本文主要是介绍LeetCode *** 210. Course Schedule II，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目：
There are a total of n courses you have to take, labeled from 0 to n - 1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.
There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
For example:
2, [[1,0]]
There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1]
4, [[1,0],[2,0],[3,1],[3,2]]
There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].
Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
分析：
跑了1484ms。。。。。
代码：
class Solution {
public:vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {vector<int> res;unordered_map<int,set<int>> record;for(auto req:prerequisites)record[req.first].insert(req.second);queue<int> que;for(int i=0;i<numCourses;++i)if(record.find(i)==record.end())que.push(i);while(!que.empty()){int tmp=que.front();que.pop();res.push_back(tmp);for(unordered_map<int,set<int>>::iterator it=record.begin();it!=record.end();++it)if(((*it).second).find(tmp)!=((*it).second).end()){((*it).second).erase(tmp);if(((*it).second).empty())que.push((*it).first);}}return res.size()==numCourses?res:vector<int>();}
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