# 690 Employee Importance

### Problem

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

Note:

1. One employee has at most one direct leader and may have several subordinates.
2. The maximum number of employees won't exceed 2000.

### Solutions

```java
/*
// Employee info
class Employee {
    // It's the unique id of each node;
    // unique id of this employee
    public int id;
    // the importance value of this employee
    public int importance;
    // the id of direct subordinates
    public List<Integer> subordinates;
};
*/
class Solution {
    public int getImportance(List<Employee> employees, int id) {
        HashMap<Integer, Employee> map = new HashMap<>();
        for (Employee emp:employees) {
            map.put(emp.id, emp);
        }
        if (!map.containsKey(id)) {
            return 0;
        }
        Queue<Employee> q = new LinkedList<>();
        q.add(map.get(id));
        int res = 0;
        while (!q.isEmpty()) {
            Employee emp = q.poll();
            res += emp.importance;
            for (Integer empId:emp.subordinates) {
                q.add(map.get(empId));
            }
        }
        return res;
    }
}
```
