690. Employee Importance

内容：

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.

思路：

根据给出的领导id，在员工集合中找到该领导，并把其所有的直系下属找出来，直系下属再通过调用同一个函数来寻找直系下属，并把重要值进行相加。

class Solution {public int getImportance(List<Employee> employees, int id) {int importance = 0;List<Integer> subordinates = new ArrayList<Integer>();for(Employee e : employees){if(e.id == id){importance = e.importance;subordinates = e.subordinates;for(Integer i : subordinates)importance = importance + getImportance(employees,i);break;}}return importance;}
}

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