Map集合中是通过什么方法判断键值是否相同的呢？ - 黑马程序员技术交流社区

import java.util.*;
class MyMap
{
public static void main(String[] args)
{
Map<String,String> mymap = new HashMap<String,String>();//定义一个Map集合
//向集合中添加元素
/*
1.通过put方法存元素的时候，返回值是该键之前对应的值
2.在HashMap中null可以当作键值或者Value值存在。
*/
sop(mymap.put("01","Zhanghongmei"));
sop(mymap.put("01","Zhanghongmei1"));
mymap.put("02","ZHM");
mymap.put("03","meimei");
mymap.put("05","ZHM");
mymap.put("05","zh");
mymap.put("04",null);
mymap.put(null,"MM");
//判断集合中是否存在指定键或者指定值
sop("mymap是否存在键值03："+mymap.containsKey("03"));
sop("mymap是否存在值ZHM："+mymap.containsValue("ZHM"));
sop("mymapz中是否存在键值003："+mymap.containsKey("003"));
sop("mymap中是否存在值ZHMx："+mymap.containsValue("ZHMx"));
//通过values方法，把集合中value值，取出。并存储在一个Collection集合中，返回该集合。
sop("mymap中所有的值是："+mymap.values());
}
public static<E> void sop(E e)
{
System.out.println(e);
}
}

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import java.util.*;
class Student implements Comparable<Student>
{
private String name;
private int age;
Student(String name,int age)
{
this.name = name;
this.age = age;
}
public int compareTo(Student s)
{
int num = new Integer(this.age).compareTo(new Integer(s.age));
if(num==0)
return this.name.compareTo(s.name);
return num;
}
<font color="#ff0000"> public int hashCode() //复写hashCode方法</font>
{
return 1;
// return name.hashCode()+age*34;
}
public boolean equals(Object obj)
{
if(!(obj instanceof Student))
throw new ClassCastException("类型不匹配");
Student s = (Student)obj;
return this.name.equals(s.name) && this.age==s.age;
}
public String getName()
{
return name;
}
public int getAge()
{
return age;
}
public String toString()
{
return name+":"+age;
}
}
class MapTest
{
public static void main(String[] args)
{
HashMap<Student,String> hm = new HashMap<Student,String>();
hm.put(new Student("lisi1",21),"beijing");
hm.put(new Student("lisi1",21),"tianjin");
hm.put(new Student("lisi2",22),"shanghai");
hm.put(new Student("lisi3",23),"nanjing");
hm.put(new Student("lisi4",24),"wuhan");
//第二种取出方式 entrySet
Set<Map.Entry<Student,String>> entrySet = hm.entrySet();
Iterator<Map.Entry<Student,String>> iter = entrySet.iterator();
while(iter.hasNext())
{
Map.Entry<Student,String> me = iter.next();
Student stu = me.getKey();
String addr = me.getValue();
System.out.println(stu+"........."+addr);
}
}
}

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/*
需求：
每一个学生都有对应的归属地。
学生student，地址string。
学生属性：姓名，年龄。
注意：姓名和年龄相同的视为同一个学生。保证学生的唯一性。
*/
import java.util.*;
class Student implements Comparable<Student>
{
private String name;
private int age;
Student(String name,int age)
{
this.name = name;
this.age = age;
}
public int compareTo(Student s)
{
int num = new Integer(this.age).compareTo(new Integer(s.age));
if(num==0)
return this.name.compareTo(s.name);
return num;
}
public int hashCode()//散列算法
{
return this.name.hashCode()+this.age*34;
}
public boolean equals(Object obj)//必须要有2个元素才具备自动调用该方法的前提,并且这里不能指定泛型
{
if(!(obj instanceof Student))//若传进来的不是student类
throw new ClassCastException("类型转换异常"); //直接报异常
Student s = (Student)obj;
return this.name.equals(s.name) && this.age==s.age;
}
public String getName()
{
return name;
}
public int getAge()
{
return age;
}
public String toString()
{
//return this.getName()+"..."+this.getAge();
return this.name+".."+this.age;
}
}
//取出方式1：keySet()
class MapTestKS
{
public static void main(String[] args)
{
HashMap<Student,String> map = new HashMap<Student,String>();
map.put(new Student("haha-1",1),"beijing");
//System.out.println(map.put(new Student("haha-1",1),"nanjing"));//put返回被覆盖的值 beijing
map.put(new Student("haha-1",1),"nanjing");//Map映射的特性，键相同，新值覆盖旧值
map.put(new Student("haha-3",3),"tianjing");
Set<Student> s = map.keySet();//导出所有的key存入一个set集合
Iterator<Student> it = s.iterator();
while(it.hasNext())
{
Student key = it.next();//逐个迭代当前键
String addr = map.get(key);//取出当前键所对应的值
System.out.println(key+"......"+addr);//输出
}
}
}
//取出方式2：Map.Entry
class MapTestME
{
public static void main(String[] args)
{
//建立一个映射集合容器
HashMap<Student,String> hm = new HashMap<Student,String>();
//存入映射元素，夫妻
hm.put(new Student("hehe-1",21),"yunnan");
hm.put(new Student("hehe-1",21),"guizhou");//重复键，覆盖
hm.put(new Student("hehe-3",23),"sichuan");
//通过hm.entrySet(),将这些夫妻的结婚证存入一个容器
Set<Map.Entry<Student,String>> sme = hm.entrySet();
//迭代,类型为结婚证
Iterator<Map.Entry<Student,String>> it = sme.iterator();
while(it.hasNext())
{
Map.Entry<Student,String> me = it.next();//将来结婚证一个一个的拿出来
System.out.println(me.getKey());//取出结婚证里的丈夫
System.out.println(me.getValue()); //取出结婚证里的妻子
}
}
}

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public V put(K key, V value) {
if (key == null)
return putForNullKey(value);
int hash = hash(key.hashCode());
int i = indexFor(hash, table.length);
for (Entry<K,V> e = table[i]; e != null; e = e.next) {
Object k;
if (e.hash == hash && ((k = e.key) == key || key.equals(k))) { //判断在这里。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。
V oldValue = e.value;
e.value = value;
e.recordAccess(this);
return oldValue;
}
}
modCount++;
addEntry(hash, key, value, i);
return null;
}

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