690. Employee Importance

Question 690

https://leetcode.com/problems/employee-importance/description/

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.

Answer

v1. Recursive

v2. Dict + queue

1. 將 employees 存成 dictionary，以 id 當作 key

2. 建立一個 list 用 queue 的概念去想它

3. 有 subordinate 時就將其加入 list 中

4. 一直相加 list 中的員工 important 值，相加後 pop 該員工出去

5. 最後 list 會是空的，結束。

v1

'''
class Employee(object): 
    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(object):
    def getImportance(self, employees, id): 
    for employee in employees:
        if employee.id == id:
            if employee.subordinates != []:
                for subordinate in employee.subordinates:
                    employee.importance += self.getImportance(employees,subordinate)
                return employee.importance
            return employee.importance
    return 0

v2

'''
class Employee(object): 
    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(object):
    def getImportance(self, employees, id): 
        DictE = { employee.id:employee for employee in employees }
        SelectL = [id]
        Total = 0
        while SelectL:
            select = SelectL.pop()
            Total += DictE[select].importance
            for sub in DictE[select].subordinates:
                SelectL.append(sub)
        return Total