# 690 Employee Importance

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.

**(1) BFS**

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);

}

Queue<Employee> queue = new LinkedList<>();

queue.add(map.get(id));

int res = 0;

while (!queue.isEmpty()) {

Employee curEmployee = queue.remove();

res += curEmployee.importance;

for (Integer employeeId : curEmployee.subordinates) {

queue.add(map.get(employeeId));

}

}

return res;

}

}

**(2) DFS**

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 = new int[1];

getImportanceByDFS(map, id, res);

return res[0];

}

public void getImportanceByDFS(Map<Integer, Employee> map, int id, int[] res) {

res[0] += map.get(id).importance;

for (Integer subId : map.get(id).subordinates) {

getImportanceByDFS(map, subId, res);

}

}

}

BFS: 时间复杂度O(n), 空间复杂度O(n),因为用了HashMap

DFS: 时间复杂度O(n), 空间复杂度O(n),因为用了HashMap

Last modified 3yr ago