技术难题：带嵌套泛型的TreeSet如何实现自然排序？ - 黑马程序员技术交流社区

package blog9;
import java.util.TreeSet;
/**
* 泛型使用实例:
* 1、定义一个表示点的类，包含横坐标和纵坐标值，且值具有三种形式，string、Integer、double。
* 2、通过实现带泛型的comparable接口排序
* 难题：如何让Point<T>实现comparable<T>接口，因为这里带了泛型，那么TreeSet的自然排序如何实现
*/
public class TestGeneric {
public static void main(String[] args) {
//坐标为String类型
TreeSet<Point<String>> spSet = new TreeSet<Point<String>>();
spSet.add(new Point<String>("10","20"));
spSet.add(new Point<String>("5","30"));
spSet.add(new Point<String>("100","20"));
System.out.println(spSet);
//输出：[Point [x=5, y=30], Point [x=10, y=20], Point [x=100, y=20]]
TreeSet<Point<Integer>> ipSet = new TreeSet<Point<Integer>>();
ipSet.add(new Point<Integer>(40,60));
ipSet.add(new Point<Integer>(20,60));
ipSet.add(new Point<Integer>(330,60));
ipSet.add(new Point<Integer>(5,60));
System.out.println(ipSet);
//输出：[Point [x=5, y=60], Point [x=20, y=60], Point [x=40, y=60], Point [x=330, y=60]]
TreeSet<Point<Double>> dpSet = new TreeSet<Point<Double>>();
dpSet.add(new Point<Double>(5.5,62.1));
dpSet.add(new Point<Double>(200.5,30.54));
dpSet.add(new Point<Double>(0.7,6.0));
dpSet.add(new Point<Double>(500.54,20.4));
System.out.println(dpSet); //这里是以y升序的，在compareTo方法中实现
//[Point [x=0.7, y=6.0], Point [x=500.54, y=20.4], Point [x=200.5, y=30.54], Point [x=5.5, y=62.1]]
}
}
class Point<T> implements Comparable<Point<T>>{
private T x;
private T y;
public Point(T x,T y) {
this.x = x;
this.y = y;
}
@Override
public String toString() {
return "Point [x=" + x + ", y=" + y + "]";
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((x == null) ? 0 : x.hashCode());
result = prime * result + ((y == null) ? 0 : y.hashCode());
return result;
}
@SuppressWarnings("rawtypes")
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Point other = (Point) obj;
if (x == null) {
if (other.x != null)
return false;
} else if (!x.equals(other.x))
return false;
if (y == null) {
if (other.y != null)
return false;
} else if (!y.equals(other.y))
return false;
return true;
}
@Override
public int compareTo(Point<T> o) {
if(o.x instanceof String){
String s = (String)this.x;
return Integer.parseInt(s) - Integer.parseInt((String)o.x);
//感觉好麻烦，两个强转，失去了泛型的意义了
}
if(o.x instanceof Integer){
Integer i = (Integer)this.x;
return i.compareTo((Integer)o.x);
}
if(o.x instanceof Double){//当Point的类型为double时，用y值排序
Double d = (Double)this.y;
return d.compareTo((Double)o.y);
}
return 0;
}
}

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