690. Employee Importance_给定一个保存员工信息的数据结构

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.
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Note:
One employee has at most one direct leader and may have several subordinates.
The maximum number of employees won’t exceed 2000.

/*
// 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) {
        Map<Integer, Employee> data = new HashMap<>();  
        for (Employee employee : employees) {  
            data.put(employee.id, employee);  
        }  
        Stack<Integer> s = new Stack<>();  
        s.add(id);  
        int re = 0;  
        while (!s.isEmpty()) {  
            int cur = s.pop();  
            re += data.get(cur).importance;  
            s.addAll(data.get(cur).subordinates);  
        }  
        return re;  
    }
}
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