在遍历列表时按索引从列表中删除项目

问题描述

public boolean isTwoPair() {
    boolean isTwoPair = false;
    Collections.sort(deck);
    List<Card> cards = new LinkedList<Card>(deck);
    System.out.println(cards);
    for (int i = 0; i < cards.size()-1; i++) {
        for (int j = i + 1; j < cards.size()-1; j++) {
            if (deck.get(i).equals(deck.get(j))) {
                cards.remove(i);
                cards.remove(j);
                System.out.println(cards);
            }
        }

    }
    return isTwoPair;

}

我认为我的问题出在我的cards.remove().当我删除卡时，下次将其删除时，会将其从更改后的列表中删除.有两种方法可以从列表中删除索引号相同的两个项目?

I think my problem is with my cards.remove(). When I remove a card, the next time the card is removed it removes it from an altered list. Is there a way to remove two items from a list while they have the same index numbers?

如果我必须删除索引0,1，因为它们都是像这样的对:

If I have to remove index 0,1 because they are both pairs like so:

[Ace,Ace,Three,Four,Four]

此代码将其删除(删除索引0)

the code removes it like this(removing index 0)

[Ace,Three,Four,Four]

而不是从第一个列表中删除索引1(Ace)，而不是从第二个列表中删除索引，因此

than instead of removing the index 1(Ace) from the first list it removes it from the second list so

[Ace,Four,Four]

它从第二个列表(三个)中删除了索引1.

It removed the index 1 from the second list which was three.

这就是我所期望的

[Three,Four,Four]

目前，我希望我的循环能够拾取并删除四"和四"

At this point I expect my loop to pick up and remove Four and Four

    public boolean isTwoPair() {
    boolean isTwoPair = false;
    Collections.sort(deck);
    List<Card> cards = new LinkedList<Card>(deck);
    System.out.println(cards);
    for (int cardOne = 0; cardOne < cards.size(); cardOne++) {
        for (int cardTwo = cardOne + 1; cardTwo < cards.size(); cardTwo++) {
            if (deck.get(cardOne).equals(deck.get(cardTwo))) {
                cards.remove(cardOne);
                cards.remove(cardTwo-1);
                System.out.println(cards);
                for(int cardThree = 0; cardThree < cards.size(); cardThree++){
                    for(int cardFour = cardThree+1; cardFour < cards.size(); cardFour++){
                        if(cards.get(cardThree).equals(cards.get(cardFour))){
                            cards.remove(cardThree);
                            cards.remove(cardFour-1);
                            System.out.println(cards);
                            isTwoPair = true;
                        }
                    }
                }
            }
        }

    }
    return isTwoPair;

}

这就是我现在正在使用的，如果没有必要，我真的不想创建一个新变量，所以我决定不删除对象

this is what I am using now, I didn't really want to make a new variable if I didn't have to so I decided to not remove Object

推荐答案

如果您知道j始终大于i(由于for循环)，则如果先删除索引为i的元素，则您可以删除索引为j - 1的元素，以得到预期的结果.

If you know j is always greater than i (because of the for loop), if you delete the element with index i first then you can delete the element with index j - 1 getting the expected result.

cards.remove(i);
cards.remove(j-1);

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