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.

题目大意 #

给定一个保存员工信息的数据结构,它包含了员工 唯一的 id ,重要度 和 直系下属的 id 。比如,员工 1 是员工 2 的领导,员工 2 是员工 3 的领导。他们相应的重要度为 15 , 10 , 5 。那么员工 1 的数据结构是 [1, 15, [2]] ,员工 2的 数据结构是 [2, 10, [3]] ,员工 3 的数据结构是 [3, 5, []] 。注意虽然员工 3 也是员工 1 的一个下属,但是由于 并不是直系 下属,因此没有体现在员工 1 的数据结构中。现在输入一个公司的所有员工信息,以及单个员工 id ,返回这个员工和他所有下属的重要度之和。

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
}

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