You are given a data structure of employee information, which includes the employee's unique id, their importance value and their 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 their 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.
A:
-------------------就是用HashMap ---------记得map[E->id] = E; 而不是map[E->id] = E->importance -------------
/* // Definition for Employee. class Employee { public: int id; int importance; vector<int> subordinates; }; */ class Solution { public: int getImportance(vector<Employee*> employees, int id) { unordered_map<int,Employee*> map; // id to importance for(auto E: employees) map[E->id] = E; unordered_set<int> FoundSet; vector<int> V{id}; int res =0; while(! V.empty()){ int curID = V.back(); V.pop_back(); FoundSet.insert(curID); res += map.find(curID)->second->importance; for(auto tt: map.find(curID)->second->subordinates){ V.push_back(tt); } } return res; } };
学到了,
哎, map.find(curID)->second->importance
记得是 ->second
No comments:
Post a Comment