LeetCode Employee Importance

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 Graph, Nested 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(employees.size()). 全部员工生成的map.

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(employees.size()). 全部员工生成的map.

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 }

 

转载于:https://www.cnblogs.com/Dylan-Java-NYC/p/7624915.html