---------------------- android培训、java培训、期待与您交流! ----------------------1.集合(Map概述)
2.集合(Map子类对象特点)
3.集合(Map共性方法)
/*
Map集合:该集合存储键值对.一对一对往里存.而且要保证键的唯一性.
1.添加.
put(K key,V value)
putAll(Map<? extends K,? extends V> m)
2.删除.
clear()
remove(Object key)
3.判断.
containValue(Object value)
containKey(Object key)
isEmpty()
4.获取
get(Object key)
size()
values()
entrySet()
keySet()
Map
|--Hashtable:底层是哈希表数据结构,不可以存入null键null值.
该集合是线程同步的.jdk1.0效率低.
|--HashMap:底层是哈希表数据结构,允许使用null键null值.
该集合是不同步的.将hashtable替代,jdk1.2效率高.
|--TreeSet:底层是二叉树数据结构.线程不同步.可以用于给map集合中的键进行排序.
和Set很像.
其实大家,Set底层就是使用的Map集合.
*/
import java.util.*;
class MapDemo
{
public static void main(String[] args)
{
Map<String,String> map = new HashMap<String,String>();
//添加元素,如果出现添加时,相同的键.那么后添加的值会覆盖原有键对应值.
//并用put方法返回被覆盖的值.
System.out.println("put:"+map.put("01","zhangsan1"));
System.out.println("put:"+map.put("01","wnagwu"));
map.put("02","zhangsan2");
map.put("03","zhangsan3");
System.out.println("containsKey:"+map.containsKey("002"));
//System.out.println("remove:"+map.remove("02"));
System.out.println("get:"+map.get("023"));
map.put("04",null);
System.out.println("get:"+map.get("04"));
//可以通过get方法的返回值来判断一个键是否存在.通过返回null来判断.
//获取map集合中的所有值.
Collection<String> coll = map.values();
System.out.println(coll);
System.out.println(map);
}
}
4.集合(Map-keySet)
5.集合(Map-entrySet)
/*
map集合的两种取出方式:
1.Set<k> keySet:将map中所有的键存入到Set集合.因为set具备迭代器.
所以可以迭代方式取出所有的键,根据get方法.获取每一个键对应的值.
Map集合的取出原理:将map集合转成set集合.通过迭代器取出.
2.Set<Map.Entry<k,v>> entrySet:将map集合中的映射关系存入到了set集合中,
而这个关系的数据类型就是Map.Entry
Entry其实就是Map中的一个static内部接口.
为什么要定义在内部呢?
因为只有有了Map集合,有了键值对,才会有键值的映射关系.
关系属于Map集合中的一个内部事物.
而且该事物在直接访问Map集合中元素.
*/
import java.util.*;
class MapDemo2
{
public static void main(String[] args)
{
Map<String,String> map = new HashMap<String,String>();
map.put("02","zhangsan2");
map.put("03","zhangsan3");
map.put("01","zhangsan1");
map.put("04","zhangsan4");
//将Map集合中的映射关系取出.存入到Set集合中.
Set<Map.Entry<String,String>> entrySet = map.entrySet();
Iterator<Map.Entry<String,String>> it = entrySet.iterator();
while(it.hasNext())
{
Map.Entry<String,String> me = it.next();
String key = me.getKey();
String value = me.getValue();
System.out.println(key+":"+value);
}
//先获取map集合的所有键的Set集合,keySet();
Set<String> keySet = map.keySet();
//有了Set集合.就可以获取其迭代器.
Iterator<String> it = keySet.iterator();
while(it.hasNext())
{
String key = it.next();
//有了键可以通过map集合的get方法获取其对应的值.
String value = map.get(key);
System.out.println("key:"+key+",value:"+value);
}
}
}
/*
Map.Entry 其实也是一个接口,它是Map接口中的一个内部接口.
interface Map
{
public static interface Entry
{
public abstract Object getKey();
public abstract Object getValue();
}
}
class HashMap implements Map
{
class Hash implements Map.Entry
{
public Object getKey(){}
public Object getValue(){}
}
*/
6.集合(Map练习)
/*
每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.
思路:
1.描述学生.
2.定义map容器.将学生作为键,地址作为值.存入.
3.获取map集合中的元素.
*/
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 hasCode()
{
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("lisi",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");
//第一种取出方式. keySet
Set<Student> keySet = hm.keySet();
Iterator<Student> it = keySet.iterator();
while(it.hasNext())
{
Student stu = it.next();
String addr = hm.get(stu);
System.out.println(stu+".."+addr);
}
//第二种取出方式.entrySet
Set<Map.Entry<Student,String>> entrySet = hm.entrySet();
Iterator<Map.Entry<Student,String>> it1 = entrySet.iterator();
while(it1.hasNext())
{
Map.Entry<Student,String> me = it1.next();
Student stu = me.getKey();
String addr = me.getValue();
System.out.println(stu+"......."+addr);
}
}
}
7.集合(TreeMap练习)
/*
需求:对学生对象的名字进行升序排序.
因为数据时以键值对形式存在的.
所以要使用可以排序的Map集合.TreeMap.
*/
import java.util.*;
class StuNameComparator implements Comparator<Student>
{
public int compare(Student s1,Student s2)
{
int num = s1.getName().compareTo(s2.getName());
if(num==0)
return new Integer(s1.getAge()).compareTo(s2.getAge());
return num;
}
}
class MapTest2
{
public static void main(String[] args)
{
TreeMap<Student,String> tm = new TreeMap<Student,String>(new StuNameComparator());
tm.put(new Student("blisi3",23),"nanjing");
tm.put(new Student("lisi1",21),"beijing");
tm.put(new Student("alisi4",24),"wuhan");
tm.put(new Student("lisi1",21),"tianjin");
tm.put(new Student("lisi2",22),"shanghai");
Set<Map.Entry<Student,String>> entrySet = tm.entrySet();
Iterator<Map.Entry<Student,String>> it = entrySet.iterator();
while(it.hasNext())
{
Map.Entry<Student,String> me = it.next();
Student s = me.getKey();
String addr = me.getValue();
System.out.println(s+"::"+addr);
}
}
}
8.集合(TreeSet练习-字母出现的次数)
/*
练习:
"sdfgzxcvasdfxcvdf"获取该字符串中的字母出现的次数.
希望打印结果:a(1)c(2)....
通过结果发现,每一个字母都有对应的次数.
说明字母和次数之间都有映射关系.
注意了,当发现有映射关系时,可以选择map集合.
因为map集合中存放就是映射关系.
什么时候使用map集合呢?
当数据之间存在着映射关系时,就要先想map集合.
思路:
1.将字符串转换成字符数组.因为要对每一个字母进行操作.
2.定义一个map集合,因为打印结果的字母有顺序,所以使用TreeMap集合.
3.遍历字符数组.
将每一个字母作为键去查map集合.
如果返回null,将该字母和1存入到map集合中.
如果返回不是null,说明该字母在map集合存在并有对应次数.
那么就获取该次数并进行自增,然后将该字母和自增后的次数存入集合中.
覆盖调用原键所对应的值.
4.将map集合中的数据变成制定的字符串形式返回.
*/
import java.util.*;
class MapTest3
{
public static void main(String[] args)
{
String s = charCount("ak+abAf1c,dCkaAbc-defa");
System.out.println(s);
}
public static String charCount(String str)
{
char[] chs = str.toCharArray();
TreeMap<Character,Integer> tm = new TreeMap<Character,Integer>();
int count = 0;
for(int x=0; x<chs.length; x++)
{
if(!(chs[x]>='a' && chs[x]<='z' || chs[x]>='A' && chs[x]<='Z'))
continue;
Integer value = tm.get(chs[x]);
if(value!=null)
count = value;
count++;
tm.put(chs[x],count);//直接往集合中存储字符和数字,为什么可以,因为自动装箱.
/*
if(value==null)
{
tm.put(chs[x],1);
}
else
{
value = value+1;
tm.put(chs[x],value);
}
*/
}
StringBuilder sb = new StringBuilder();
Set<Map.Entry<Character,Integer>> entrySet = tm.entrySet();
Iterator<Map.Entry<Character,Integer>> it = entrySet.iterator();
while(it.hasNext())
{
Map.Entry<Character,Integer> me = it.next();
Character ch = me.getKey();
Integer value = me.getValue();
sb.append(ch+"("+value+")");
}
return sb.toString();
}
}
9.集合(Map扩展)
/*
map扩展知识.
map集合被使用是因为 具备映射关系.
"yureban" Student("01","zhangsan");
"yureban" Student("02","lisi");
"jiuyeban" "01" "wangwu";
"jiuyeban" "02" "zhouliu";
一个学校有多个教室.每一个教师都有名称
*/
import java.util.*;
class Student
{
private String id;
private String name;
Student(String id,String name)
{
this.id = id;
this.name = name;
}
public String toString()
{
return id+":"+name;
}
}
class MapDemo3
{
public static void demo()
{
HashMap<String,List<Student>> czbk = new HashMap<String,List<Student>>();
List<Student> reyu = new ArrayList<Student>();
List<Student> jiuye = new ArrayList<Student>();
czbk.put("yureban",reyu);
czbk.put("jiuyeban",jiuye);
reyu.add(new Student("01","zhangsan"));
reyu.add(new Student("04","wangwu"));
jiuye.add(new Student("01","zhouqi"));
jiuye.add(new Student("02","zhaoli"));
Iterator<String> it = czbk.keySet().iterator();
while(it.hasNext())
{
String roomName = it.next();
List<Student> room = czbk.get(roomName);
System.out.println(roomName);
getInfos(room);
}
}
public static void getInfos(List<Student> list)
{
Iterator<Student> it = list.iterator();
while(it.hasNext())
{
Student s = it.next();
System.out.println(s);
}
}
public static void main(String[] args)
{
demo();
/*
HashMap<String,List<Stduent>> czbk = new HashMap<String,List<Student>>();
HashMap<String,String> yure = new HashMap<String,String>();
HashMap<String,String> jiuye = new HashMap<String,String>();
czbk.put("yureban",yure);
czbk.put("jiuyeban",jiuye);
yure.put("01","zhangsan");
yure.put("02",lisi);
jiuye.put("01","zhaliu");
jiuye.put("02","wangwu");
//遍历czbk集合.获取所有的教室.
Iterator<String> it = czbk.keySet().iterator();
while(it.hasNext())
{
String roomName = it.next();
HashMap<String,String> room = cabk.get(roomName);
System.out.println(room);
}
getStudentInfo(jiuye);
getStudentInfo(yure);
*/
}
public static void getStudentInfo(HashMap<String,String> roomMap)
{
Iterator<String> it = roomMap.keySet().iterator();
while(it.hasNext())
{
String id = ir.next();
String name = roomMap.get(id);
System.out.println(id+":"+name);
}
}
}
---------------------- android培训、 java培训、期待与您交流! ----------------------详细请查看: http://edu.youkuaiyun.com/heima