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LeetCode Employee Importance

时间:2017-10-04 12:38:39      阅读:230      评论:0      收藏:0      [点我收藏+]

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原题链接在这里:https://leetcode.com/problems/employee-importance/description/

题目:

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:

  1. One employee has at most one direct leader and may have several subordinates.
  2. The maximum number of employees won‘t exceed 2000.

题解:

类似Clone GraphNested List Weight Sum.

可以采用BFS. 每层的Employee挨个加进去.

Time Complexity: O(n). n是root id的所有下属个数, 包括直系下属和非直系下属.

Space: O(n). 有可能都是直系下属.

AC Java:

 1 /*
 2 // Employee info
 3 class Employee {
 4     // It‘s the unique id of each node;
 5     // unique id of this employee
 6     public int id;
 7     // the importance value of this employee
 8     public int importance;
 9     // the id of direct subordinates
10     public List<Integer> subordinates;
11 };
12 */
13 class Solution {
14     public int getImportance(List<Employee> employees, int id) {
15         int res = 0;
16         
17         HashMap<Integer, Employee> hm = new HashMap<Integer, Employee>();
18         for(Employee employee : employees){
19             hm.put(employee.id, employee);
20         }
21         
22         LinkedList<Employee> que = new LinkedList<Employee>();
23         que.add(hm.get(id));
24         while(!que.isEmpty()){
25             Employee cur = que.poll();
26             res += cur.importance;
27             for(int subId : cur.subordinates){
28                 que.add(hm.get(subId));
29             }
30         }
31         
32         return res;
33     }
34 }

DFS.逐层往深dfs.

Time Complexity: O(n). n是root id的所有下属个数, 包括直系下属和非直系下属.

Space: O(n). stack space, 最深的树.

AC Java:

 1 /*
 2 // Employee info
 3 class Employee {
 4     // It‘s the unique id of each node;
 5     // unique id of this employee
 6     public int id;
 7     // the importance value of this employee
 8     public int importance;
 9     // the id of direct subordinates
10     public List<Integer> subordinates;
11 };
12 */
13 class Solution {
14     public int getImportance(List<Employee> employees, int id) {
15         HashMap<Integer, Employee> hm = new HashMap<Integer, Employee>();
16         for(Employee employee : employees){
17             hm.put(employee.id, employee);
18         }
19         
20         return dfs(hm, id);
21     }
22     
23     private int dfs(HashMap<Integer, Employee> hm, int id){
24         int res = 0;
25         Employee cur = hm.get(id);
26         
27         res += cur.importance;
28         for(int subId : cur.subordinates){
29             res += dfs(hm, subId);
30         }
31         return res;
32     }
33 }

 

LeetCode Employee Importance

标签:ati   exit   res   ems   poll   nat   return   seve   includes   

原文地址:http://www.cnblogs.com/Dylan-Java-NYC/p/7624915.html

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