我有两个 ArrayList
。
一种是 Double
类型并包含距离。另一种是 String
类型,包含城市类型。
这些是列表:
private ArrayList<Double> distance_list = new ArrayList<Double>();
distance.add(123.43);
distance.add(450.43);
distance.add(230.65);
private ArrayList<String> city_list = new ArrayList<String>();
city_list.add("Munich");
city_list.add("Berlin");
city_list.add("Frankfurt");
Collections.sort(distance_list); //to get the closest distance first
如何按照与匹配距离相同的顺序对 city_list
进行排序?
答案 0 :(得分:0)
NavigableMap
/SortedMap
改用 Map
。映射以键值关系将两个项目配对在一起。请求访问其价值合作伙伴的密钥。
使用 NavigableMap
或 SortedMap
保持键的顺序。
NavigableMap< Double , String > map = new TreeMap<>() ;
map.add( 123.43d , "Munich" ) ;
…
然后您可以循环该地图。请参阅:How to Loop in NavigableMap in Java;
for( Double key : map.keySet() )
{
String city = map.get( key ) ;
System.out.println( city + " = " + key ) ;
}
这是我制作的图表图形,作为与 Java 11 捆绑在一起的 Map
实现的概述。您可以看到 SortedMap
接口首先被添加,然后由 NavigableMap
接口扩展。
答案 1 :(得分:0)
您需要创建一个将两个值配对并实现 Comparable 接口的类,例如:
public class City implements Comparable<City> {
private Double distance;
private String name;
// constructor, getters, setters
@Override
public int compareTo(City city) {
return (double)(this.distance - city.getDistance());
}
}
然后构建一个 Treemap 或其他一些保持排序的 Map 类型。我假设您希望按距离排序。
//Create TreeMap
TreeMap<Double, String> cities = new TreeMap();
//Add cities
cities.put(123.43, "Munich");
项目将按距离值排序!
如果您需要更多帮助,请尝试并发布您的代码。
答案 2 :(得分:0)
你可以分两步完成
class DistanceCity implements Comparable<DistanceCity> {
private Double distance;
private String city;
public DistanceCity(Double distance, String city) {
this.distance = distance;
this.city = city;
}
@Override
public int compareTo(DistanceCity o) {
return distance.compareTo(o.distance);
}
@Override
public String toString() {
return "DistanceCity{" + "distance=" + distance + ", city='" + city + "'}";
}
}
private ArrayList<DistanceCity> liste;
liste = new ArrayList<>();
liste.add(new DistanceCity(123.43,"Munich"));
liste.add(new DistanceCity(450.43,"Berlin"));
liste.add(new DistanceCity(230.65,"Frankfurt"));
System.out.println(liste); // unordered
//[DistanceCity{distance=123.43, city='Munich'}, DistanceCity{distance=450.43, city='Berlin'}, DistanceCity{distance=230.65, city='Frankfurt'}]
Collections.sort(liste);
System.out.println(liste); // ordered
// [DistanceCity{distance=123.43, city='Munich'}, DistanceCity{distance=230.65, city='Frankfurt'}, DistanceCity{distance=450.43, city='Berlin'}]
答案 3 :(得分:0)
class City implements Comparable<City>{
public String name = null;
public Double distance = 0.0;
public City(String name, Double distance){
this.name = name;
this.distance = distance;
}
@Override
public int compareTo(City o){
return this.distance.compareTo( o.distance );
}
@Override
public String toString(){
return this.name;
}
}
public class CitiesDemo{
public static void main (String[] args ){
City munich = new City("Munich", 123.43);
City berlin = new City("Berlin", 450.43);
City frankfurt = new City("Frankfurt", 230.65);
java.util.List<City> cities = new java.util.ArrayList<City>();
cities.add( munich );
cities.add( berlin );
cities.add( frankfurt );
java.util.Collections.sort( cities );
System.out.println( cities );
}
}
将返回:
~/ (main)$ javac CitiesDemo.java
~/ (main)$ java CitiesDemo
[Munich, Frankfurt, Berlin]
~/ (main)$
或者,您可以在此处进行测试:https://ideone.com/fbbLv6