问题描述：简单型，BFS解法
You are given a data structure of employee information, which includes the employee’s unique id, his importance value and his directsubordinates’ 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.
40/108
"""
# 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 # 是个列表，列表中的数据类型是Employee
:type id: int
:rtype: int
"""
weight_sum = 0
ids = [id] # 用于存储相关的id
while employees != []:
employee = employees.pop(0) # 排在第一的不一定是最高领导，乱序
[cur_id, weight, sub_lst] = employee.id,employee.importance, employee.subordinates
if cur_id in ids:
weight_sum += weight
ids += sub_lst
return weight_sum
93/108
"""
# 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 # 是个列表，列表中的数据类型是Employee
:type id: int
:rtype: int
"""
weight_sum = 0
ids = [id] # 用于存储相关的id
temp = employees[:] # 深度复制
# 第一遍遍历加入所有的id
while temp != []:
temp_employee = temp.pop(0) # 排在第一的不一定是最高领导，乱序
[cur_id, weight, sub_lst] = temp_employee.id,temp_employee.importance, temp_employee.subordinates
if cur_id in ids:
ids += sub_lst
# 第二遍遍历计算weight_sum
while employees != []:
employee = employees.pop(0)
[cur_id, weight, sub_lst] = employee.id, employee.importance, employee.subordinates
if cur_id in ids:
weight_sum += weight
return weight_sum
这种都是思想含有漏洞的解法，正确、全面的思路应当是结合map数据结构进行。
想到用字典/map结构来进行数据处理，问题将变得非常简单：
"""
# 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 # 是个列表，列表中的数据类型是Employee
:type id: int
:rtype: int
"""
dic = {} # 用于列表变字典
total = 0
for e in employees:
dic[e.id] = [e.importance, e.subordinates]
ids = [id] # 用于模拟queue
while ids != [] :
cur_id = ids.pop(0)
total += dic[cur_id][0]
ids += dic[cur_id][1]
return total
一旦列表编程以id作为键，重要性和下属作为值以后，就可以顺藤摸瓜，再用队列来走一遍，此时字典已经就位，就可以按照直接下属关系来游走了。
开始想建立一棵树，但是很麻烦，也没有想清楚细节，结合字典来处理，建树也变得简单起来。
END.