java Map练习
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2022-05-18 18:57:46
描述学生，map容器，学生为键，地址为值，获取map中的元素。 public class MapDemo { public static void main(String[] args) { HashMap hm = new HashMap
描述学生，map容器，学生为键，地址为值，获取map中的元素。
public class mapdemo {
public static void main(string[] args) {
hashmap<student, string> hm = new hashmap<student, string>();
hm.put(new student("lisi1", 1), "北京");
hm.put(new student("lisi2", 2), "上海");
hm.put(new student("lisi3", 3), "天津");
hm.put(new student("lisi4", 4), "武汉");
//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>> i = entryset.iterator();
while (i.hasnext()) {
map.entry<student, string> stu = i.next();
system.out.println(stu.getkey() + "--" + stu.getvalue());
}
}
}
class student implements comparable<student> {
private string name;
private int age;
public student(string name, int age) {
this.name = name;
this.age = age;
}
@override
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;
}
@override
public int hashcode() {
return name.hashcode() + age * 34;
}
public string getname() {
return name;
}
public void setname(string name) {
this.name = name;
}
public int getage() {
return age;
}
public void setage(int age) {
this.age = age;
}
@override
public string tostring() {
return "student{" +
"name='" + name + '\'' +
", age=" + age +
'}';
}
@override
public int compareto(student o) {
int num = new integer(this.age).compareto(new integer(o.age));
if (num == 0) {
return this.name.compareto(o.name);
}
return num;
}
}
排序年龄：
public class mapdemo {
public static void main(string[] args) {
treemap<student, string> tm = new treemap<student, string>();
tm.put(new student("lisi4", 1), "武汉");
tm.put(new student("lisi1", 4), "北京");
tm.put(new student("lisi3", 2), "天津");
tm.put(new student("lisi2", 3), "上海");
set<map.entry<student, string>> entryset = tm.entryset();
iterator<map.entry<student, string>> it = entryset.iterator();
while (it.hasnext()) {
map.entry<student, string> stu = it.next();
system.out.println(stu.getkey() + "--" + stu.getvalue());
}
}
}
class student implements comparable<student> {
private string name;
private int age;
public student(string name, int age) {
this.name = name;
this.age = age;
}
@override
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;
}
@override
public int hashcode() {
return name.hashcode() + age * 34;
}
public string getname() {
return name;
}
public void setname(string name) {
this.name = name;
}
public int getage() {
return age;
}
public void setage(int age) {
this.age = age;
}
@override
public string tostring() {
return "student{" +
"name='" + name + '\'' +
", age=" + age +
'}';
}
@override
public int compareto(student o) {
int num = new integer(this.age).compareto(new integer(o.age));
if (num == 0) {
return this.name.compareto(o.name);
}
return num;
}
}
姓名排序：
public class mapdemo {
public static void main(string[] args) {
treemap<student, string> tm = new treemap<student, string>(new stucomparator());
tm.put(new student("lisi4", 1), "武汉");
tm.put(new student("lisi1", 4), "北京");
tm.put(new student("lisi3", 2), "天津");
tm.put(new student("lisi2", 3), "上海");
set<map.entry<student, string>> entryset = tm.entryset();
iterator<map.entry<student, string>> it = entryset.iterator();
while (it.hasnext()) {
map.entry<student, string> stu = it.next();
system.out.println(stu.getkey() + "--" + stu.getvalue());
}
}
}
class student implements comparable<student> {
private string name;
private int age;
public student(string name, int age) {
this.name = name;
this.age = age;
}
@override
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;
}
@override
public int hashcode() {
return name.hashcode() + age * 34;
}
public string getname() {
return name;
}
public void setname(string name) {
this.name = name;
}
public int getage() {
return age;
}
public void setage(int age) {
this.age = age;
}
@override
public string tostring() {
return "student{" +
"name='" + name + '\'' +
", age=" + age +
'}';
}
@override
public int compareto(student o) {
int num = new integer(this.age).compareto(new integer(o.age));
if (num == 0) {
return this.name.compareto(o.name);
}
return num;
}
}
class stucomparator implements comparator<student> {
@override
public int compare(student o1, student o2) {
int num = o1.getname().compareto(o2.getname());
if (num == 0)
return new integer(o1.getage()).compareto(new integer(o2.getage()));
return num;
}
}
获取字符串("aabbbcccccdfff")中字母出现的次数：
希望打印结果：a(1)c(2)......
有映射关系先想到map集合。
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map集合，打印结果有顺序，使用treemap
3.遍历字符数组，将每一个字母作为键去查map集合，返回null，就将该字母和1存入map集合中，返回不是null，获取次数加1，在存入集合，覆盖原来的键值对。
4.将map集合的数据变成指定的字符串形式返回。
public class demo {
public static void main(string[] args) {
string s = charcount("aa+++___bbbcccccdfff");
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 i = 0; i < chs.length; i++) {
if (!(chs[i]>='a' && chs[i]<='z' || chs[i]>='a' && chs[i]<='z'))
continue;
integer value = tm.get(chs[i]);
if (value != null)
count = value;
count++;
tm.put(chs[i], count);
count = 0;
}
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();
}
}
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