题目:
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:
- One employee has at most one direct leader and may have several subordinates.
- The maximum number of employees won't exceed 2000.
分析:
有一个保存员工信息的数据结构,包含id和重要度,以及一个其下属的id列表。现在给定一个这样的员工数组,求员工和其所有下属的重要度之和。
首先遍历数组,将员工的id和其对应的员工存入map中,方便我们直接获取到它的信息。
然后深度优先搜索,从所给的id开始求,递归求解他们的和即可。也可以使用bfs,将每一个员工的下属加入到队列中,将重要度累加,知道队列为空即可。
程序:
C++
/* // Employee info class Employee { public: // It's the unique ID of each node. // unique id of this employee int id; // the importance value of this employee int importance; // the id of direct subordinates vector<int> subordinates; }; */ class Solution { public: int getImportance(vector<Employee*> employees, int id) { unordered_map<int, Employee*> m; for(auto em:employees){ m.insert({em->id, em}); } return dfs(id, m); } private: int dfs(int id, unordered_map<int, Employee*> &m){ int sum = m[id]->importance; for(auto i:m[id]->subordinates){ sum += dfs(i, m); } return sum; } };
Java
/* // Employee info class Employee { // It's the unique id of each node; // unique id of this employee public int id; // the importance value of this employee public int importance; // the id of direct subordinates public List<Integer> subordinates; }; */ class Solution { public int getImportance(List<Employee> employees, int id) { HashMap<Integer, Employee> m = new HashMap<>(); for(Employee e:employees){ m.put(e.id, e); } return dfs(id, m); } private int dfs(int id, HashMap<Integer, Employee> m){ int sum = m.get(id).importance; for(Integer i:m.get(id).subordinates) sum += dfs(i, m); return sum; } }