# 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.
``````

``````class Solution {
public:
int getImportance(vector<Employee*> employees, int id) {
unordered_map<int, Employee*> m;
for (auto e : employees) m[e->id] = e;
return helper(id, m);
}
int helper(int id, unordered_map<int, Employee*>& m) {
int res = m[id]->importance;
for (int num : m[id]->subordinates) {
res += helper(num, m);
}
return res;
}
};
``````

``````class Solution {
public:
int getImportance(vector<Employee*> employees, int id) {
int res = 0;
queue<int> q{{id}};
unordered_map<int, Employee*> m;
for (auto e : employees) m[e->id] = e;
while (!q.empty()) {
auto t = q.front(); q.pop();
res += m[t]->importance;
for (int num : m[t]->subordinates) {
q.push(num);
}
}
return res;
}
};
``````

Nested List Weight Sum

https://leetcode.com/problems/employee-importance/

https://leetcode.com/problems/employee-importance/discuss/112587/Java-HashMap-bfs-dfs

LeetCode All in One 题目讲解汇总(持续更新中…)

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