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 directsubordinates' 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.
/* // 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 { public int getImportance(List<Employee> employees, int id) { return get(employees, id); } private int get(List<Employee> employees, int id) { int total = 0; for (Employee e : employees) { if (id == e.id) { total = e.importance; for (int subId : e.subordinates) { total += get(employees, subId); } break; } } return total; } }
作者:Pickle
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