1 """ 2 There are a total of n courses you have to take, labeled from 0 to n-1. 3 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] 4 Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses? 5 Example 1: 6 Input: 2, [[1,0]] 7 Output: true 8 Explanation: There are a total of 2 courses to take. 9 To take course 1 you should have finished course 0. So it is possible. 10 Example 2: 11 Input: 2, [[1,0],[0,1]] 12 Output: false 13 Explanation: There are a total of 2 courses to take. 14 To take course 1 you should have finished course 0, and to take course 0 you should 15 also have finished course 1. So it is impossible. 16 """ 17 """ 18 此题是判断有向无环图 19 即拓扑排序 20 提供两种思路 21 第一种思路是维护一个indegrees数组 22 每次删除度数为0的结点，并将其指向结点度数减1 23 N次把所有结点取完 即返回True 24 """ 25 class Solution1(object): 26 def canFinish(self, numCourses, prerequisites): 27 from collections import defaultdict 28 graph = defaultdict(list) #defaultdict(<class 'list'>) 29 indegrees = defaultdict(int) #defaultdict(<class 'int'>) 30 for u, v in prerequisites: #构建图{初始结点1:[目的结点1, 目的结点2, ...], ...} 31 graph[v].append(u) #结点度数{结点1:度数, 结点2:度数, ...} 32 indegrees[u] += 1 33 N = numCourses 34 for i in range(N): #每次循环将度为0的结点削掉，共N次 35 zeroDegree = False 36 for j in range(N): 37 if indegrees[j] == 0: #找到度为0的结点 38 zeroDegree = True 39 break 40 if not zeroDegree: 41 return False 42 indegrees[j] -= 1 #将度为0的结点减为-1 43 for node in graph[j]: #将其所连的结点各减1 44 indegrees[node] -= 1 45 return True 46 47 """ 48 第二种思路用DFS，递归，没有掌握 49 传送门：https://blog.csdn.net/fuxuemingzhu/article/details/82951771 50 """ 51 class Solution2(object): 52 def canFinish(self, numCourses, prerequisites): 53 from collections import defaultdict 54 graph = defaultdict(list) # defaultdict(<class 'list'>) 55 for u, v in prerequisites: #构造图 56 graph[v].append(u) 57 N = numCourses 58 visited = [0] * N 59 # 0 = Unknown, 1 = visiting, 2 = visited 60 for i in range(N): #分别从N个结点出发调用dfs判断是否有环 61 if not self.dfs(graph, visited, i): #有环dfs返回True 62 return False 63 return True 64 def dfs(self, graph, visited, i): #这个函数云里雾里。。没懂 65 if visited[i] == 1: 66 return False 67 if visited[i] == 2: 68 return True 69 visited[i] = 1 70 for j in graph[i]: 71 if not self.dfs(graph, visited, j): 72 return False 73 visited[i] = 2 74 return True