Question

我想多次使用Eq，这样第一个连音符的第二个元素来自另一个类型而不是其余的

错了，但是，这就是我想要的想法

例如。 [（a，a）] - ＆gt; [（a，a）] - ＆gt; Bool ----＆gt; [（a，b）] - ＆gt; [（a，a）] - ＆gt; 布尔

代码

canColor ::  Eq a => [(a, a)] -> [(a, a)] -> Bool
canColor _ [] = True
canColor xs ((x,y):rest) =
    if findNeighbour xs x == findNeighbour xs y
    then False
    else canColor xs rest

findNeighbour :: Eq a => [(a, a)] -> a -> Maybe a
findNeighbour [] _ = Nothing
findNeighbour ((x,y):rest) z =
    if x == z
    then Just y
    else findNeighbour rest z

输入数据和期望值

Main> canColor [('a',"purple"),('b',"green"),('c',"blue")] [('a','b'),('b','c'),('c','a')]
True

Main> canColor [('a',"purple"),('b',"green"),('c',"purple")] [('a','b'),('b','c'),('c','a')]
False

Main> canColor [('1',"purple"),('2',"green"),('3',"blue")] [('1','2'),('2','3'),('3','1')]
True

**Main> canColor [('a', 4),('b',5),('c', 6 )] [('a','b'),('b','c'),('c','a')]
True

Main> colors [('a', 4),('b', 4 ),('c', 5 )] [('a','b'),('b','c'),('c','a')]
False**

Answer 1

只需为它们提供不同的类型变量，并且两者都需要Eq。我想你正在寻找这个代码：

canColor ::  (Eq a, Eq b) => [(a, b)] -> [(a, a)] -> Bool
canColor _ [] = True
canColor xs ((x,y):rest) =
    if findNeighbour xs x == findNeighbour xs y
    then False
    else canColor xs rest

findNeighbour :: Eq a => [(a, b)] -> a -> Maybe b
findNeighbour [] _ = Nothing
findNeighbour ((x,y):rest) z =
    if x == z
    then Just y
    else findNeighbour rest z

或者这个更简洁和惯用的代码：

canColor ::  (Eq a, Eq b) => [(a, b)] -> [(a, a)] -> Bool
canColor xs = all (\(x,y) -> lookup x xs /= lookup y xs)

Haskell，如何使用多个Eq（Eq a，Eq b）=＆gt;

1 个答案: