210. 课程表 II

该题用到「拓扑排序」的算法思想，关于拓扑排序，直观地说就是，让你把⼀幅图「拉平」，⽽且这个「拉平」的图⾥⾯，所有箭头⽅向都是⼀致的，⽐如上图所有箭头都是朝右的。
很显然，如果⼀幅有向图中存在环，是⽆法进⾏拓扑排序的，因为肯定做不到所有箭头⽅向⼀致；反过来，如果⼀幅图是「有向⽆环图」，那么⼀定可以进⾏拓扑排序。

class TopologicalSorting:"""拓扑排序算法课程表II，给出可能的课程安排顺序https://leetcode.cn/problems/course-schedule-ii/"""def __init__(self):self.hascycle = Falseself.postorder = []def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:"""dfs:param numCourses::param prerequisites::return:"""graph = self.buildGraph(numCourses, prerequisites)self.visited = [False] * numCoursesself.onPath = [False] * numCoursesfor i in range(numCourses):self.dfs(graph, i)if self.hascycle:return []# 对后序遍历进行反转res = self.postorder[::-1]return resdef dfs(self, graph, i):if self.onPath[i]:self.hascycle = Truereturnif self.visited[i]:return# 前序遍历位置self.onPath[i] = Trueself.visited[i] = Truefor t in graph[i]:self.dfs(graph, t)# 后序遍历位置self.postorder.append(i)self.onPath[i] = Falsedef buildGraph(self, numCourses: int, prerequisites: List[List[int]]) -> List[List[int]]:# 注意这两种新建对象的区别，前者是传的引用，后者是拷贝一个新的变量# graph = [[]] * numCoursesgraph = [[] for _ in range(numCourses)]for edge in prerequisites:src = edge[1]dst = edge[0]graph[src].append(dst)return graphdef findOrder2(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:"""BFS 实现借助 indegree 数组实现visited数组的作用，只有入度为0的节点才能入队，不会出现死循环:param numCourses::param prerequisites::return:"""graph = self.buildGraph(numCourses, prerequisites)self.indegree = [0] * numCoursesfor edge in prerequisites:dst = edge[0]self.indegree[dst] += 1queue = []for i in range(numCourses):if self.indegree[i] == 0:queue.append(i)res = [0] * numCourses# 记录遍历节点的顺序count = 0while queue:cur = queue.pop(0)# 弹出节点的顺序即为拓扑排序结果res[count] = curcount += 1for neighbor in graph[cur]:self.indegree[neighbor] -= 1if self.indegree[neighbor] == 0:queue.append(neighbor)# 存在环if count != numCourses:return []return res