690 Employee Importance

690. Employee Importance

1. Question

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:
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Input:
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[[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
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Output:
5
11
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Explanation:
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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:
  1. 1.
    One employee has at most one direct leader and may have several subordinates.
  2. 2.
    The maximum number of employees won't exceed 2000.

2. Implementation

(1) BFS
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class Solution {
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public int getImportance(List<Employee> employees, int id) {
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Map<Integer, Employee> map = new HashMap<>();
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for (Employee employee : employees) {
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map.put(employee.id, employee);
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}
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Queue<Employee> queue = new LinkedList<>();
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queue.add(map.get(id));
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int res = 0;
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while (!queue.isEmpty()) {
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Employee curEmployee = queue.remove();
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res += curEmployee.importance;
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for (Integer employeeId : curEmployee.subordinates) {
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queue.add(map.get(employeeId));
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}
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}
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return res;
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}
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}
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(2) DFS
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class Solution {
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public int getImportance(List<Employee> employees, int id) {
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Map<Integer, Employee> map = new HashMap<>();
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for (Employee employee : employees) {
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map.put(employee.id, employee);
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}
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int[] res = new int[1];
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getImportanceByDFS(map, id, res);
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return res[0];
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}
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public void getImportanceByDFS(Map<Integer, Employee> map, int id, int[] res) {
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res[0] += map.get(id).importance;
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for (Integer subId : map.get(id).subordinates) {
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getImportanceByDFS(map, subId, res);
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}
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}
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}
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3. Time & Space Complexity

BFS: 时间复杂度O(n), 空间复杂度O(n),因为用了HashMap
DFS: 时间复杂度O(n), 空间复杂度O(n),因为用了HashMap