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  • [leetcode-690-Employee Importance]

    You are given a data structure of employee information, which includes the employee's unique id, his importance value and his directsubordinates' 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
    Explanation:
    Employee 1 has importance value 5, and he has two direct subordinates: employee 2 and employee 3. They both have importance value 3. So the total importance value of employee 1 is 5 + 3 + 3 = 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. 

    思路:

    图的DFS遍历,但是感觉写的太耗时了,没必要用两个map。。待优化。

    
    

    // 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;
    };


    void
    DFS(int id,map<int,vector<int>>&mp,map<int,int>&mp2,int& sum) { sum += mp2[id]; for(int i =0;i<mp[id].size();i++) { DFS(mp[id][i],mp,mp2,sum); } } int getImportance(vector<Employee*> employees, int id) { map<int,vector<int>>mp; map<int,int>mp2;//importance for(int i =0;i<employees.size();i++) { mp2[employees[i]->id] = employees[i]->importance; mp[employees[i]->id] = employees[i]->subordinates; } int sum = 0; DFS(id,mp,mp2,sum); return sum; }
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  • 原文地址:https://www.cnblogs.com/hellowooorld/p/7636319.html
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