如何在scala列表中找到可重复元素的数量 [英] How to find count of repeatable elements in scala list

问题描述

假设您有一个List(1,1,1,4,4,1)，并且必须计算连续重复出现在列表开头的元素多少次.在上面的示例中，方法应返回3.在此方法中，我们只关心第一个元素.

Suppose you have a List(1,1,1,4,4,1) and have to calculate how many times is element that is a head of the list consecutively repeated. In the example above, method should return 3. In this method we only care about the first element.

我走到了这么远，被卡住了.给定第一个不可重复的角色，我想休息一下，怎么办?

I got this far and got stuck. Given a the first non repeatable character, i'd like to break, how how?

 def firstRepeated [X] (xs: List[X]) : Int = xs match {
     case Nil    => 0
     case y::ys  => ys match {
       case Nil   => 0
       case z::zs => if (y == z) 1 + firstRepeated(zs) else // break
     }
 }

此外，在上面的代码中，我认为我没有正确处理列表为z::Nil

Also, in the code above, i don't think i am properly handling a case when list is z::Nil

任何指针将不胜感激

推荐答案

在

There is also a method for this in GenSeqLike (which List inherit from): prefixLength. That makes a very short answer:

s.prefixLength(_==s.head)

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