java - リストをアルファベット順に並べ替えるにはどうすればよいですか?

Question

List<String>国名を含むオブジェクトがあります。このリストをアルファベット順に並べ替えるにはどうすればよいですか?

Answer 1

それらが文字列であると仮定すると、便利な静的メソッドを使用してsort…</p>

 java.util.Collections.sort(listOfCountryNames)

Answer 2

Solution with Collections.sort

If you are forced to use that List, or if your program has a structure like

• Create List
• Add some country names
• sort them once
• never change that list again

then Thilos answer will be the best way to do it. If you combine it with the advice from Tom Hawtin - tackline, you get:

java.util.Collections.sort(listOfCountryNames, Collator.getInstance());

Solution with a TreeSet

If you are free to decide, and if your application might get more complex, then you might change your code to use a TreeSet instead. This kind of collection sorts your entries just when they are inserted. No need to call sort().

Collection<String> countryNames = 
    new TreeSet<String>(Collator.getInstance());
countryNames.add("UK");
countryNames.add("Germany");
countryNames.add("Australia");
// Tada... sorted.

Side note on why I prefer the TreeSet

This has some subtle, but important advantages:

• It's simply shorter. Only one line shorter, though.
• Never worry about is this list really sorted right now becaude a TreeSet is always sorted, no matter what you do.
• You cannot have duplicate entries. Depending on your situation this may be a pro or a con. If you need duplicates, stick to your List.
• An experienced programmer looks at TreeSet<String> countyNames and instantly knows: this is a sorted collection of Strings without duplicates, and I can be sure that this is true at every moment. So much information in a short declaration.
• Real performance win in some cases. If you use a List, and insert values very often, and the list may be read between those insertions, then you have to sort the list after every insertion. The set does the same, but does it much faster.

Using the right collection for the right task is a key to write short and bug free code. It's not as demonstrative in this case, because you just save one line. But I've stopped counting how often I see someone using a List when they want to ensure there are no duplictes, and then build that functionality themselves. Or even worse, using two Lists when you really need a Map.

Don't get me wrong: Using Collections.sort is not an error or a flaw. But there are many cases when the TreeSet is much cleaner.

Answer 3

Java 8 Stream または Guava を使用して、新しい並べ替えられたコピーを作成できます。

// Java 8 version
List<String> sortedNames = names.stream().sorted().collect(Collectors.toList());
// Guava version
List<String> sortedNames = Ordering.natural().sortedCopy(names); 

別のオプションは、コレクション API を介してその場でソートすることです。

Collections.sort(names);

Answer 4

of の 2 つの引数を使用しますCollections.sort。Comparatorを介して取得できるような、大文字と小文字を適切に処理する (つまり、UTF16 順序付けではなくレキシカルを行う) 適切なものが必要になりますjava.text.Collator.getInstance。

Answer 5

を使用するCollections.sort()と、リストを並べ替えることができます。

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);
        }
    }
}

出力：

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

Answer 6

Java 8 、

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

String の compareToがニーズに合わない場合は、他のコンパレータを提供できます。

Answer 7

//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;

}
}