【LeetCode690】Employee Importance

/*
// 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) {
//         if((employees[id]->subordinates).empty())
//             return employees[id]->importance;
//         int sum = employees[id]->importance;
//         for(int i = 0; i < (employees[id]->subordinates).size(); ++i){
//             sum += getImportance(employees, (employees[id]->subordinates)[i]);
//         }
//         return sum;
//     }
    
// };
// class Solution {
// public:
//     unordered_map<int, int> temp;
//     int getImportance(vector<Employee*> employees, int id) {
//         //we have to find out which one in the vector employees has the correct id.        
//         for(int i =0; i < employees.size(); ++i){
//             temp[employees[i]->id] = i; 
//         }
//         return help_getImportance(employees, id);
//     }
//     int help_getImportance(vector<Employee*> employees, int id) {
//         Employee* p_employ = employees[temp[id]]; 
//         if((p_employ->subordinates).empty())
//             return p_employ->importance;
//         int sum = p_employ->importance;
//         for(int i = 0; i < (p_employ->subordinates).size(); ++i){
//             sum += help_getImportance(employees, (p_employ->subordinates)[i]);
//         }
//         return sum;
//     }    
// };
class Solution {
public:
    unordered_map<int, Employee*> temp;
    int getImportance(vector<Employee*> employees, int id) {
        //we have to find out which one in the vector employees has the correct id.        
        for(int i =0; i < employees.size(); ++i){
            temp[employees[i]->id] = employees[i]; 
        }
        return help_getImportance(temp, id);
    }
    int help_getImportance(unordered_map<int, Employee*> temp, int id) {
        Employee* p_employ = temp[id]; 
        // if((p_employ->subordinates).empty())
        //     return p_employ->importance;
        int sum = p_employ->importance;
        // for(int i = 0; i < (p_employ->subordinates).size(); ++i){
        //     sum += help_getImportance(temp, (p_employ->subordinates)[i]);
        // }
        for(int id_sub : (p_employ->subordinates)){
            sum += help_getImportance(temp, id_sub);
        }
        return sum;
    }    
};

第一种方法是错的，理解题目错了，由于给定的id并非对应employees的索引，须要本身建哈希表，把id与该id对应的Employee*型指针造成映射（或者在employees中的索引），抛弃方法二中的employees的索引。其实后两种时间差很少。

更新：

class Solution {
public:
    unordered_map<int, Employee*> temp;
    int getImportance(vector<Employee*> employees, int id) {
        //we have to find out which one in the vector employees has the correct id.        
        for(Employee* e : employees){
            temp[e->id] = e; 
        }
        return help_getImportance(id);
    }
    int help_getImportance(int id) {
        Employee* p_employ = temp[id]; 
        // if((p_employ->subordinates).empty())
        //     return p_employ->importance;
        int sum = p_employ->importance;
        // for(int i = 0; i < (p_employ->subordinates).size(); ++i){
        //     sum += help_getImportance(temp, (p_employ->subordinates)[i]);
        // }
        for(int id_sub : (p_employ->subordinates)){
            sum += help_getImportance(id_sub);
        }
        return sum;
    }    
};

这里对help函数，再也不传入哈希表做为实参，从而大大提高了速度，也大大下降了堆栈溢出的风险。

Runtime: 32 ms, faster than 97.67% of C++ online submissions for Employee Importance.

Memory Usage: 15 MB, less than 75.00% of C++ online submissions for Employee Importance.