LeetCode 690. Employee Importance（员工重要度）

题目描述

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.

解法一

思路

这题挺简单的。首先需要以id找到value和下属，所以需要一个Map结构来索引；其次下属可能也有他自己的下属，并且没有一个确定的层级，所以只能用到递归。所以答案就显而易见啦。

Python

"""
# Employee info
class Employee:
    def __init__(self, id, importance, subordinates):
        # It's the unique id of each node.
        # unique id of this employee
        self.id = id
        # the importance value of this employee
        self.importance = importance
        # the id of direct subordinates
        self.subordinates = subordinates
"""


class Solution:
    def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """

        def dfs(employee, infos):
            importance = employee.importance
            for child in employee.subordinates:
                importance += dfs(infos[child], infos)
            return importance

        infos = {}
        for employee in employees:
            infos[employee.id] = employee
        return dfs(infos[id], infos)

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) {
        class Utils {
            int dfs(Employee employee, Map<Integer, Employee> infos) {
                int importance = employee.importance;
                for (int child : employee.subordinates) {
                    importance += dfs(infos.get(child), infos);
                }
                return importance;
            }
        }
        Map<Integer, Employee> infos = new HashMap<>();
        for (Employee employee : employees) {
            infos.put(employee.id, employee);
        }
        Utils utils = new Utils();
        return utils.dfs(infos.get(id), infos);
    }
}

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) {
        class Utils {
        public:
            int dfs(Employee *employee, unordered_map<int, Employee *> &infos) {
                int importance = employee->importance;
                for (int child : employee->subordinates) {
                    importance += dfs(infos[child], infos);
                }
                return importance;
            }
        };
        unordered_map<int, Employee *> infos;
        for (Employee *employee : employees) {
            infos[employee->id] = employee;
        }
        Utils utils;
        return utils.dfs(infos[id], infos);
    }
};