0690. Employee Importance

# 690. Employee Importance#

## 题目 #

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

## 解题思路 #

• 简单题。根据题意，DFS 或者 BFS 搜索找到所求 id 下属所有员工，累加下属员工的重要度，最后再加上这个员工本身的重要度，即为所求。

## 代码 #

``````package leetcode

type Employee struct {
Id           int
Importance   int
Subordinates []int
}

func getImportance(employees []*Employee, id int) int {
m, queue, res := map[int]*Employee{}, []int{id}, 0
for _, e := range employees {
m[e.Id] = e
}
for len(queue) > 0 {
e := m[queue[0]]
queue = queue[1:]
if e == nil {
continue
}
res += e.Importance
for _, i := range e.Subordinates {
queue = append(queue, i)
}
}
return res
}
``````

Nov 25, 2022