Java中Arrays.sort()和Collections.sort()
1.简单示例
sort方法的使用很是的简单明了，下面的例子中，先定义一个比较Dog大小的Comparator，而后将其实例对象做为参数传给sort方法，经过此示例，你应该可以快速掌握Arrays.sort()的使用方法。java
import java.util.Arrays;
import java.util.Comparator;
class Dog{
int size;
public Dog(int s){
size = s;
}
}
class DogSizeComparator implements Comparator<Dog>{
@Override
public int compare(Dog o1, Dog o2) {
return o1.size - o2.size;
}
}
public class ArraySort {
public static void main(String[] args) {
Dog d1 = new Dog(2);
Dog d2 = new Dog(1);
Dog d3 = new Dog(3);
Dog[] dogArray = {d1, d2, d3};
printDogs(dogArray);
//或者采用匿名内部类
/**
Arrays.sort(dogArray, new Comparator<Dog>(){
@Override
public int compare(Dog o1, Dog o2) {
return o1.size - o2.size;
}
});
*/
Arrays.sort(dogArray, new DogSizeComparator());
printDogs(dogArray);
}
public static void printDogs(Dog[] dogs)
for(Dog d: dogs) System.out.print(d.size + " " );
System.out.println();
}
}
2.使用策略模式
这是策略模式(Strategy pattern)的一个完美又简洁的示例，值得一提的是为何这种场景下适合使用策略模式。
整体来讲，策略模式容许在程序执行时选择不一样的算法。好比在排序时，传入不一样的比较器(Comparator)，就采用不一样的算法。
根据上面的例子，假设你想要根据Dog的重量来进行排序，能够像下面这样，建立一个新的比较器来进行排序：算法
class Dog{
int size;
int weight;
public Dog(int s,int w){
size = s;
weight = w;
}
}
class DogSizeComparator implements Comparator<Dog>{
@Override
public int compare(Dogo1,Dogo2){
return o1.size - o2.size;
}
}
class DogWeightComparator implements Comparator<Dog>{
@Override
public int compare(Dogo1,Dogo2){
return o1.weight - o2.weight;
}
}
public class ArraySort{
public static void main(String[] args){
Dog d1 = new Dog(2,50);
Dog d2 = new Dog(1,30);
Dog d3 = new Dog(3,40);
Dog[] dogArray = {d1,d2,d3};
printDogs(dogArray);
Arrays.sort(dogArray, new DogSizeComparator());
printDogs(dogArray);
Arrays.sort(dogArray,new DogWeightComparator());
printDogs(dogArray);
}
public static void printDogs(Dog[]dogs){
for(Dog d : dogs) System.out.print("size=" + d.size + "weight=" + d.weight + "");
System.out.println();
}
}
Comparator是一个接口，因此sort方法中能够传入任意实现了此接口的类的实例，这就是策略模式的主要思想。ide
3.为什么使用"super"
若是使用“Comparator < T > c”那是很简单易懂的，可是sort的第2个参数里面的< ? super T >意味着比较器所接受的类型能够是T或者它的超类，为何是超类呢？ 答案是：这容许使用同一个比较器对不一样的子类对象进行比较。在下面的示例中很明显地演示了这一点：this
import java.util.Arrays;
import java.util.Comparator;
class Animal{
int size;
}
class Dog extends Animal{
public Dog(int s){
size = s;
}
}
class Cat extends Animal{
public Cat(int s){
size= s;
}
}
class AnimalSizeComparator implements Comparator<Animal>{
@Override
public int compare(Animal o1, Animal o2) {
return o1.size - o2.size;
}
//in this way, all sub classes of Animal can use this comparator.
}
public class ArraySort {
public static void main(String[] args) {
Dog d1 = new Dog(2);
Dog d2 = new Dog(1);
Dog d3 = new Dog(3);
Dog[] dogArray = {d1, d2, d3};
printDogs(dogArray);
Arrays.sort(dogArray, new AnimalSizeComparator());
printDogs(dogArray);
System.out.println();
//when you have an array of Cat, same Comparator can be used.
Cat c1 = new Cat(2);
Cat c2 = new Cat(1);
Cat c3 = new Cat(3);
Cat[] catArray = {c1, c2, c3};
printDogs(catArray);
Arrays.sort(catArray, new AnimalSizeComparator());
printDogs(catArray);
}
public static void printDogs(Animal[] animals){
for(Animal a: animals) System.out.print("size="+a.size + " ");
System.out.println();
}
}
4.Collections sort(List< T >,Comparator<? super T>) 用法
public class Student implements Comparable<Student> {
String name;
int age;
public Student(String name, int age) {
this.name = name;
this.age = age;
}
@Override
public String toString() {
return name + ":" + age;
}
@Override
public int compareTo(Student o) {
return Comparators.NAME.compare(this, o);
}
public static class Comparators {
public static Comparator<Student> NAME = new Comparator<Student>() {
@Override
public int compare(Student o1, Student o2) {
return o1.name.compareTo(o2.name);
}
};
public static Comparator<Student> AGE = new Comparator<Student>() {
@Override
public int compare(Student o1, Student o2) {
return o1.age - o2.age;
}
};
public static Comparator<Student> NAMEANDAGE = new Comparator<Student>() {
@Override
public int compare(Student o1, Student o2) {
int i = o1.name.compareTo(o2.name);
if (i == 0) {
i = o1.age - o2.age;
}
return i;
}
};
}
}
//用法
List<Student> studentList = new LinkedList<>();
Collections.sort(studentList, Student.Comparators.AGE);
5.Sorting HashMap by values（HashMap值排序）
package leetcode;
/**
* Created by slyuan on 17-4-18.
*/
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
public class SortMapByValue {
public static boolean ASC = true;
public static boolean DESC = false;
public static void main(String[] args) {
// Creating dummy unsorted map
Map<String, Integer> unsortMap = new HashMap<String, Integer>();
unsortMap.put("B", 55);
unsortMap.put("A", 80);
unsortMap.put("D", 20);
unsortMap.put("C", 70);
System.out.println("Before sorting......");
printMap(unsortMap);
System.out.println("After sorting ascending order......");
Map<String, Integer> sortedMapAsc = sortByComparator(unsortMap, ASC);
printMap(sortedMapAsc);
System.out.println("After sorting descindeng order......");
Map<String, Integer> sortedMapDesc = sortByComparator(unsortMap, DESC);
printMap(sortedMapDesc);
}
private static Map<String, Integer> sortByComparator(Map<String, Integer> unsortMap, final boolean order) {
List<Entry<String, Integer>> list = new LinkedList<Entry<String, Integer>>(unsortMap.entrySet());
// Sorting the list based on values
Collections.sort(list, new Comparator<Entry<String, Integer>>() {
public int compare(Entry<String, Integer> o1, Entry<String, Integer> o2) {
if (order) {
return o1.getValue().compareTo(o2.getValue());
} else {
return o2.getValue().compareTo(o1.getValue());
}
}
});
// Maintaining insertion order with the help of LinkedList
Map<String, Integer> sortedMap = new LinkedHashMap<String, Integer>();
for (Entry<String, Integer> entry : list) {
sortedMap.put(entry.getKey(), entry.getValue());
}
return sortedMap;
}
public static void printMap(Map<String, Integer> map) {
for (Entry<String, Integer> entry : map.entrySet()) {
System.out.println("Key : " + entry.getKey() + " Value : "+ entry.getValue());
}
}
}
//另外一种实现方式
public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue( Map<K, V> map ) {
List<Map.Entry<K, V>> list = new LinkedList<Map.Entry<K, V>>( map.entrySet() );
Collections.sort( list, new Comparator<Map.Entry<K, V>>() {
public int compare( Map.Entry<K, V> o1, Map.Entry<K, V> o2 ) {
return (o1.getValue()).compareTo( o2.getValue() );
}
} );
Map<K, V> result = new LinkedHashMap<K, V>();
for (Map.Entry<K, V> entry : list) {
result.put( entry.getKey(), entry.getValue() );
}
return result;
}
5.Sorting HashMap by key
Using the TreeMap you can sort the Map， 默认状况下，TreeMap对key进行升序排序code
Map<String, String> unsortMap = new HashMap<String, String>();
Map<String, String> treeMap = new TreeMap<String, String>(unsortMap);
for (String str : treeMap.keySet()) {
System.out.println(str);
}
注意：正常状况下Map是不可使用Collections.sort()方法进行排序的，不过能够将Map转换成list以后再进行排序。对象
List<Entry<String,Integer>> list = new LinkedList<>(unsortMap.entrySet());
Collections.sort(list, new Comparator<Entry<String, Integer>>() {
@Override
public int compare(Entry<String, Integer> e1, Entry<String, Integer> e2) {
return e1.getKey().compareTo(e2.getKey());
}
});
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