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# 690.Employee Importance

**690.Employee Importance**

难度:Easy

> 给定一个保存员工信息的数据结构，它包含了员工唯一的id，重要度 和 直系下属的id。

比如，员工1是员工2的领导，员工2是员工3的领导。他们相应的重要度为15, 10, 5。那么员工1的数据结构是\[1, 15, \[2]]，员工2的数据结构是\[2, 10, \[3]]，员工3的数据结构是\[3, 5, \[]]。注意虽然员工3也是员工1的一个下属，但是由于并不是直系下属，因此没有体现在员工1的数据结构中。

现在输入一个公司的所有员工信息，以及单个员工id，返回这个员工和他所有下属的重要度之和。

```
示例 1:

输入: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
输出: 11
解释:
员工1自身的重要度是5，他有两个直系下属2和3，而且2和3的重要度均为3。因此员工1的总重要度是 5 + 3 + 3 = 11。
```

注意:

一个员工最多有一个直系领导，但是可以有多个直系下属 员工数量不超过2000。

DFS，通过递归解决。

```
/*
// 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 {
private:
    int getIndex(vector<Employee*> employees, int id)
    {
        for(int i=0;i<employees.size();i++)
            if(employees[i]->id ==id)
                return i;
        return -1;
    }
public:
    int getImportance(vector<Employee*> employees, int id) {
        int ind=getIndex(employees,id);
        int res=employees[ind]->importance;
        for(auto i: employees[ind]->subordinates)
        {
            int n=getIndex(employees,i);
            if(employees[n]->subordinates.empty())
            res+= employees[n]->importance;
            else
                res+=getImportance(employees,employees[n]->id);
            
        }

        return res;
        
    }
};
```

> 执行用时 : 56 ms, 在Employee Importance的C++提交中击败了77.53% 的用户 内存消耗 : 41 MB, 在Employee Importance的C++提交中击败了5.36% 的用户


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