The importance of LeetCode 690. Employee Importance of employees (C ++ / Java)

topic:

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.

analysis:

There is a data structure stored employee information, including the id and the degree of importance, as well as a list of their subordinates id. Now given a array of such employees, find employees and their degree of importance and all subordinate.

First through the array, the id of employees and their corresponding employees into the map to help us get the information directly to it.

Then depth-first search, from the beginning to find the id, and their recursive solution can be. You can also use bfs, each employee will be subordinate to the queue, the importance of accumulating, you can know the queue is empty.

program:

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;
    }
}