[LeetCode&Python] Problem 690. Employee Importance
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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:
- One employee has at most one direct leader and may have several subordinates.
- The maximum number of employees won't exceed 2000.
? DFS ?
""" # Employee info 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): """ :type employees: Employee :type id: int :rtype: int """ emap={e.id:e for e in employees} def dfs(sid): em=emap[sid] return em.importance+sum(dfs(sid) for sid in em.subordinates) return dfs(id)
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