Java中如何对LinkedList排序呢？

实现思路:
    方式1：
	  继承Comparable接口创建一个内部比较器
    方式2:
	  创建一个外部比较器，然后传入sort方法中或传入LinkedList构造器中

package com.java265.other;
import java.util.Collections;
import java.util.LinkedList;
import java.util.List;
class User2 implements Comparable {
	public int height;
	public String name;

	public User2() {
	}
	public User2(int height, String name) {
		this.height = height;
		this.name = name;
	}

	@Override
	public int compareTo(Object o) {
		return this.height - ((User2) o).height;
	}

	@Override
	public String toString() {
		return "User2 [height=" + height + ", name=" + name + "]";
	}
}

public class Test2 {
	/*
	 * java265.com 自定义类的排序方法分享
	 */

	@SuppressWarnings("unchecked")
	public static void main(String[] args) {
		List<User2> list = new LinkedList<>();
		list.add(new User2(10, "java265.com"));
		list.add(new User2(1, "java265.com-1"));
		list.add(new User2(20, "java265.com-20"));
		list.add(new User2(30, "java265.com-30"));

		// 排序方式1
		Collections.sort(list);
		System.out.println(list);

		System.out.println("----------");
		// 排序方式2

		Collections.sort(list,
				(o1, o2) -> {
					return ((User2) o2).height - ((User2) o1).height;
				});
		System.out.println(list);
	}
}
------运行以上代码，将输出以下信息-----
[User2 [height=1, name=java265.com-1], User2 [height=10, name=java265.com], User2 [height=20, name=java265.com-20], User2 [height=30, name=java265.com-30]]
----------
[User2 [height=30, name=java265.com-30], User2 [height=20, name=java265.com-20], User2 [height=10, name=java265.com], User2 [height=1, name=java265.com-1]]