LeetCode 690 Employee Importance 解题报告

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