[LeetCode] 690. Employee Importance

You are given a data structure of employee information, which includes the employee's unique id, their importance value and their 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 their 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.

员工重要性。

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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/employee-importance
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这是一道图论graph的题，不难，搞懂题意就能做的出来。题目给的是所有员工的id，重要度和他们知悉下属的id。既然能拿到当前员工的id，那么拿到当前员工的重要度和他的下属也就不是问题，所以无论是什么做法，我们都需要一个hashmap记住所有员工的重要度和他们各自的下属。至于做法方面，我们可以用BFS或DFS，做法都跟一般的BFS和DFS类型的题非常类似，看了代码应该能懂。

BFS

时间O(V + E)

空间O(V) - number of employees

Java实现

class Solution {
    public int getImportance(List<Employee> employees, int id) {
        int total = 0;
        HashMap<Integer, Employee> map = new HashMap<>();
        for (Employee employee : employees) {
            map.put(employee.id, employee);
        }
        Queue<Employee> queue = new LinkedList<>();
        queue.offer(map.get(id));
        while (!queue.isEmpty()) {
            Employee cur = queue.poll();
            total += cur.importance;
            for (int subordinate : cur.subordinates) {
                queue.offer(map.get(subordinate));
            }
        }
        return total;
    }
}

DFS

时间O(V + E)

空间O(V) - number of employees

Java实现

class Solution {
    public int getImportance(List<Employee> employees, int id) {
        HashMap<Integer, Employee> map = new HashMap<>();
        for (Employee employee : employees) {
            map.put(employee.id, employee);
        }
        return helper(map, id);
    }

    private int helper(HashMap<Integer, Employee> map, int id) {
        Employee root = map.get(id);
        int total = root.importance;
        for (int subordinate : root.subordinates) {
            total += helper(map, subordinate);
        }
        return total;
    }
}

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