LeetCode 690. Employee Importance

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.

求某個員工的價值程度，明顯就是一個遞迴 直接寫程式碼

public int getImportance(List<Employee> employees, int id) {
        Employee boss = getEmployee(employees, id);
        return getImportance(employees, boss);
    }

    public int getImportance(List<Employee> employees,Employee employee) {
        int importance = employee.importance;
        List<Integer> subList = employee.subordinates;
        for(int id : subList){
            Employee sb = getEmployee(employees, id);
            importance += getImportance(employees, sb);
        }
        return importance;

    }

    public Employee getEmployee(List<Employee> employees, int id){
        Employee emp = null;
        for (Employee employee : employees) {
            if (employee.id == id) {
                emp = employee;
            }
        }
        return emp;
    }