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, is it possible for you to finish all courses?
There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.
2, [[1,0],[0,1]]There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented. You may assume that there are no duplicate edges in the input prerequisites.
本题实际上是对有向无环图（DAG）的判断，解法通常有拓扑排序和DFS（或BFS），这里可以用拓扑排序。拓扑排序的判断方法是找出一个图中入度为0的点，然后删除它，再在剩下的点中找入度为0的点，直到所有的点都删除完毕，如果所有点都能删除掉，那么就完成了一次拓扑排序，这时候说明图中无环，否则则说明有环。
下面是使用C++的实现过程：
class Solution { public: bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) { vector<vector<int>> graph(numCourses); vector<int> indegree(numCourses,0);//用于记录入度 for(int i=0;i<prerequisites.size();i++){ graph[prerequisites[i].second].push_back(prerequisites[i].first); indegree[prerequisites[i].first]++; } queue<int> Q;//用于记录入度为0的点 for(int i = 0;i < numCourses;i++){ if(indegree[i] == 0) Q.push(i); } int counter = 0; while(!Q.empty()) { int u = Q.front(); Q.pop(); count++;//每删除一个点做一次记录 for(int v=0;v<graph[u].size();v++){ if(--indegree[graph[u][v]]==0) Q.push(graph[u][v]); } } return count == numCourses; } };