LeetCode 690.Employee Importance

LeetCode 690.Employee Importance

Description:

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:

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.

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分析：

这道题做的我好气，倒不是说题难，相反题简单，然而我在写main函数一直写不好Employee类，使我很烦躁，最后写好了想把代码输入写的漂亮一点，方便读者阅读，所以我采用了中文提示输入，然而sublime怎么鬼畜了，中文乱码，网上说可以用convert to UTF8或者GBK support解决，然而我早就已经安装了这两个packet了，后来我放弃了，改成英文提示输入。。。

回到这道题来，也不是很容易，因为涉及到不太熟悉的unordered_map库函数。
首先遍历所有员工把每一个id和员工对应起来放进unordered_map中，然后把该id对应的员工的importance加上，最后递归得到该id对应员工的直接下属的importance并把它们加上，即可。

代码如下：

#include <iostream>#include <vector>#include <unordered_map>using namespace std;// 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) {        unordered_map<int, Employee*> um;        for (auto e : employees) {            um[e->id] = e;        }        return getSum(um, id);    }private:    int getSum(unordered_map<int, Employee*>& um, int id) {        int ans = um[id]->importance;        for (auto e : um[id]->subordinates) {            ans += getSum(um, e);        }        return ans;    }};int main() {    int n, num;    cout << "Please input the number of Employees:";    cin >> n;    vector<Employee*> ems;    Employee em[n];    for (int i = 0; i < n; i++) {        cin >> em[i].id >> em[i].importance;        cout << "Please input the number of direct Employees:";        cin >> num;        int temp;        for (int j = 0; j < num; j++) {                 cin >> temp;            em[i].subordinates.push_back(temp);        }        ems.push_back(&em[i]);    }    // cout << ems.size() << endl;    // for (int i = 0; i < ems.size(); i++) {    //  cout << ems[i]->id << " " << ems[i]->importance << endl;    // }    int idNum;    cin >> idNum;    Solution s;    cout << s.getImportance(ems, idNum) << endl;    return 0;}/*input:Please input the number of Employees:315Please input the number of direct Employees:22 323Please input the number of direct Employees:033Please input the number of Employees:01output:11 */