【leetcode】207. Course Schedule

题目如下：

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.

解题思路：如果某一种输入课程无法被安排，那么一定存在至少这样一门课：通过BFS/DFS的方法从这门课开始，依次遍历需要安排在这门课后面的其他课，最终还会回到这门课，即组成了一个环。我们只有遍历所有的课，看看有没有哪门课会形成环路即可。

代码如下：

class Solution(object):
    def canFinish(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: bool
        """
        dic = {}
        for cou,pre in prerequisites:
            if pre not in dic:
                dic[pre] = [cou]
            else:
                dic[pre].append(cou)

        for i in range(numCourses):
            visit = [0] * numCourses
            queue = [i]
            start = None
            while len(queue) > 0:
                inx = queue.pop(0)
                if start == None:
                    start = inx
                elif inx == start:
                    return False
                if visit[inx] == 1:
                    continue
                visit[inx] = 1
                if inx in dic and len(dic[inx]) > 0:
                    queue += dic[inx]
        return True