問題描述

假設我有這個:

public class Unit<MobileSuit, Pilot> {

    ...

    List<MobileSuit> mobileSuits;
    List<Pilot> pilots;

    ...
}

并且我想在該類之外以最簡單的方式遍歷這對.我該怎么做呢?我想過這樣做:

And I would like to iterate through the pair of each in the simplest way outside of that class. How should I go about doing that? I thought about doing this:

public class Unit<MobileSuit, Pilot> {

    ...
    Iterator<MobileSuit> iteratinMechas;
    Iterator<Pilot> iteratinPeople;

    class IteratorCustom<MobileSuit, Pilot> implements Iterator {

        public boolean hasNext() {
            return iteratinMechas.hasNext() && iteratinPeople.hasNext();
        }

        public void remove() {
            iteratinMechas.remove();
            iteratinPeople.remove();
        }

        public Object next() {
            // /!
        }

    }

    public Iterator iterator() {
        return new IteratorCustom<MobileSuit, Pilot>(mobileSuits, pilots);
    }
}

類似的東西.

無論如何，問題是我不能真正從 next() 返回單個對象，而且我也不能讓迭代器采用多種類型.那么，有什么想法嗎?

Anyway, the problem is that I can't really return just a single object from next(), and I also can't have a Iterator take more than one type. So, any thoughts?

另外，我無法創建一個新課程來結合 MobileSuit 和 Pilot.我需要將它們分開，即使我一次遍歷兩者.原因是可能有沒有飛行員的機動戰士，我不確定如何通過將它們保持在同一級別來解決這個問題.這個類需要在其他地方處理，所以我必須圍繞它和很多其他東西統一一個接口.基本上，假設 MobileSuit 和 Pilot 需要分開.

Also, I can't make a new class to combine MobileSuit and Pilot. I need to keep them separate, even though I'm iterating through both at a time. The reason is that there might be Mobile Suits that have no pilots, and I'm not sure how to fix that by keeping them at the same class. This class needs to be processed in other places, so I'd have to unify a interface around that and a lot of other stuff. Basically, assume MobileSuit and Pilot need to be separated.

推薦答案

無論如何，問題是我不能真正從 next() 返回單個對象，而且我也不能讓迭代器采用多種類型.那么，有什么想法嗎?

Anyway, the problem is that I can't really return just a single object from next(), and I also can't have a Iterator take more than one type. So, any thoughts?

顯然，您將需要一個輕量級的pair"類.這大致類似于 Map.Entry 內部類.

Obviously you are going to need a light-weight "pair" class. This is roughly analogous to the Map.Entry inner class.

這是一個通用解決方案的粗略:

Here's a rough cut at a generic solution:

public class ParallelIterator <T1, T2> implements Iterator<Pair<T1, T2>> {

    public class Pair<TT1, TT2> {
        private final TT1 v1;
        private final TT2 v2;
        private Pair(TT1 v1, TT2 v2) { this.v1 = v1; this.v2 = v2; }
        ...
    }

    private final Iterator<T1> it1;
    private final Iterator<T2> it2;

    public ParallelIterator(Iterator<T1> it1, Iterator<T2> it2) { 
        this.it1 = it1; this.it2 = it2;
    }

    public boolean hasNext() { return it1.hasNext() && it2.hasNext(); }

    public Pair<T1, T2> next() {
        return new Pair<T1, T2>(it1.next(), it2.next());
    }

    ...

}

注意:這并沒有明確處理列表長度不同的情況.將會發生的情況是，較長列表末尾的額外元素將被靜默忽略.

Note: this doesn't explicitly deal with cases where the lists have different lengths. What will happen is that extra elements at the end of the longer list will be silently ignored.

這篇關于同時提供兩個列表內容的迭代器?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！