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java - 从包含键的 LinkedHashMap 中删除特定值和价值 LinkedList

问题描述

我遇到了一个令人头疼的问题，我知道出了什么问题，但无法将其放入代码中。我目前正在研究一个卡片组，我想从LinkedHashMap. 我到处搜索了针对我非常具体的问题的具体解决方案，但找不到一个。我尝试过使用 entrySet、一个迭代器和一个逻辑解决方案，但该remove()操作似乎删除了所有记录，而value忽略了key. 下面的示例展示了我尝试移除 5 颗红心的尝试。

import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;

public class Main {

    public static void main(String[] args) {
        Deck deck = new Deck();
        System.out.println(deck.deck);

        //ENTRYSET
        for (Map.Entry<String, LinkedList<String>> entry : deck.deck.entrySet()) {
            System.out.println(entry.getValue());
            System.out.println(entry.getKey());
            if (entry.getKey().contains("5")) {
                if (entry.getValue().contains("Hearts")) {
                    entry.getValue().remove("Hearts");
                    break;
                }
            }
        }
        System.out.println(deck.deck);

        //LOGIC SOLUTION
        if (! deck.deck.get("5").isEmpty()) {
            deck.deck.get("King").remove("Hearts");
        }

        //ITERATOR
        for (Iterator<Map.Entry<String, LinkedList<String>>> it = deck.deck.entrySet().iterator(); it.hasNext(); ) {
            Map.Entry<String, LinkedList<String>> entry = it.next();
            LinkedList<String> list = entry.getValue();
            System.out.println(list);
            if (entry.getKey().equals("5")) {


                for (int i = 0; i < list.size(); i++) {
                    if (list.get(i).equals("Hearts")) {
                        list.remove(i);
                        break;
                    }
                }
                if (list.isEmpty())
                    it.remove();
            }
        }
        System.out.println(deck.deck);
    }
}

输出入口集：

{2=[Clubs, Diamonds, Hearts, Spades], 3=[Clubs, Diamonds, Hearts, Spades], 4=[Clubs, Diamonds, Hearts, Spades], 5=[Clubs, Diamonds, Hearts, Spades], 6=[Clubs, Diamonds, Hearts, Spades], 7=[Clubs, Diamonds, Hearts, Spades], 8=[Clubs, Diamonds, Hearts, Spades], 9=[Clubs, Diamonds, Hearts, Spades], 10=[Clubs, Diamonds, Hearts, Spades], Jack=[Clubs, Diamonds, Hearts, Spades], Queen=[Clubs, Diamonds, Hearts, Spades], King=[Clubs, Diamonds, Hearts, Spades], Ace=[Clubs, Diamonds, Hearts, Spades]}
[Clubs, Diamonds, Hearts, Spades]
2
[Clubs, Diamonds, Hearts, Spades]
3
[Clubs, Diamonds, Hearts, Spades]
4
[Clubs, Diamonds, Hearts, Spades]
5
{2=[Clubs, Diamonds, Spades], 3=[Clubs, Diamonds, Spades], 4=[Clubs, Diamonds, Spades], 5=[Clubs, Diamonds, Spades], 6=[Clubs, Diamonds, Spades], 7=[Clubs, Diamonds, Spades], 8=[Clubs, Diamonds, Spades], 9=[Clubs, Diamonds, Spades], 10=[Clubs, Diamonds, Spades], Jack=[Clubs, Diamonds, Spades], Queen=[Clubs, Diamonds, Spades], King=[Clubs, Diamonds, Spades], Ace=[Clubs, Diamonds, Spades]}

输出逻辑解决方案：

{2=[Clubs, Diamonds, Hearts, Spades], 3=[Clubs, Diamonds, Hearts, Spades], 4=[Clubs, Diamonds, Hearts, Spades], 5=[Clubs, Diamonds, Hearts, Spades], 6=[Clubs, Diamonds, Hearts, Spades], 7=[Clubs, Diamonds, Hearts, Spades], 8=[Clubs, Diamonds, Hearts, Spades], 9=[Clubs, Diamonds, Hearts, Spades], 10=[Clubs, Diamonds, Hearts, Spades], Jack=[Clubs, Diamonds, Hearts, Spades], Queen=[Clubs, Diamonds, Hearts, Spades], King=[Clubs, Diamonds, Hearts, Spades], Ace=[Clubs, Diamonds, Hearts, Spades]}
{2=[Clubs, Diamonds, Spades], 3=[Clubs, Diamonds, Spades], 4=[Clubs, Diamonds, Spades], 5=[Clubs, Diamonds, Spades], 6=[Clubs, Diamonds, Spades], 7=[Clubs, Diamonds, Spades], 8=[Clubs, Diamonds, Spades], 9=[Clubs, Diamonds, Spades], 10=[Clubs, Diamonds, Spades], Jack=[Clubs, Diamonds, Spades], Queen=[Clubs, Diamonds, Spades], King=[Clubs, Diamonds, Spades], Ace=[Clubs, Diamonds, Spades]}

输出迭代器：

{2=[Clubs, Diamonds, Hearts, Spades], 3=[Clubs, Diamonds, Hearts, Spades], 4=[Clubs, Diamonds, Hearts, Spades], 5=[Clubs, Diamonds, Hearts, Spades], 6=[Clubs, Diamonds, Hearts, Spades], 7=[Clubs, Diamonds, Hearts, Spades], 8=[Clubs, Diamonds, Hearts, Spades], 9=[Clubs, Diamonds, Hearts, Spades], 10=[Clubs, Diamonds, Hearts, Spades], Jack=[Clubs, Diamonds, Hearts, Spades], Queen=[Clubs, Diamonds, Hearts, Spades], King=[Clubs, Diamonds, Hearts, Spades], Ace=[Clubs, Diamonds, Hearts, Spades]}
[Clubs, Diamonds, Hearts, Spades]
[Clubs, Diamonds, Hearts, Spades]
[Clubs, Diamonds, Hearts, Spades]
[Clubs, Diamonds, Hearts, Spades]
[Clubs, Diamonds, Spades]
[Clubs, Diamonds, Spades]
[Clubs, Diamonds, Spades]
[Clubs, Diamonds, Spades]
[Clubs, Diamonds, Spades]
[Clubs, Diamonds, Spades]
[Clubs, Diamonds, Spades]
[Clubs, Diamonds, Spades]
[Clubs, Diamonds, Spades]
{2=[Clubs, Diamonds, Spades], 3=[Clubs, Diamonds, Spades], 4=[Clubs, Diamonds, Spades], 5=[Clubs, Diamonds, Spades], 6=[Clubs, Diamonds, Spades], 7=[Clubs, Diamonds, Spades], 8=[Clubs, Diamonds, Spades], 9=[Clubs, Diamonds, Spades], 10=[Clubs, Diamonds, Spades], Jack=[Clubs, Diamonds, Spades], Queen=[Clubs, Diamonds, Spades], King=[Clubs, Diamonds, Spades], Ace=[Clubs, Diamonds, Spades]}

我希望地图包含的内容：

{2=[Clubs, Diamonds, Hearts, Spades], 3=[Clubs, Diamonds, Hearts, Spades], 4=[Clubs, Diamonds, Hearts, Spades], 5=[Clubs, Diamonds, Spades], 6=[Clubs, Diamonds, Hearts, Spades], 7=[Clubs, Diamonds, Hearts, Spades], 8=[Clubs, Diamonds, Hearts, Spades], 9=[Clubs, Diamonds, Hearts, Spades], 10=[Clubs, Diamonds, Hearts, Spades], Jack=[Clubs, Diamonds, Hearts, Spades], Queen=[Clubs, Diamonds, Hearts, Spades], King=[Clubs, Diamonds, Hearts, Spades], Ace=[Clubs, Diamonds, Hearts, Spades]}

那么我将如何指定移除西装的钥匙呢？我不希望删除“Hearts”的所有实例（如本例中）。非常感谢帮助和线索！

编辑：

西装的数量使键指向地图中内存中的一个列表。

前：

public Deck() {
        for (int i = 2; i <= 10; i++) {
            deck.put(String.valueOf(i), suits));
        }
        for (int i = 0; i <= 3; i++) {
            deck.put(highRank.get(i), suits));
        }

    }

后：

public Deck() {
        for (int i = 2; i <= 10; i++) {
            deck.put(String.valueOf(i), new LinkedList<String>(Arrays.asList("Clubs", "Diamonds", "Hearts", "Spades")));
        }
        for (int i = 0; i <= 3; i++) {
            deck.put(highRank.get(i), new LinkedList<String>(Arrays.asList("Clubs", "Diamonds", "Hearts", "Spades")));
        }

    }

标签： javalistlinked-listhashmap

解决方案

问题出在地图人口上，而不是在删除时：

如果您有以下情况：

Map<String, List> myMap = ...
List<String> myList = ...
myMap.put("1", myList);
myMap.put("2", myList);

然后，当您删除时，您将同时从两个地图条目中删除：

myMap.get("1").remove("Hearts");  // removes both from myMap.get("1") and MyMap.get("2")

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