Posted on 2012-01-17 10:58
yuhaibo736 閱讀(213)
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集合的一個(gè)很重要的操作---遍歷，學(xué)習(xí)了三種遍歷方法，三種方法各有優(yōu)缺點(diǎn)~~
/*
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package cn.tsp2c.liubao;
import java.util.Collection;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Set;
import java.util.TreeMap;
/**
*
* @author Administrator
*/
public class TestMap {
public static void main(String[] args) {
Map<String, Student> map = new HashMap<String, Student>();
Student s1 = new Student("宋江", "1001", 38);
Student s2 = new Student("盧俊義", "1002", 35);
Student s3 = new Student("吳用", "1003", 34);
map.put("1001", s1);
map.put("1002", s2);
map.put("1003", s3);
Map<String, Student> subMap = new HashMap<String, Student>();
subMap.put("1008", new Student("tom", "1008", 12));
subMap.put("1009", new Student("jerry", "1009", 10));
map.putAll(subMap);
work(map);
workByKeySet(map);
workByEntry(map);
}
//最常規(guī)的一種遍歷方法，最常規(guī)就是最常用的，雖然不復(fù)雜，但很重要，這是我們最熟悉的，就不多說了?。?br /> public static void work(Map<String, Student> map) {
Collection<Student> c = map.values();
Iterator it = c.iterator();
for (; it.hasNext();) {
System.out.println(it.next());
}
}
//利用keyset進(jìn)行遍歷，它的優(yōu)點(diǎn)在于可以根據(jù)你所想要的key值得到你想要的 values，更具靈活性??！
public static void workByKeySet(Map<String, Student> map) {
Set<String> key = map.keySet();
for (Iterator it = key.iterator(); it.hasNext();) {
String s = (String) it.next();
System.out.println(map.get(s));
}
}
//比較復(fù)雜的一種遍歷在這里，呵呵~~他很暴力哦，它的靈活性太強(qiáng)了，想得到什么就能得到什么~~
public static void workByEntry(Map<String, Student> map) {
Set<Map.Entry<String, Student>> set = map.entrySet();
for (Iterator<Map.Entry<String, Student>> it = set.iterator(); it.hasNext();) {
Map.Entry<String, Student> entry = (Map.Entry<String, Student>) it.next();
System.out.println(entry.getKey() + "--->" + entry.getValue());
}
}
}
class Student {
private String name;
private String id;
private int age;
public Student(String name, String id, int age) {
this.name = name;
this.id = id;
this.age = age;
}
@Override
public String toString() {
return "Student{" + "name=" + name + "id=" + id + "age=" + age + '}';
}
}