[leetcode]690. Employee Importance

[leetcode]690. Employee Importance

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Analysis

ummmm~—— [ummmm~]

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.
DFS 递归解决~

Implement

/*
// 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) {
        int len = employees.size();
        if(len == 0)
            return 0;
        Employee* target;
        for(auto e:employees){
            if(e->id == id){
                target = e;
                break;
            }
        }
        int sum = target->importance;
        for(auto s:target->subordinates){
            sum += getImportance(employees, s);
        }
        return sum;
    }
};