Leetcode 207. Course Schedule

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?

Example 1:

Input: 2, [[1,0]] 
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: 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.

Answer:

from collections import defaultdict 
class Solution( ):
    def dfs(self,node,graph):
        self.visited[node]=True
        self.track[node]=True
        
        for nextnode in graph[node]:
            if not self.visited[nextnode]:
                if self.dfs(nextnode,graph):
                    return True
            elif self.track[nextnode]:
                return True
        
        self.track[node]=False
        return False
                
    def canFinish(self, numCourses, prerequisites):
        \"\"\"
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: bool
        \"\"\"
        self.visited=[False]*numCourses
        self.track=[False]*numCourses
        graph=defaultdict(list)
        for p in prerequisites:
            graph[p[0]].append(p[1])
        
        
        
        for start in graph.keys():
            if self.visited[start]==False:
                if self.dfs(start,graph):
                    return False
        return True
        
        

dfs

如果有cycle存在，则一定有back edge存在，就是有edge从子node连到父级node

https://www.geeksforgeeks.org/detect-cycle-in-a-graph/