Question

我遇到了麻烦，我知道哪里出了问题，但是无法将其放入代码中。我目前正在纸牌上工作，我想从LinkedHashMap的纸牌等级中删除特定的纸牌套装。我到处都在搜索针对我的特定问题的特定解决方案，但是找不到。我尝试使用entrySet，迭代器和逻辑解决方案，但是remove()操作似乎删除了value的所有记录，而无视key。下面的示例演示了我去除5颗心的尝试。

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entrySet：

{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]}

我希望地图包含以下内容：

{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，我将如何指定去除诉讼的钥匙？我不希望删除“心脏”的所有实例（如本例所示）。帮助和线索非常感谢！

编辑：

填充西装使关键点仅指向地图内存中的一个列表。

之前：

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));
        }

    }

之后：

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")));
        }

    }

Answer 1

问题出在地图上，而不是删除时：

如果您具有以下条件：

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

然后，当您将其删除时，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")

从LinkedHashMap中删除包含键<string>和值LinkedList <string>的特定值

1 个答案: