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.


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:

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#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
// 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) {
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:3
1
5
Please input the number of direct Employees:2
2 3

2
3
Please input the number of direct Employees:0

3
3
Please input the number of Employees:0

1

output:
11
*/
------本文结束感谢阅读------