LeetCode 207. Course Schedule--有向图找环--面试算法题--DFS递归，拓扑排序迭代--Python

题目地址：Course Schedule - LeetCode

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 shouldalso have finished course 1. So it is impossible.

Note:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.

这道题目是我在XX公司面试时遇到的算法题，当时就看出来这是判断有向图（DAG）是否存在环。
一时想不起算法，于是就自己实现了一个时间复杂度为O(V^2)的算法，面试官自然不满意，于是当时现场又用递归的做法实现了时间复杂度为O(N+V)的算法(dfs)，这个做法只是口述了一遍，并没有真的实现。现在想想实现起来还是非常复杂的。

刚刚我实现了O(V^2)的算法，发现算法有bug，尴尬了。
然后实现自己的DFS算法，花了至少20分钟，现场做肯定做不出来。

DFS-Python解法如下：

class Solution:def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:s = {}res = set()flag = 0def helper(begin=-1, find=set()):nonlocal flag, res, sfor i in s[begin]:if i not in res:if i in find:flag = 1returnfind.add(i)helper(i, find=find)res.add(begin)for i in range(0, numCourses):s[i] = set()for i in prerequisites:s[i[0]].add(i[1])for i in range(0, numCourses):helper(begin=i, find={i})if flag == 1:return Falsereturn True

代码不好，别人写的优化过的代码如下：

class Solution(object):def canFinish(self, numCourses, prerequisites):""":type numCourses: int:type prerequisites: List[List[int]]:rtype: bool"""graph = [[] for _ in range(numCourses)]visited = [0 for _ in range(numCourses)]# create graphfor pair in prerequisites:x, y = pairgraph[x].append(y)# visit each nodefor i in range(numCourses):if not self.dfs(graph, visited, i):return Falsereturn Truedef dfs(self, graph, visited, i):# if ith node is marked as being visited, then a cycle is foundif visited[i] == -1:return False# if it is done visted, then do not visit againif visited[i] == 1:return True# mark as being visitedvisited[i] = -1# visit all the neighboursfor j in graph[i]:if not self.dfs(graph, visited, j):return False# after visit all the neighbours, mark it as done visitedvisited[i] = 1return True

随后看到另外一位大佬用OOP的思想重写了代码，虽然速度没有提升，但思路清晰多了：

class Solution:class Course(object):def __init__(self):self.being_visit, self.visit_done = False, Falseself.pre_course = []def is_cyclic(self):if self.visit_done is True:return Falseif self.being_visit is True:return Trueself.being_visit = Truefor course in self.pre_course:if course.is_cyclic():return Trueself.being_visit = Falseself.visit_done = Truereturn Falsedef canFinish(self, num_courses, prerequisites) -> bool:l = [self.Course() for _ in range(0, num_courses)]for (course1, course2) in prerequisites:l[course1].pre_course.append(l[course2])for i in range(0, num_courses):if l[i].is_cyclic():return Falsereturn True

标准的更正式更快的解法是拓扑排序（Topological Sorting），可以参考：Kahn’s algorithm for Topological Sorting - GeeksforGeeks

class Solution:def canFinish(self, num_courses, prerequisites) -> bool:vertices = num_coursesadj_list = [[] for _ in range(0, vertices)]in_degree = [0 for _ in range(0, vertices)]for (course1, course2) in prerequisites:adj_list[course1].append(course2)in_degree[course2] += 1q = [i for i in range(0, vertices) if in_degree[i] == 0]count = 0while q:front = q.pop(0)for i in adj_list[front]:in_degree[i] -= 1if in_degree[i] < 0:return Falseelif in_degree[i] == 0:q.append(i)count += 1if count == vertices:return Trueelse:return False

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LeetCode 207. Course Schedule--有向图找环--面试算法题--DFS递归，拓扑排序迭代--Python

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