Java集合排序及java集合类详解(Collection、List、Ma(20)

java的集合排序

public static void main(String[] args) {

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map1.putAll(map2);//将map2全部元素放入map1中 map2.clear();//清空map2 System.out.println("map1 IsEmpty?="+map1.isEmpty()); System.out.println("map2 IsEmpty?="+map2.isEmpty()); // 根据键 "1" 移除键值对"1"-"aaa1" System.out.println("map1.remove(\"1\")="+map1.remove Map map1 = new HashMap(); Map map2 = new HashMap(); map1.put("1","aaa1"); map1.put("2","bbb2"); map2.put("10","aaaa10"); map2.put("11","bbbb11"); //根据键 "1" 取得值："aaa1" System.out.println("map1.get(\"1\")="+map1.get("1("1")); System.out.println("map1.get(\"1\")="+map1.get("1 System.out.println("map1 中的键值对的个数size = "+map1.size());

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System.out.println("entrySet="+map1.entrySet()); System.out.println("KeySet="+map1.keySet());//set System.out.println("values="+map1.values());//Collec

System.out.println("map1 是否包含键：11 = "+map1.containsKey("11"));

System.out.println("map1 是否包含值：aaa1 = "+map1.containsValue("aaa1"));

}

运行输出结果为：

map1.get("1")=aaa1

map1.remove("1")=aaa1

map1.get("1")=null

map1 IsEmpty?=false

map2 IsEmpty?=true

map1 中的键值对的个数size = 3

KeySet=[10, 2, 11]

values=[aaaa10, bbb2, bbbb11]

entrySet=[10=aaaa10, 2=bbb2, 11=bbbb11]

map1 是否包含键：11 = true

map1 是否包含值：aaa1 = false

}