690. Employee Importance

You are given a data structure of employee information, which includes the employee's unique id, his importance value and his directsubordinates' id.

For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is not direct.

Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all his subordinates.

Example 1:

Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
Output: 11
Explanation:

Employee 1 has importance value 5, and he has two direct subordinates: employee 2 and employee 3. They both have importance value 3. So the total importance value of employee 1 is 5 + 3 + 3 = 11.
题目含义：规定了一个员工信息的数据结构，包含了雇员的id(唯一)、重要值和他的直系下属的雇员id。一个员工至多只有一个直系领导，但是可以有多个下属。求出给定一个员工集合及领导的id，求出领导及其所有下属的重要值总和。

    public int getImportance(List<Employee> employees, int id) {
        int res = 0;
        HashMap<Integer, Employee> hm = new HashMap<Integer, Employee>();
        for(Employee employee : employees){
            hm.put(employee.id, employee);
        }
        LinkedList<Employee> que = new LinkedList<Employee>();
        que.add(hm.get(id));
        while(!que.isEmpty()){
            Employee cur = que.poll();
            res += cur.importance;
            for(int subId : cur.subordinates){
                que.add(hm.get(subId));
            }
        }
        return res; 
    }