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?

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 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.

Note:

- 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.

Hints:

- This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
- Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
- Topological sort could also be done via BFS.

**[Thoughts]**

首先，用一个hashmap来保存课程之间的dependence。然后做DFS检查课程中是否有环存在，如果有环，则不可能。

**[Code]**

```
1: bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
2: unordered_map<int, vector<int>> deps(numCourses);
3:
4: for(auto pair : prerequisites) {
5: deps[pair.first].push_back(pair.second);
6: }
7:
8: vector<int> visited(numCourses, 0);
9: for(int i =0 ; i< numCourses; i++) {
10: if(isCycle(i, deps, visited)) return false;
11: }
12: return true;
13: }
14:
15: bool isCycle(int start, unordered_map<int, vector<int>>& deps, vector<int>& visited) {
16: if(visited[start] == 1) return true;
17:
18: visited[start] =1;
19:
20: for(auto edge: deps[start]) {
21: if(isCycle(edge, deps, visited)) return true;
22:
23: }
24: visited[start] =0;
25: return false;
26: }
```

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