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
使用Streams的Java 8解决方案:
long count = animals.stream().filter(animal -> "bat".equals(animal)).count();
其他回答
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}
所以用传统的方法自己卷吧:
Map<String, Integer> instances = new HashMap<String, Integer>();
void add(String name) {
Integer value = instances.get(name);
if (value == null) {
value = new Integer(0);
instances.put(name, value);
}
instances.put(name, value++);
}
package traversal;
import java.util.ArrayList;
import java.util.List;
public class Occurrance {
static int count;
public static void main(String[] args) {
List<String> ls = new ArrayList<String>();
ls.add("aa");
ls.add("aa");
ls.add("bb");
ls.add("cc");
ls.add("dd");
ls.add("ee");
ls.add("ee");
ls.add("aa");
ls.add("aa");
for (int i = 0; i < ls.size(); i++) {
if (ls.get(i) == "aa") {
count = count + 1;
}
}
System.out.println(count);
}
}
输出:4
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();
}
}
