210. 课程表 II

 1 class Solution
 2 {
 3     vector<int> res;
 4 public:
 5     vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites)
 6     {
 7         vector<vector<int>> graph(numCourses);
 8         vector<int> in_degree(numCourses, 0);
 9         for (int i = 0; i < prerequisites.size(); i++)
10         {
11             in_degree[prerequisites[i][0]]++;
12             graph[prerequisites[i][1]].push_back(prerequisites[i][0]);
13         }
14
15         queue<int> q;
16         vector<bool> vis(numCourses, false);
17
18         for (int i = 0; i < numCourses; i++)
19             if (in_degree[i] == 0)
20                 q.push(i);
21         while (!q.empty())
22         {
23             int sta = q.front();
24             q.pop();
25             vis[sta] = true;
26             res.push_back(sta);
27
28             //有哪些邻边
29             for (int i = 0; i < graph[sta].size(); i++)
30             {
31                 in_degree[graph[sta][i]]--;// 入度-1
32                 if (in_degree[graph[sta][i]] == 0) //入度如果为0，加入队列
33                     q.push(graph[sta][i]);
34             }
35         }
36
37         //0->1->2
38         for (int i = 0; i < numCourses; i++)
39             if (vis[i] == false)//只要有false，则返回{}
40                 return {};//有环
41         return res;
42     }
43 };

原文地址：https://www.cnblogs.com/yuhong1103/p/12633615.html