LN : leetcode 690 Employee Importance

lc 690 Employee Importance

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690 Employee Importance

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

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.

BFS Accepted##

很经典的可以用BFS算法快速解决的问题。唯一需要注意的是push进队列的应该为id-1，而不是id，因为它是vector的下标。

/*
// 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.size()) return 0;
        int total_value = 0;
        queue<int> q;
        q.push(id-1);
        while (!q.empty()) {
            auto employee = employees[q.front()];
            q.pop();
            total_value += employee->importance;
            for (auto i : employee->subordinates)   q.push(i-1);
        }
        return total_value;
    }
};