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chenwenhang
AlgorithmPractice
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array_and_matrix
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ImportanceOfEmployees.java
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ImportanceOfEmployees.java
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package array_and_matrix;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
/**
* @Author: Wenhang Chen
* @Description:给定一个保存员工信息的数据结构,它包含了员工唯一的id,重要度 和 直系下属的id。
* <p>
* 比如,员工1是员工2的领导,员工2是员工3的领导。他们相应的重要度为15, 10, 5。那么员工1的数据结构是[1, 15, [2]],员工2的数据结构是[2, 10, [3]],员工3的数据结构是[3, 5, []]。注意虽然员工3也是员工1的一个下属,但是由于并不是直系下属,因此没有体现在员工1的数据结构中。
* <p>
* 现在输入一个公司的所有员工信息,以及单个员工id,返回这个员工和他所有下属的重要度之和。
* <p>
* 示例 1:
* <p>
* 输入: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
* 输出: 11
* 解释:
* 员工1自身的重要度是5,他有两个直系下属2和3,而且2和3的重要度均为3。因此员工1的总重要度是 5 + 3 + 3 = 11。
* @Date: Created in 21:00 8/5/2020
* @Modified by:
*/
public class ImportanceOfEmployees {
class Employee {
public int id;
public int importance;
public List<Integer> subordinates;
}
Map<Integer, Employee> emap;
public int getImportance(List<Employee> employees, int queryid) {
emap = new HashMap();
for (Employee e : employees) emap.put(e.id, e);
return dfs(queryid);
}
public int dfs(int eid) {
Employee employee = emap.get(eid);
int ans = employee.importance;
for (Integer subid : employee.subordinates)
ans += dfs(subid);
return ans;
}
}
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