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.

思路：

1. 需要根据职员id找到职员信息，在list中查找时间复杂度很高，所以可以先把职员信息存储在hashmap中。
2. 从要求的第一个职员开始查找，价值总和等于他的价值和他所有下属价值之和，所以total总价值加上他的价值之后，递归再查找他的下属的总价值。
3. 当查找到某个职员没有下属时，这一条上下级线就走到底可以直接返回了。

public class EmployeeImportance690 {
    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 subordinates;
    };

    public int getImportance(List employees, int id) {
        if (employees == null || employees.size() == 0) {
            return 0;
        }

        Map map = new HashMap<>();
        for (Employee e : employees) {
            map.put(e.id, e);
        }

        return helper(map, id);
    }

    private int helper(Map map, int id) {
        if (!map.containsKey(id)) {
            return 0;
        }

        int total = map.get(id).importance;
        for (int subId : map.get(id).subordinates) {
            total += helper(map, subId);
        }
        return total;
    }
}