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:
One employee has at most one direct leader and may have several subordinates.
The maximum number of employees won't exceed 2000.
这题,读题时间比解题长。其实就是带权重的树遍历。可以BFS或者DFS。图的遍历,T:O(e+v),所以是O(n),S:O(n),队列长度。但其实每次还得O(n)地找,是不是其实是n方?啊!看了solution发现,leetcode太狡猾了,用了O(n)先把东西存到hashmap里,然后查找起来就不用每次再O(n)了
/*
// Definition for Employee.
class Employee {
public int id;
public int importance;
public List<Integer> subordinates;
};
*/
class Solution {
public int getImportance(List<Employee> employees, int id) {
if (employees == null || employees.size() == 0 || id < 0) {
return 0;
}
Map<Integer, Employee> map = new HashMap<>();
for (Employee e : employees) {
map.put(e.id, e);
}
Queue<Employee> queue = new LinkedList<>();
Employee root = map.get(id);
// employee not in the list
if (root == null) {
return 0;
}
queue.offer(root);
int sum = 0;
while (!queue.isEmpty()) {
Employee cur = queue.poll();
sum += cur.importance;
for (Integer sub : cur.subordinates) {
queue.offer(map.get(sub));
}
}
return sum;
}
}
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