[每日一题] leetcode 690. 员工的重要性

/*
// Definition for Employee.
class Employee {
public:
    int id;
    int importance;
    vector<int> subordinates;
};
*/

class Solution {
public:
    int vis[10010];
    int dfs(Employee *p, vector<Employee*> employees)
    {
        if(p == NULL) return 0;
       // cout << 1111 << endl;
        int sum = 0;
        sum += p->importance;
      int n = (p->subordinates).size();
        for(int i = 0; i < n; i++)
        {
            sum += dfs(employees[vis[p->subordinates[i]]], employees);
        }
        return sum;

    }

    int getImportance(vector<Employee*> employees, int id) {
        
        int n = employees.size();
        memset(vis, 0, sizeof(vis));
        for(int i = 0; i < n; i++)
            vis[employees[i]->id] = i;

        return dfs(employees[vis[id]], employees);
        
    }
};

 

posted @ 2021-05-02 17:23  WTSRUVF  阅读(47)  评论(0编辑  收藏  举报