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690employeeimportance.py
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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.
"""
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
"""
# 树的层次遍历, depth-first
emp_dict = {}
for e in employees:
emp_dict[e.id] = e
emp = emp_dict[id]
queue = [emp]
s = 0
while len(queue) > 0:
new_queue = []
for emp in queue:
s += emp.importance
for id in emp.subordinates:
new_queue.append(emp_dict[id])
queue = new_queue
return s
if __name__ == '__main__':
s = Solution()
a = [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]]
b = 1
c = s.getImportance(a, b)
print(c)