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

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.

LeetCode：連結

對於每一對課程的順序關係，把它看做是一個有向邊，邊是由兩個端點組成的，用兩個點來表示邊，所有的課程關係即構成一個有向圖，問題相當於判斷有向圖中是否有環。判斷有向圖是否有環的方法是拓撲排序。

拓撲排序：維護一張表記錄所有點的入度，移出入度為0的點並更新其他點的入度，重複此過程直到沒有點的入度為0。如果原有向圖有環的話，此時會有剩餘的點且其入度不為0；否則沒有剩餘的點。

圖的拓撲排序可以DFS或者BFS。遍歷所有邊，計算點的入度；將入度為0的點移出點集，並更新剩餘點的入度；重複步驟2，直至沒有剩餘點或剩餘點的入度均大於0。

import collections
class Solution(object):
    def canFinish(self, N, prerequisites):
        """
        :type N,: int
        :type prerequisites: List[List[int]]
        :rtype: bool
        """
        # 記錄圖的位置關係
        graph = collections.defaultdict(list)
        # 記錄圖上節點的度
        indegrees = collections.defaultdict(int)
        for u, v in prerequisites:
            graph[v].append(u)
            indegrees[u] += 1
        # 正常取點N次
        for i in range(N):
            zeroDegree = False
            for j in range(N):
                # 每次都找一個入度為0的點
                if indegrees[j] == 0:
                    zeroDegree = True
                    break
            if not zeroDegree: 
                return False
            # 把它的入度變為-1
            indegrees[j] = -1
            # 把從這個點指向的點的入度都-1
            for node in graph[j]:
                indegrees[node] -= 1
        return True