递归方法的逻辑 [英] Logic on a recursive method

问题描述

我的一项练习要求我编写一个递归方法，在该方法中给出一个列表，并且该方法返回仅包含所有其他元素的相同列表.

One of my exercises requires me to write a recursive method in which a list is given, and it returns the same list with only every other element on it.

例如:列表{"a"，"b"，"c"}将返回 列表{"a"，"c"}

for example : List {"a", "b", "c"} would return List{"a","c"}

我正在用Scala编写程序，我知道它已内置在库中，但我不应该使用它们.我只能使用if/else，辅助方法和模式.

I am writing in scala, and I understand that it has built in library but I am not supposed to use those. I can only use if/else, helper methods,and patterns.

我如何仅使用头和尾来解析列表?

How could I parse thru a list using head and tail only?

到目前为止，我有这个:

so far I have this:

def removeLetter(list:List[String]):List[String]=list match{

 case Nil => Nil
 case n::rest=>  

  if (n == rest){  // I understand that this doesn't quite work.
     tail
   }
  else
     head::removeLetter(tail)
  }
   }

我在寻找逻辑而不是代码.

I am looking for the logic and not code.

推荐答案

使用模式匹配，您还可以以与n::rest构造相同的方式来解构其前两个元素上的列表.请记住，还要考虑长度不均匀的列表.

Using pattern matching, you can also deconstruct a list on it's first two elements in the same way you're doing with your n::rest construction. Just remember to also take lists with uneven length into account.

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