leetcode-690. Employee Importance

690. Employee Importance

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.

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:

1. One employee has at most one direct leader and may have several subordinates.
2. The maximum number of employees won't exceed 2000.

解法：就是类似树的遍历，BFS和DFS都可以写。

/*
// 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) {
        int sum =0;
        Stack <Integer> stack = new Stack<>();
        Map<Integer, Employee> data = new HashMap<>();
        for (int i = 0; i < employees.size(); i++) {
            if(employees.get(i).id==id){
                stack.addAll(employees.get(i).subordinates);
                sum+=employees.get(i).importance;
            }
            //sum+=employees.get(i).importance;
            data.put(employees.get(i).id, employees.get(i));//用map查找应该会更快吧
        }
        while(!stack.empty()){
            int cur = stack.pop();
            sum+=data.get(cur).importance;
            stack.addAll(data.get(cur).subordinates);//查找对应的employee
        }
        return sum;
    }
}