Employee Importance

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题目

leetcode题目

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, []]], 1Output: 11Explanation: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 one direct leader and may have several subordinates.
2. The maximum number of employees won’t exceed 2000.

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解决

由于给的是Employee*类型，所以我们需要判断Employee* employee->id == id是否成立。
并且我们需要注意，a的直系下属的直系下属同样是a的下属，我们需要留意a的直系下属是否还有自己的直系下属。

1. 原始想法

/*// Employee infoclass 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) {        int result = 0;        // 将id和importance相互绑定        unordered_map<int, int> im;         // 将id和subordinates相互绑定        unordered_map<int, vector<int>> su;        int num = employees.size();        for (int i = 0; i < num; i++) {            im[employees[i]->id] = employees[i]->importance;            su[employees[i]->id] = employees[i]->subordinates;        }        queue<int> sid; // 使用队列记录所有下属的id        sid.push(id);        while (!sid.empty()) {            int i = sid.front();            sid.pop();            // 将"id = i"的employee的直系下属添加到队列中            for (int k = 0; k < su[i].size(); k++) {                sid.push(su[i][k]);            }            result += im[i];        }        return result;    }};

2. 优化

/*// Employee infoclass 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) {        int result = 0;        unordered_map<int, Employee*> e;        int num = employees.size();        for (int i = 0; i < num; i++) {            e[employees[i]->id] = employees[i];        }        queue<int> sid;        sid.push(id);        while (!sid.empty()) {            int i = sid.front();            sid.pop();            result += e[i]->importance;            for (vector<int>::iterator it = e[i]->subordinates.begin(); it != e[i]->subordinates.end(); it++) {                sid.push(*it);            }        }        return result;    }};