如何删除字符串中的每第 n 个元素?
我猜你会使用drop
以某种方式发挥作用。
就像这样删除第一个 n,你怎么能改变它,只删除第 n 个,然后是后面的第 n 个,依此类推,而不是全部?
dropthem n xs = drop n xs
简单的。取 (n-1) 个元素,然后跳过 1 个,冲洗并重复。
dropEvery _ [] = []
dropEvery n xs = take (n-1) xs ++ dropEvery n (drop n xs)
或者为了效率而采用 showS 风格
dropEvery n xs = dropEvery' n xs $ []
where dropEvery' n [] = id
dropEvery' n xs = (take (n-1) xs ++) . dropEvery n (drop n xs)
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