----測試類

package TreeSetTest;

import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;
import java.util.TreeSet;


public class TreeSetTest {
	
	
	public static void main(String[] args) {
		/*map迴圈輸出*/
		Map<String,Object> map = new HashMap<String,Object>();
		map.put("1", "a");
		map.put("2", "a");
		map.put("3", "a");
		map.put("4", "a");
		Set<Entry<String, Object>> set = map.entrySet();//map中可以看做是多個map.entry的介面物件
		System.out.println(set);//[3=a, 2=a, 1=a, 4=a]
		Iterator<Entry<String, Object>>  it = set.iterator();//Collection實現了Iterable<T> 介面，故可以該介面提供的迭代方法
		while (it.hasNext()) {
			Entry<String, Object> entry = it.next();
			System.out.println(entry.getKey() + "==" + entry.getValue());//再利用Map.Entry提供的方法，遍歷，map主要用來查詢，集合用來輸出hash都是無序的
		}
		/*TreeSet 實現物件比較,必須實現Comparable介面,自定義  排序和過濾重複的資訊*/
		Set<Student> setStu = new TreeSet<Student>();
		setStu.add(new Student("胡明明","24"));
		setStu.add(new Student("胡明明","24"));
		setStu.add(new Student("胡明明1","25"));
		setStu.add(new Student("胡明明2","25"));
		setStu.add(new Student("肖體秀","23"));
		setStu.add(new Student("王二小","22"));
		System.out.println(setStu);
		/**
		 *  雖然TreeSet能實現重複元素的判斷，但它只適用於排序類操作的環境下，而其它子類需要消除重複元素該怎麼做？？比如hashSet，hashmap
		 *  依靠Object類提供的兩個方法hashcode和equals  假設equals和hashcode返回的結果都一致則重複
		 */
		Set<Book> hashSet = new HashSet<Book>();
		hashSet.add( new Book("java", 78));
		hashSet.add( new Book("java", 78));
		hashSet.add(new Book("php", 78));
		System.out.println(hashSet);//[Book [titie=java, price=78], Book [titie=php, price=78]]
		//為了確保map中key的唯一性，也可以重寫equals和hashcode方法
		Map<Book,Integer> mmp = new HashMap<Book, Integer>();
		mmp.put( new Book("js", 78),1);
		mmp.put( new Book("js", 78),1);
		mmp.put(new Book("html", 78),1);
		System.out.println(mmp);//{Book [titie=js, price=78]=1, Book [titie=html, price=78]=1}
	}
	 
}

----學生類

package TreeSetTest;

public class Student implements Comparable<Student>{
		private String name;
		private String age;
		
		public Student(String name, String age) {
			super();
			this.name = name;
			this.age = age;
		}
		public String getName() {
			return name;
		}
		public void setName(String name) {
			this.name = name;
		}
		public String getAge() {
			return age;
		}
		public void setAge(String age) {
			this.age = age;
		}
		@Override
		//備註：比較物件時TreeSet<Objetc>，假設類中存了5個屬性，但是我們只比較了3個，並且這3個屬性還相同，則TreeSet會認為這個物件也相同，從而會新的替換舊的（有序），造成資料遺失
		public int compareTo(Student o) {
			//如何比較自己自定義吧，我這裡就暫時認定為假設名字相同和年齡相同則認為元素重複，在比較年齡，按大小排序
			if(this.age.equals(o.age) && this.name.equals(o.name)){
				return 0;
			}else if(Integer.parseInt(this.age )>=Integer.parseInt(o.age)){
				return 1;
			}else{
				return -1;
			}
		}
		@Override
		public String toString() {
			return name +":" + age;
		}
		
	
}

----Book類


package TreeSetTest;

public class Book {
	private String titie;
	private int price;

	public Book(String titie, int price) {
		super();
		this.titie = titie;
		this.price = price;
	}
	public String getTitie() {
		return titie;
	}
	public void setTitie(String titie) {
		this.titie = titie;
	}
	public int getPrice() {
		return price;
	}
	public void setPrice(int price) {
		this.price = price;
	}
	@Override
	public String toString() {
		return "Book [titie=" + titie + ", price=" + price + "]";
	}
	@Override
	public int hashCode() {
		final int prime = 31;
		int result = 1;
		result = prime * result + price;
		result = prime * result + ((titie == null) ? 0 : titie.hashCode());
		return result;
	}
	@Override
	public boolean equals(Object obj) {
		if (this == obj)
			return true;
		if (obj == null)
			return false;
		if (getClass() != obj.getClass())
			return false;
		Book other = (Book) obj;
		if (price != other.price)
			return false;
		if (titie == null) {
			if (other.titie != null)
				return false;
		} else if (!titie.equals(other.titie))
			return false;
		return true;
	}
	
}