题目地址：https://leetcode.com/problems/employee-importance/description/open in new window
题目描述
Youare given a data structure of employee information, which includes the employee's unique id, his importance value and his direct subordinates' id.
Forexample, 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.
Nowgiven 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:
1、 Oneemployeehasatmostonedirectleaderandmayhaveseveralsubordinates.；
2、 Themaximumnumberofemployeeswon'texceed2000.；
题目大意
给的数据结构是[1, 5, [2, 3]]表示的是1号员工的重要性是5，有两个下属2和3。
输入一个员工的Id,求它自己和它所有的下属的重要性之和。
解题方法
方法一：DFS
题目意思是找出每个节点与其子节点的所有重要性之和。
为了快速查询每个节点的id与其对应，建立了map。
然后采用dfs遍历。当某个子节点不再有子节点的时候会自动终止该分支的遍历。
我觉得这个题应该背下来。
"""
# 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
"""
employee_dict = {employee.id : employee for employee in employees}
def dfs(id):
return employee_dict[id].importance + sum(dfs(id) for id in employee_dict[id].subordinates)
return dfs(id)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 二刷，换了一个写法，没有新定义dfs，而是直接使用了题目给的函数。效率竟然提高了不少。
"""
# 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
"""
emap = {employee.id : employee for employee in employees}
res = emap[id].importance
for sub in emap[id].subordinates:
res += self.getImportance(employees, sub)
return res
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