77 lines
2.7 KiB
C++
77 lines
2.7 KiB
C++
/*/*******************************************************************************
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** **
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** Jiedi(China nanjing)Ltd. **
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** 创建：丁宋涛 夏曹俊，此代码可用作为学习参考 **
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*******************************************************************************/
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/*****************************FILE INFOMATION***********************************
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**
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** Project : 算法设计与编程实践---基于leetcode的企业真题库
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** Contact : xiacaojun@qq.com
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** 博客 : http://blog.csdn.net/jiedichina
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** 视频课程 : 网易云课堂 http://study.163.com/u/xiacaojun
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腾讯课堂 https://jiedi.ke.qq.com/
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csdn学院 https://edu.csdn.net/course/detail/25037
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** 51cto学院 http://edu.51cto.com/lecturer/index/user_id-100013755.html
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** 老夏课堂 http://www.laoxiaketang.com
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**
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** 算法设计与编程实践---基于leetcode的企业真题库 课程群 ：296249312 加入群下载代码和交流
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** 微信公众号 : jiedi2007
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** 头条号 : 夏曹俊
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**
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*****************************************************************************
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//！！！！！！！！！ 算法设计与编程实践---基于leetcode的企业真题库 课程 QQ群：296249312 下载代码和交流*/
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/*
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* @lc app=leetcode.cn id=690 lang=cpp
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*
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* [690] 员工的重要性
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*/
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/*
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// Employee info
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class Employee {
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public:
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// It's the unique ID of each node.
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// unique id of this employee
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int id;
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// the importance value of this employee
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int importance;
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// the id of direct subordinates
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vector<int> subordinates;
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};
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*/
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#include <vector>
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#include <queue>
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using namespace std;
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class Solution {
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public:
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int getImportance(vector<Employee*> employees, int id) {
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int res = 0;
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for(auto k :employees){
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if(k->id == id){
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BFS(employees,k,res);
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}
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}
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return res;
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}
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void BFS(vector<Employee*> employees,Employee* k,int& res){
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queue<Employee*> q;
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q.push(k);
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while(!q.empty()){
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Employee* top = q.front();
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q.pop();
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res += top->importance;
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if(top->subordinates.size()!=0){
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for(auto k:top->subordinates){
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for(int i=0;i<employees.size();i++){
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if(k==employees[i]->id){
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q.push(employees[i]);
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break;
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}
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}
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}
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}
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}
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}
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};
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