[Leetcode] Course Schedule, Solution

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:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

click to show more hints.

Hints:

1. 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.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. 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:    }