LinkedList：迭代并删除元素 [英] LinkedList: Iterate and remove element

问题描述

在Scala中，在迭代LinkedList的元素时，我希望有一些方法remove（）删除当前元素，（非常重要）使迭代器指向下一个元素（或者如果是当前元素是最后一个;如果没有更多元素，则为null或者某事。）

In Scala, while iterating through the elements of a LinkedList, I would like to have some method remove() which removes the current element and (very important) makes the iterator point to the next element (or to the first if the current element is the last one; to null or something if there are no more elements).

推荐答案

不确定这是不是你的意思想但是怎么样：

Not sure if this what you want but how about:

@annotation.tailrec
def round(xs: Set[(Int, Int)]) = {
    // here you might check if items are exhausted and possibly don't go through recursion deeper


    val suitable = xs.filter {
        case (x, i) => 
            println("Element "+x+" on position "+i)
            x != 0
    }

    // do something with suitable items

    round(xs.diff(suitable)) // next round with items that doesn't succeed in current round
}

val list = List(3,4,5)
round(list.zipWithIndex.toSet)

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