作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/


题目地址:https://leetcode.com/problems/employee-importance/description/

题目描述

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:

  1. One employee has at most one direct leader and may have several subordinates.
  2. The maximum number of employees won’t exceed 2000.

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)

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

日期

2018 年 1 月 17 日
2018 年 11 月 10 日 —— 这么快就到双十一了??

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