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  • 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;
        }
    }
    

      

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