我该如何按字母顺序对列表进行排序？

Question

我该如何按字母顺序对列表进行排序？

javalistsortingcollectionsalphabetical

224

我有一个包含国家名称的 List<String> 对象，如何按字母顺序对此列表进行排序？

- Lennie

16个回答

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- ASR · Answer 1

通过使用Collections.sort()，我们可以对列表进行排序。

public class EmployeeList {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        List<String> empNames= new ArrayList<String>();

        empNames.add("sudheer");
        empNames.add("kumar");
        empNames.add("surendra");
        empNames.add("kb");

        if(!empNames.isEmpty()){

            for(String emp:empNames){

                System.out.println(emp);
            }

            Collections.sort(empNames);

            System.out.println(empNames);
        }
    }
}

输出：

sudheer
kumar
surendra
kb
[kb, kumar, sudheer, surendra]

- norbDEV · Answer 2

降序字母表：

List<String> list;
...
Collections.sort(list);
Collections.reverse(list);

- Sabir Khan · Answer 3

Java 8，

countries.sort((country1, country2) -> country1.compareTo(country2));

如果 String's compareTo 不适合您的需求，您可以提供任何其他比较器。

- betelgeuse · Answer 4

public static void sortByAlphabetCountry(List<Employee> listCountry) {
    listCountry.sort((o1, o2) -> {
        return o1.getName().compareTo(o2.getName());
    });
}

- Vijay · Answer 5

在JAVA 8中同样适用：

//Assecnding order
listOfCountryNames.stream().sorted().forEach((x) -> System.out.println(x));

//Decending order
listOfCountryNames.stream().sorted((o1, o2) -> o2.compareTo(o1)).forEach((x) -> System.out.println(x));

- user4423251 · Answer 6

//Here is sorted List alphabetically with syncronized
package com.mnas.technology.automation.utility;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Iterator;
import java.util.List;

import org.apache.log4j.Logger;
/**
* 
* @author manoj.kumar
*/
public class SynchronizedArrayList {
static Logger log = Logger.getLogger(SynchronizedArrayList.class.getName());
@SuppressWarnings("unchecked")
public static void main(String[] args) {

List<Employee> synchronizedList = Collections.synchronizedList(new ArrayList<Employee>());
synchronizedList.add(new Employee("Aditya"));
synchronizedList.add(new Employee("Siddharth"));
synchronizedList.add(new Employee("Manoj"));
Collections.sort(synchronizedList, new Comparator() {
public int compare(Object synchronizedListOne, Object synchronizedListTwo) {
//use instanceof to verify the references are indeed of the type in question
return ((Employee)synchronizedListOne).name
.compareTo(((Employee)synchronizedListTwo).name);
}
}); 
/*for( Employee sd : synchronizedList) {
log.info("Sorted Synchronized Array List..."+sd.name);
}*/

// when iterating over a synchronized list, we need to synchronize access to the synchronized list
synchronized (synchronizedList) {
Iterator<Employee> iterator = synchronizedList.iterator();
while (iterator.hasNext()) {
log.info("Sorted Synchronized Array List Items: " + iterator.next().name);
}
}

}
}
class Employee {
String name;
Employee (String name) {
this.name = name;

}
}