在三元组中生成所有可能的数字组合？ [英] Generating all possible combinations of numbers in a triplet?

问题描述

举个例子，我想要构造一个三元组，在三元组中取1到100之间的数字的每个组合;即：

  [（0,0,0），（0,0,1），（0,1,1） ，（1,1,1），（0,0,2），（0,1,2），（0,2,2），（1,2,2）] 
 
解决方案
我认为您的描述最符合列表理解来表达您想要做的事情： 
 
 
  [（a，b，c）| c < -  [0..100]，
b < -  [0..c]，
a < -  [0..b]] 
  
 
Say for example I want to construct a triplet, taking in every combination of numbers from 1..100 in a triplet; i.e:
[(0,0,0),(0,0,1),(0,1,1),(1,1,1),(0,0,2),(0,1,2),(0,2,2),(1,2,2)]
..etc etc, up until a given bound (i.e: 100: giving us a final triplet of (100,100,100)); is there any reasonable way of doing this within haskell, or would I be best off writing a method that in short held a boundary pointer, and recursively increased each number until it was equal to the number to its right?
解决方案
I think your description best fits a list comprehension to express what you want to do:
[(a, b, c) | c <- [0..100],
             b <- [0..c],
             a <- [0..b] ]


                        
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