我有一个数组列表,一个Java的集合类,如下所示:

ArrayList<String> animals = new ArrayList<String>();
animals.add("bat");
animals.add("owl");
animals.add("bat");
animals.add("bat");

如您所见,animals数组列表由3个bat元素和1个owl元素组成。我想知道在Collection框架中是否有返回蝙蝠出现次数的API,或者是否有另一种方法来确定出现次数。

我发现谷歌的集合Multiset确实有一个API,返回一个元素的总出现次数。但是这只与JDK 1.5兼容。我们的产品目前是JDK 1.6,所以我不能使用它。


当前回答

我不想让这种情况变得更困难,所以使用了两个迭代器 我有一个HashMap与LastName -> FirstName。我的方法应该删除具有dulicate FirstName的项目。

public static void removeTheFirstNameDuplicates(HashMap<String, String> map)
{

    Iterator<Map.Entry<String, String>> iter = map.entrySet().iterator();
    Iterator<Map.Entry<String, String>> iter2 = map.entrySet().iterator();
    while(iter.hasNext())
    {
        Map.Entry<String, String> pair = iter.next();
        String name = pair.getValue();
        int i = 0;

        while(iter2.hasNext())
        {

            Map.Entry<String, String> nextPair = iter2.next();
            if (nextPair.getValue().equals(name))
                i++;
        }

        if (i > 1)
            iter.remove();

    }

}

其他回答

实际上,Collections类有一个名为:frequency(Collection c, Object o)的静态方法,它返回你正在搜索的元素的出现次数,顺便说一下,这将为你完美地工作:

ArrayList<String> animals = new ArrayList<String>();
animals.add("bat");
animals.add("owl");
animals.add("bat");
animals.add("bat");
System.out.println("Freq of bat: "+Collections.frequency(animals, "bat"));
List<String> lst = new ArrayList<String>();

lst.add("Ram");
lst.add("Ram");
lst.add("Shiv");
lst.add("Boss");

Map<String, Integer> mp = new HashMap<String, Integer>();

for (String string : lst) {

    if(mp.keySet().contains(string))
    {
        mp.put(string, mp.get(string)+1);

    }else
    {
        mp.put(string, 1);
    }
}

System.out.println("=mp="+mp);

输出:

=mp= {Ram=2, Boss=1, Shiv=1}
Map<String,Integer> hm = new HashMap<String, Integer>();
for(String i : animals) {
    Integer j = hm.get(i);
    hm.put(i,(j==null ? 1 : j+1));
}
for(Map.Entry<String, Integer> val : hm.entrySet()) {
    System.out.println(val.getKey()+" occurs : "+val.getValue()+" times");
}
List<String> list = Arrays.asList("as", "asda", "asd", "urff", "dfkjds", "hfad", "asd", "qadasd", "as", "asda",
        "asd", "urff", "dfkjds", "hfad", "asd", "qadasd" + "as", "asda", "asd", "urff", "dfkjds", "hfad", "asd",
        "qadasd", "as", "asda", "asd", "urff", "dfkjds", "hfad", "asd", "qadasd");

方法1:

Set<String> set = new LinkedHashSet<>();
set.addAll(list);

for (String s : set) {

    System.out.println(s + " : " + Collections.frequency(list, s));
}

方法2:

int count = 1;
Map<String, Integer> map = new HashMap<>();
Set<String> set1 = new LinkedHashSet<>();
for (String s : list) {
    if (!set1.add(s)) {
        count = map.get(s) + 1;
    }
    map.put(s, count);
    count = 1;

}
System.out.println(map);

我不想让这种情况变得更困难,所以使用了两个迭代器 我有一个HashMap与LastName -> FirstName。我的方法应该删除具有dulicate FirstName的项目。

public static void removeTheFirstNameDuplicates(HashMap<String, String> map)
{

    Iterator<Map.Entry<String, String>> iter = map.entrySet().iterator();
    Iterator<Map.Entry<String, String>> iter2 = map.entrySet().iterator();
    while(iter.hasNext())
    {
        Map.Entry<String, String> pair = iter.next();
        String name = pair.getValue();
        int i = 0;

        while(iter2.hasNext())
        {

            Map.Entry<String, String> nextPair = iter2.next();
            if (nextPair.getValue().equals(name))
                i++;
        }

        if (i > 1)
            iter.remove();

    }

}