690. Employee Importance

Problem:

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:

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

思路：

Solution (C++):

int getImportance(vector<Employee*> employees, int id) {
    unordered_map<int, Employee*> hash;
    for (auto x : employees)  hash[x->id] = x;
    int sum = 0;
    deque<Employee*> que;
    que.push_back(hash[id]);
    while (!que.empty()) {
        auto p = que.front();
        que.pop_front();
        for (auto x : p->subordinates)  que.push_back(hash[x]);
        sum += p->importance;
    }
    return sum;
}

性能：

Runtime: 32 ms  Memory Usage: 12.6 MB

思路：

Solution (C++):

性能：

Runtime: ms  Memory Usage: MB