690. Employee Importance*

690. Employee Importance*

​​https://leetcode.com/problems/employee-importance/​​

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.

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 onedirectleader and may have several subordinates.
2. The maximum number of employees won’t exceed​​2000​​.

C++ 实现 1

首先弄明白题意, 题目要求获取 ​​id​​ 对应的所有直接或非直接的雇员的 importance. 另外注意每个雇员只有最多一个 leader.

本质上这题是树类型的题. 使用 DFS 解决. ​​C++ 实现 2​​ 使用层序遍历来解决.

/*
// 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 subordinates;
};
*/
class Solution {
private:
    unordered_map<int, Employee*> record;
    void dfs(Employee* &leader, auto &importance) {
        if (leader->subordinates.empty()) return;
        for (auto &id : leader->subordinates) {
            auto sub_leader = record[id];
            importance += sub_leader->importance;
            dfs(sub_leader, importance);
        }
    }
public:
    int getImportance(vector<Employee*> employees, int id) {
        for (auto &e : employees) record[e->id] = e;
        auto leader = record[id];
        auto importance = leader->importance;
        dfs(leader, importance);
        return importance;
    }
};

C++ 实现 2

层序遍历.

/*
// 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 subordinates;
};
*/
class Solution {
public:
    int getImportance(vector<Employee*> employees, int id) {

        queue<Employee*> Queue;
        unordered_map<int, Employee*> record;
        for (const auto &man : employees)
            record[man->id] = man;
        Queue.push(record[id]);

        int importance = 0;
        while (!Queue.empty()) {
            auto man = Queue.front();
            Queue.pop();
            importance += man->importance;
            if (!man->subordinates.empty())
                for (const auto &subid : man->subordinates)
                    Queue.push(record[subid]);
        }
        return importance;
    }
};