690-Employee Importance

[难度] easy
[分类] array

1.题目描述

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.

2.测试样例

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.

3.算法分析

采用递归算法，对于子数组中的每个id，递归计算它们的sum，最终获得总的sum，基本思路比较简单，直接看代码就能够理解。

4.代码实现

/* // Employee info class Employee { public: // It's the unique ID of each node. // unique id of this employee int id; // the importance value of this employee int importance; // the id of direct subordinates vector<int> subordinates; }; */
class Solution {
public:
    int getImportance(vector<Employee*> employees, int id) {
        Employee* temp = employees[id - 1];
        int result = temp->importance;
        vector<int> sub = temp->subordinates;
        int n = sub.size();
        for (int i = 0; i < n; ++i) {
            // 递归实现
            result += getImportance(employees, sub[i]);
        }
        return result;
    }
};