> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/search/employee-importance.md).

# Employee Importance

<https://leetcode.com/problems/employee-importance/description/>

> 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.

## Thoughts

思路很直白，从给定的员工往他的手下一直遍历下去。由于找员工只给了id, 因此还需要一个id和员工的map.

## Code

```
/*
// 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) {
        Map<Integer, Employee> map = new HashMap<>();
        for (Employee employee : employees) {
            map.put(employee.id, employee);
        }

        int res = 0;
        Queue<Employee> queue = new LinkedList<>();
        queue.offer(map.get(id));
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                Employee employee = queue.poll();
                res += employee.importance;
                for (Integer sub : employee.subordinates) {
                    queue.offer(map.get(sub));
                }
            }
        }

        return res;
    }
}
```

## Analysis

时间复杂度和空间复杂度都是O(n).