Title

the problem link

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: One employee has at most one direct leader and may have several subordinates. The maximum number of employees won’t exceed 2000.

利用了一个map存储id到数组索引的关系

Solution

//cpp:
class Solution {
public:
	int getImport(unordered_map<int, int> &map1,vector<Employee*> &employees, int id) {
		Employee *e = employees[map1[id]];
		int ans = e->importance;
		for (int i = 0; i < e->subordinates.size(); i++) {
			ans += getImport(map1, employees, e->subordinates[i]);
		}
		return ans;
	}
	int getImportance(vector<Employee*> employees, int id) {
		unordered_map<int, int> map1;
		for (int i = 0; i < employees.size(); i++) {
			map1[employees[i]->id] = i;
		}
		int ans = getImport(map1, employees, id);
		return ans;
	}
};