do
               if j < k then
                  list1 :: [i,j,k]

unfinished forloop has compile error

forloop :: Integer -> Integer -> [a]
forloop n m
    | n == 1 = 1
    | otherwise =  if n > m 
                     then [(forloop n-1 m)] ++ [n]
                     else []

expect a recursive version of function which can be saved in redis like Action type of .net framework

解决方案

Suppose you have a list of range end for each level of nesting. Consider how you are going to generate the lists, if you have a solution for the shorter list of ranges.

solve :: [Int] -> Int -> [[Int]]
solve [] _ = [[]]
solve (r:t) i = [j:s | j <- [i+1..r], s <- solve t j]

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