2020/3/31学习笔记-day30

java-day30

集合的使用总结

怎么往集合中放数据
怎么从集合中删除数据
怎么从集合中找数据
怎么循环遍历集合中所有数据

List接口的采用实现：

ArrayList 实现类

LinkedList 实现类

Vector 实现类

Set集合的实现类

HashSet直接实现了Set接口

Set还有一个子接口SortedSet（可以进行数据排序）

SortedSet下有一个常用的实现类TreeSet

Set/List接口的特点：

Set接口的特点	List接口的特点
无序不可重复	有序可重复
没有定义额外的抽象方法，都是从父接口中继承过来的	有自己新的方法

public class SetTest {
	
	public static void main(String[] args){
		SetTest x = new SetTest();
		x.test1();
	}

	public void test1(){
		Set set1 = new HashSet();
		set1.add("set1");
		set1.add("set2");
		set1.add("set3");
		set1.add("set4");
		set1.add("set5");
		this.showCollection(set1);
	}

	private void showCollection(Collection c){
		Iterator it = c.iterator();
		System.out.println("--------begin-----------");
		while(it.hasNext()){
			Object obj = it.next();
			System.out.println("\t"+obj);	
		}
		System.out.println("---------end------------");	
	}
} 

Set是无重复的，且内部按照hashCode值来排序的（一般我们认为无序是因为hashCode不确定的）

//重写hashCode按照age
public void test2(){
    Set set =null;
    set = new HashSet();
    set.add(new Student(1L,"TOM",21));
    set.add(new Student(2L,"JACK",24));
    set.add(new Student(3L,"MARRY",22));
    set.add(new Student(4L,"JAMES",23));
    set.add(new Student(5L,"CURRY",20));
    set.add(new Student(5L,"CURRY",20));
    this.showCollection(set);
}
public int hashCode(){
    return age;
}

//重写hashCode按照id
public void test2(){
    Set set =null;
    set = new HashSet();
    set.add(new Student(1L,"TOM",21));
    set.add(new Student(2L,"JACK",24));
    set.add(new Student(3L,"MARRY",22));
    set.add(new Student(4L,"JAMES",23));
    set.add(new Student(5L,"CURRY",20));
    set.add(new Student(5L,"CURRY",20));
    this.showCollection(set);
}
public int hashCode(){
    return (int)this.id;
}

String内部也重写了hashCode方法

public boolean equals(Object anObject) {
    if (this == anObject) {
        return true;
    }
    if (anObject instanceof String) {
        String anotherString = (String)anObject;
        int n = value.length;
        if (n == anotherString.value.length) {
            char v1[] = value;
            char v2[] = anotherString.value;
            int i = 0;
            while (n-- != 0) {
                if (v1[i] != v2[i])
                    return false;
                i++;
            }
            return true;
        }
    }
    return false;
}

public int hashCode() {
    int h = hash;
    if (h == 0 && value.length > 0) {
        char val[] = value;

        for (int i = 0; i < value.length; i++) {
            h = 31 * h + val[i];
        }
        hash = h;
    }
    return h;
}