【java】使用lambda和函数接口Comparator

public class Demo {
/*
* 实现分组
*/
@Test
public void testWhat() {
　　　　 //注：Employee的属性有：{id,name,age,salary,deptId}
List<Employee> list = Arrays.asList(
new Employee(0, "赵", 35, 5555,1),
new Employee(1, "钱", 44, 5555,5),
new Employee(2, "孙", 22, 5555,2),
new Employee(3, "李", 26, 5555,1),
new Employee(4, "周", 15, 5555,5),
new Employee(5, "吴", 45, 5555,5)
);
//需求1:按年龄排序
Collections.sort(list,(emp1,emp2)->emp1.getAge()>emp2.getAge()?1:-1);
//升序
Collections.sort(list,(emp1,emp2)->emp1.getAge()>emp2.getAge()?-1:1);
//降序

list.forEach(System.out::println);
//需求2按部门分组
new What<Employee>().grouping(list,(e1,e2)->e1.getDepId()==e2.getDepId()?1:-1);
}
public class What<T>{
public void grouping(List<T>list,Comparator<T> comp) {
List<List<T>> groups = new ArrayList<List<T>>();
for(T t:list) {
boolean isGroupExists=false;
for(List<T> group:groups) {
if(comp.compare(t,group.get(0))>=0) {
group.add(t);
isGroupExists=true;
break;
}
}
if(!isGroupExists) {
List<T> group=new ArrayList<T>();
group.add(t);
groups.add(group);
}
}
}
}
}

 

 

　　注：Employee的属性有：{id,name,age,salary,deptId}

List<Employee> list = Arrays.asList(
    new Employee(0, "赵", 55, 5555,1),
    new Employee(0, "钱", 55, 5555,5),
    new Employee(0, "孙", 55, 5555,2),
    new Employee(0, "李", 55, 5555,1),
    new Employee(0, "周", 55, 5555,5),
    new Employee(0, "吴", 55, 5555,5)
    );

//需求1实现排序