对HashMap进行排序,同时保持重复

Question

我试图以HashMap两种方式排序.默认方式:按字母顺序排列,第二种方式:按键数字,数字越高,位于顶部.我已经四处寻找,但找不到任何关于这个主题的内容,我发现的内容也不起作用.如果不能对它们进行排序(我希望顶部有最高键的人,随着人们拥有较低的键而减少,则按字母顺序对所有其余键进行排序(将0作为键的人).

这是我到目前为止所尝试的:

private HashMap<String, Integer> userGains = new HashMap<String, Integer>();

public void sortGains(int skill, int user) {
    userGains.put(users.get(user).getUsername(), users.get(user).getGainedExperience(skill));
    HashMap<String, Integer> map = sortHashMap(userGains);
    for (int i = 0; i < map.size(); i++) {
        Application.getTrackerOutput().getOutputArea(skill).append(users.get(user).getUsername() + " gained " + map.get(users.get(user).getUsername()) + "  experience in " + getSkillName(skill) + ".\n");
    }
}

public LinkedHashMap<String, Integer> sortHashMap(HashMap<String, Integer> passedMap) {
    List<String> mapKeys = new ArrayList<String>(passedMap.keySet());
    List<Integer> mapValues = new ArrayList<Integer>(passedMap.values());
    LinkedHashMap<String, Integer> sortedMap = new LinkedHashMap<String, Integer>();

    Collections.sort(mapValues);
    Collections.sort(mapKeys);

    Iterator<Integer> it$ = mapValues.iterator();
    while (it$.hasNext()) {
        Object val = it$.next();
        Iterator<String> keyIt = mapKeys.iterator();
        while (keyIt.hasNext()) {
            Object key = keyIt.next();
            String comp1 = passedMap.get(key).toString();
            String comp2 = val.toString();
            if (comp1.equals(comp2)) {
                passedMap.remove(key);
                mapKeys.remove(key);
                sortedMap.put((String) key, (Integer) val);
                break;
            }
        }
    }
    return sortedMap;
}

既然你不能在这里运行SSCCE:

private HashMap<String, Integer> userGains = new HashMap<String, Integer>();

private Object[][] testUsers = { { "Test user", 15 }, { "Test", 25 }, { "Hello", 11 }, { "I'm a user", 21 }, { "No you're not!", 14 }, { "Yes I am!", 45 }, { "Oh, okay.  Sorry about the confusion.", 0 }, { "It's quite alright.", 0 } };

public static void main(String[] arguments) {
    new Sorting().sortGains();
}

public void sortGains() {
    for (Object[] test : testUsers) {
        userGains.put((String) test[0], (Integer) test[1]);
    }
    HashMap<String, Integer> map = sortHashMap(userGains);
    for (int i = 0; i < map.size(); i++) {
        System.out.println(testUsers[i][0] + " gained " + map.get(testUsers[i][0]) + "  experience.");
    }
}

public LinkedHashMap<String, Integer> sortHashMap(HashMap<String, Integer> passedMap) {
    List<String> mapKeys = new ArrayList<String>(passedMap.keySet());
    List<Integer> mapValues = new ArrayList<Integer>(passedMap.values());
    LinkedHashMap<String, Integer> sortedMap = new LinkedHashMap<String, Integer>();

    Collections.sort(mapValues);
    Collections.sort(mapKeys);

    Iterator<Integer> it$ = mapValues.iterator();
    while (it$.hasNext()) {
        Object val = it$.next();
        Iterator<String> keyIt = mapKeys.iterator();
        while (keyIt.hasNext()) {
            Object key = keyIt.next();
            String comp1 = passedMap.get(key).toString();
            String comp2 = val.toString();
            if (comp1.equals(comp2)) {
                passedMap.remove(key);
                mapKeys.remove(key);
                sortedMap.put((String) key, (Integer) val);
                break;
            }
        }
    }
    return sortedMap;
}

该计划的输出目前是:

Test user gained 15  experience.
Test gained 25  experience.
Hello gained 11  experience.
I'm a user gained 21  experience.
No you're not! gained 14  experience.
Yes I am! gained 45  experience.
Oh, okay.  Sorry about the confusion. gained 0  experience.
It's quite alright. gained 0  experience.

当我需要它时:

Yes I am! gained 45  experience. // start numeric sorting here, by highest key.
Test gained 25  experience.
I'm a user gained 21  experience.
Test user gained 15  experience.
No you're not! gained 14  experience.
Hello gained 11  experience.
It's quite alright. gained 0  experience. // start alphabetical sorting here, if possible.
Oh, okay.  Sorry about the confusion. gained 0  experience.

任何见解？

Answer 1

这是不可能的排序HashMap 在所有.根据定义,a HashMap中的键是无序的.如果您想要Map订购您的钥匙,请使用TreeMap带有适当Comparator物品的钥匙.如果要以多种方式访问​​相同的数据,可以TreeMaps使用不同的Comparators 创建多个.