690. Employee Importance 员工重要性

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.

计算该员工及其所有下属的总体重要性。

2. # Employee info
3. class Employee:
4. def __init__(self, id, importance, subordinates):
5. # It's the unique id of each node.
6. # unique id of this employee
7. self.id = id
8. # the importance value of this employee
9. self.importance = importance
10. # the id of direct subordinates
11. self.subordinates = subordinates
13. class Solution:
14. def getImportance(self, employees, id):
15. """
16. :type employees: Employee
17. :type id: int
18. :rtype: int
19. """
20. val = 0
21. employDict = dict()
22. for i in employees:
23. employDict.update({i.id:i})
25. emp = employDict[id]
26. if emp :
27. val += emp.importance
29. sub = emp.subordinates
30. while sub:
31. item = employDict[sub.pop(0)]
32. if item:
33. val += item.importance
34. if item.subordinates:
35. sub += item.subordinates
37. return val

来自为知笔记(Wiz)

转载于:https://www.cnblogs.com/xiejunzhao/p/7613047.html