Employee Importance

You are given a data structure of employee information, which includes the employee'sunique id, hisimportance valueand hisdirectsubordinates' 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 isnot 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:

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

分析

员工建map。dfs子员工下去

"""
# 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(e):
            a,b,c = e.id,e.importance,e.subordinates
            if len(c) == 0:
                return b

            return b + sum([dfs(mm[i]) for i in c])

        mm = collections.defaultdict()    
        for e in employees:
            mm[e.id] = e
        return dfs(mm[id])