# 207. Course Schedule 课程表

## # 题目描述：

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.

## # 方法一：拓扑排序，BFS

``````class Solution(object):
def canFinish(self, N, prerequisites):
"""
:type N,: int
:type prerequisites: List[List[int]]
:rtype: bool
"""
graph = collections.defaultdict(list)
indegrees = collections.defaultdict(int)
for u, v in prerequisites:
graph[v].append(u)
indegrees[u] += 1
for i in range(N):
zeroDegree = False
for j in range(N):
if indegrees[j] == 0:
zeroDegree = True
break
if not zeroDegree: return False
indegrees[j] = -1
for node in graph[j]:
indegrees[node] -= 1
return True
``````

### # 方法二：拓扑排序，DFS

findOrder函数中的for循环是怎么回事呢？这个和BFS循环次数不是同一个概念，这里的循环就是看从第i个节点开始能否到达合理结果。这个节点可能没有出度了，那就把它直接放到path里；也可能有出度，那么就把它后面的节点都进行一次遍历，如果满足条件的节点都放到path里，同时把这次遍历的所有节点都标记成了已经遍历；如果一个节点已经被安全的访问过，那么就放过它，继续遍历下个节点。

``````class Solution(object):
def canFinish(self, N, prerequisites):
"""
:type N,: int
:type prerequisites: List[List[int]]
:rtype: bool
"""
graph = collections.defaultdict(list)
for u, v in prerequisites:
graph[u].append(v)
# 0 = Unknown, 1 = visiting, 2 = visited
visited = [0] * N
for i in range(N):
if not self.dfs(graph, visited, i):
return False
return True

# Can we add node i to visited successfully?
def dfs(self, graph, visited, i):
if visited[i] == 1: return False
if visited[i] == 2: return True
visited[i] = 1
for j in graph[i]:
if not self.dfs(graph, visited, j):
return False
visited[i] = 2
return True
``````

## # 日期

2018 年 10 月 6 日 —— 努力看书