用值替换空列表

Question

我有一个返回如下列表的函数:

[ [1, 2, 3], [], [5], [5,6], []]

但我想用0替换空列表,所以它看起来像这样

[ [1, 2, 3], [0], [5], [5,6], [0]]

到目前为止,我曾尝试filter和map收效甚微.有人可以指点我正确的方向吗？

这是有问题的代码:

knightPlace:: [Int] -> [[Int]]
knightPlace n = makeboard n

    where
    makeboard n =  [x | i<-[0..(length n -1 )], x <- [checkPos i]]
    -- checkPos add zero
    checkPos i = [j+1 | j<-[0..(length n -1 )], queenFilter n i j]
    -- filters all rows, cols and diags, also knights
    queenFilter n i j = and [n!!(i) == 0 && n!!(k) /=(j+1) && (n!!(k)==0 || (abs(n!!(k)-(j+1))/=abs(i-k))) && (n!!(k)==0 ||not( ((abs(i-k)==2)&& (abs(n!!(k)-(j+1))==1)) ||((abs(i-k)==1)&& (abs(n!!(k)-(j+1))==2))))       | k<-[0..(length n - 1)] ]

叫做像

 knightPlace [0, 0, 6, 0, 0, 4, 0, 0]

Answer 1

想想看,如果你有一个功能foo,可以转化[]到[0]并返回所有其他名单不变,map foo就是你想要的最终功能.