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:
- One employee has at most one direct leader and may have several subordinates.
- The maximum number of employees won't exceed 2000.
题目标签:HashTable
题目给了我们一个 employees list, 和 一个 id,让我们找到这个 id 的员工的手下所有员工的 importance 累加,包括他自己的。
首先把 employees 存入 HashMap, id 为 key, Employee 为value。
然后建立一个 dfs function:
当员工的 subordinates 的 size 等于 0 的时候, 说明没有必要继续递归了,返回员工的重要值;
如果 size 大于0,那么遍历 subordinates,把每一个 员工id 递归,累加重要值。
Java Solution:
Runtime beats 71.9%
完成日期:11/16/2017
关键词:HashMap, DFS
关键点:把employee 信息存入map,id 为 key,employee 为 value,便于dfs 直接调取 员工信息
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 { 15 public int getImportance(List<Employee> employees, int id) 16 { 17 HashMap<Integer, Employee> map = new HashMap<>(); 18 19 for(Employee e: employees) 20 map.put(e.id, e); 21 22 return dfs(id, map); 23 } 24 25 private int dfs(int id, HashMap<Integer, Employee> map) 26 { 27 Employee e = map.get(id); 28 29 if(e.subordinates.size() == 0) 30 return e.importance; 31 32 int imp = e.importance; 33 34 for(int sub: e.subordinates) 35 imp += dfs(sub, map); 36 37 38 return imp; 39 } 40 }
参考资料:N/A
LeetCode 题目列表 - LeetCode Questions List
题目来源:https://leetcode.com/