LeetCode 690. Employee Importance （职员的重要值）

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:

1. One employee has at most one direct leader and may have several subordinates.
2. The maximum number of employees won't exceed 2000.

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 题目标签：HashTable

　　题目给了我们一个 employees list， 和 一个 id，让我们找到这个 id 的员工的手下所有员工的 importance 累加，包括他自己的。

　　首先把 employees 存入 HashMap， id 为 key， Employee 为value。

　　然后建立一个 dfs function：

　　　　当员工的 subordinates 的 size  等于 0 的时候， 说明没有必要继续递归了，返回员工的重要值；

　　　　如果 size 大于0，那么遍历 subordinates，把每一个 员工id 递归，累加重要值。

Java Solution:

Runtime beats 71.9% 

完成日期：11/16/2017

关键词：HashMap， DFS

关键点：把employee 信息存入map，id 为 key，employee 为 value，便于dfs 直接调取 员工信息

/*
// 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 
{
    public int getImportance(List<Employee> employees, int id) 
    {
        HashMap<Integer, Employee> map = new HashMap<>();
        
        for(Employee e: employees)
            map.put(e.id, e);
             
        return dfs(id, map);
    }
    
    private int dfs(int id, HashMap<Integer, Employee> map)
    {
        Employee e = map.get(id);
        
        if(e.subordinates.size() == 0)
            return e.importance;
        
        int imp = e.importance;
        
        for(int sub: e.subordinates)
            imp += dfs(sub, map);
    
        
        return imp;
    }
}

参考资料：N/A

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