690. Employee Importance 员工重要性

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







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