做者: 负雪明烛
id: fuxuemingzhu
我的博客: http://fuxuemingzhu.cn/node
题目地址:https://leetcode.com/problems/employee-importance/description/python
You are given a data structure of employee information, which includes the employee’s unique id, his importance value and his direct subordinates’ id.web
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.数据结构
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.svg
Example 1:函数
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:ui
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。this
输入一个员工的Id,求它本身和它全部的下属的重要性之和。spa
题目意思是找出每一个节点与其子节点的全部重要性之和。code
为了快速查询每一个节点的id与其对应,创建了map。
而后采用dfs遍历。当某个子节点再也不有子节点的时候会自动终止该分支的遍历。
我以为这个题应该背下来。
""" # 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。效率居然提升了很多。
""" # 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
2018 年 1 月 17 日 2018 年 11 月 10 日 —— 这么快就到双十一了??