LeetCode 690 Employee Importance 解题报告
作者：互联网
题目要求
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.
题目分析及思路
定义了一个含有employee信息的数据结构，该结构包括了employee的唯一id、他的重要值以及他的直接下属的id。现给定一个公司的employee的信息，以及一个employee的id，要求返回这个employee以及他所有直接下属的重要值的总和。可以使用递归的方法，跳出递归的条件是当前employee没有直接下属，否则则遍历当前employee的所有下属获得重要值。
python代码
"""
# 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
"""
ids = [employee.id for employee in employees]
leader = employees[ids.index(id)]
employees.pop(ids.index(id))
if leader.subordinates == []:
return leader.importance
else:
im = 0
for id in leader.subordinates:
im += self.getImportance(employees, id)
return im + leader.importance
标签：690,Importance,subordinates,employees,leader,importance,employee,Employee,id 来源： https://www.cnblogs.com/yao1996/p/10619315.html