lc 690 Employee Importance

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:

One employee has at most one direct leader and may have several

subordinates.
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;

    }

};

巴特西

LN : leetcode 690 Employee Importance

lc 690 Employee Importance

BFS Accepted##

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