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.
給你一個員工信息的數據結構,包括員工的唯一id,他的重要性值和他的直接下屬的id。
例如,employee 1是employee 2的leader, employee 2是employee 3的leader。它們的重要性值分別為15、10和5。那麽employee 1的數據結構為[1,15,[2]],employee 2的數據結構為[2,10,[3]],employee 3的數據結構為[3,5,[]]。請註意,盡管employee 3也是employee 1的下屬,但是關系並不直接。
現在,給定一個公司的員工信息和一個員工id,您需要返回該員工及其所有下屬的總重要性值。
/* // Employee info class Employee { // It‘s the unique id of each node; // unique id of this employee public int id; // the importance value of this employee public int importance; // the id of direct subordinates public List<Integer> subordinates; }; */ class Solution { Map<Integer,Employee> emap; public int getImportance(List<Employee> employees, int id) { emap=new HashMap(); for(Employee e:employees) emap.put(e.id,e); return dfs(id); } public int dfs(int eid) { Employee e=emap.get(eid); int ans=e.importance; for(Integer subid:e.subordinates) ans+=dfs(subid); return ans; } }
690. Employee Importance