You are given a data structure of employee information, which includes the employee's unique id, his importance value and his directsubordinates' 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:
- One employee has at most one direct leader and may have several subordinates.
- The maximum number of employees won't exceed 2000.
计算员工的总价值
C++(115ms):
1 /* 2 // Employee info 3 class Employee { 4 public: 5 // It's the unique ID of each node. 6 // unique id of this employee 7 int id; 8 // the importance value of this employee 9 int importance; 10 // the id of direct subordinates 11 vector<int> subordinates; 12 }; 13 */ 14 class Solution { 15 public: 16 int getImportance(vector<Employee*> employees, int id) { 17 unordered_map<int,Employee*> map ; 18 for (auto e : employees){ 19 map[e->id] = e ; 20 } 21 22 return dfs(map , id) ; 23 } 24 25 int dfs(unordered_map<int,Employee*> map , int id){ 26 int sum = map[id]->importance ; 27 for (auto sub_id : map[id]->subordinates){ 28 sum += dfs(map,sub_id) ; 29 } 30 return sum ; 31 } 32 };