LeetCode 690. Employee Importance 员工的重要性(C++/Java)

题目：

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.

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.

分析：

有一个保存员工信息的数据结构，包含id和重要度，以及一个其下属的id列表。现在给定一个这样的员工数组，求员工和其所有下属的重要度之和。

首先遍历数组，将员工的id和其对应的员工存入map中，方便我们直接获取到它的信息。

然后深度优先搜索，从所给的id开始求，递归求解他们的和即可。也可以使用bfs，将每一个员工的下属加入到队列中，将重要度累加，知道队列为空即可。

程序：

C++

/*
// Employee info
class Employee {
public:
    // It's the unique ID of each node.
    // unique id of this employee
    int id;
    // the importance value of this employee
    int importance;
    // the id of direct subordinates
    vector<int> subordinates;
};
*/
class Solution {
public:
    int getImportance(vector<Employee*> employees, int id) {
        unordered_map<int, Employee*> m;
        for(auto em:employees){
            m.insert({em->id, em});
        }
        return dfs(id, m);
    }
private:
    int dfs(int id, unordered_map<int, Employee*> &m){
        int sum = m[id]->importance;
        for(auto i:m[id]->subordinates){
            sum += dfs(i, m);
        }
        return sum;
    }
};

Java

/*
// 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> m = new HashMap<>();
        for(Employee e:employees){
            m.put(e.id, e);
        }
        return dfs(id, m);
    }
    private int dfs(int id, HashMap<Integer, Employee> m){
        int sum = m.get(id).importance;
        for(Integer i:m.get(id).subordinates)
            sum += dfs(i, m);
        return sum;
    }
}