[Java]LeetCode690. 员工的重要性 | Employee Importance

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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:

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

---

给定一个保存员工信息的数据结构，它包含了员工唯一的id，重要度 和 直系下属的id。

比如，员工1是员工2的领导，员工2是员工3的领导。他们相应的重要度为15, 10, 5。那么员工1的数据结构是[1, 15, [2]]，员工2的数据结构是[2, 10, [3]]，员工3的数据结构是[3, 5, []]。注意虽然员工3也是员工1的一个下属，但是由于并不是直系下属，因此没有体现在员工1的数据结构中。

现在输入一个公司的所有员工信息，以及单个员工id，返回这个员工和他所有下属的重要度之和。

示例 1:

输入: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
输出: 11
解释:
员工1自身的重要度是5，他有两个直系下属2和3，而且2和3的重要度均为3。因此员工1的总重要度是 5 + 3 + 3 = 11。

注意:

1. 一个员工最多有一个直系领导，但是可以有多个直系下属
2. 员工数量不超过2000。

---

4ms

/*
// 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 {
    HashMap<Integer,Employee> h;
    public int getImportance(List<Employee> employees, int id) {
        h = new HashMap<Integer,Employee>();
        for(Employee emp: employees)
            h.put(emp.id,emp);
        return dfs(id);
    }
    
    public int dfs(int id){
        Employee e = h.get(id);
        int ans = e.importance;
        for(Integer sub:e.subordinates)
            ans += dfs(sub);
        return ans;
    }
}

---

5ms

class Solution {
    
    private int getSum(HashMap<Integer, Employee> map, int id){
        int sum = map.get(id).importance;
        for(int sub : map.get(id).subordinates){
            sum += getSum(map, sub);
        }
        return sum;
    }
    
    public int getImportance(List<Employee> employees, int id) {
        HashMap<Integer, Employee> map = new HashMap<Integer, Employee>();
        
        for(Employee e : employees){
            map.put(e.id, e);
        }
        
        return getSum(map, id);
    }
}

---

6ms

class Solution {
    public int getImportance(List<Employee> employees, int id) {
        HashMap<Integer, Integer> map = new HashMap<>();
        HashMap<Integer, List<Integer>> rela = new HashMap<>();
        for (Employee emp : employees) {
            rela.put(emp.id, emp.subordinates);
            map.put(emp.id, emp.importance);
        }
        int sum = 0;
        List<Integer> list = new ArrayList<>();
        list.add(id);
        while (list.size() > 0) {
            List<Integer> next = new ArrayList<>();
            for (int i : list) {
                sum += map.get(i);
                if (rela.containsKey(i)) {
                    next.addAll(rela.get(i));
                }
            }
            list = next;
        }
        return sum;
    }
}