【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