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.

给你一个员工信息的数据结构，包括员工的唯一id，他的重要性值和他的直接下属的id。

例如，employee 1是employee 2的leader, employee 2是employee 3的leader。它们的重要性值分别为15、10和5。那么employee 1的数据结构为[1,15，[2]]，employee 2的数据结构为[2,10，[3]]，employee 3的数据结构为[3,5，[]]。请注意，尽管employee 3也是employee 1的下属，但是关系并不直接。

现在，给定一个公司的员工信息和一个员工id，您需要返回该员工及其所有下属的总重要性值。

/*
// 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

原文：https://www.cnblogs.com/longlyseul/p/9918311.html