[LeetCode] 207 Course Schedule_Medium tag: BFS, DFS
2018-07-05 06:08 Johnson_强生仔仔 阅读(263) 评论(0) 编辑 收藏 举报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:
- 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.
我的理解是这个题就是问, 如果有两门课互为prerequisite课, 那么就False, 否则True, 注意的是这里的互为有可能是中间隔了几门课, 而不是直接的prerequisite, 例如: [(0,1),(2,0),(1,2)], 这里 1 是0 的前置课, 0 是2 的前置课, 所以1是2 的间接前置课, 但是最后一个input说2 是1 的前置课, 所以就矛盾, 不可能完成, return False. 所以思路为, 建一个dictionary, 分别将input 的每个pair(c1, c2)放入dictionary里面, 前置课(c2)为key, 后置课(c1)为value, 不过放入之前要用bfs 判断c1 是否为c2 的前置课, 如果是, 那么矛盾, return False. 否则一直判断到最后的pair, 返回Ture.
12/05/2019 Update: 这个题目实际上是有向图里面找是否有环的问题。用dfs去遍历每个graph的点,可以参考Directed Graph Loop detection and if not have, path to print all path. T: O(n) S: O(n)
1. Constraints:
1) 实际这里的n对我这个做法没有什么用处, 因为课程id 是unique的.
2. Ideas
BFS: T: O(n) number of nodes, S: O(n^2)
1) init dictionary, d
2) for pair(c1,c2) in prerequisites, use bfs to see if c1 is a prerequisity of c2, if so , return False, else, d[c2].add(c1), and until all pairs been checked. return True
3) bfs: use queue and visited to check whether there is a path from source to target.
3. Code
1 class Solution: 2 def courseSchedule(self, numCourse, prerequisites): 3 def bfs(d, source, target): 4 if source not in d: return False 5 queue, visited = collections.deque([source]), set([source]) 6 while queue: 7 node = queue.popleft() 8 if node == target: return True 9 for each in d[node]: 10 if each not in visited: 11 queue.append(each) 12 visited.add(each) 13 return False 14 15 d = collections.defaultdict(set) 16 for c1, c2 in prerequisites: 17 if bfs(d,c1, c2): return False 18 d[c2].add(c1) 19 return True
Using DFS:
class Solution: def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool: visited, graph = collections.Counter(), collections.defaultdict(set) for c1, c2 in prerequisites: graph[c1].add(c2) for i in range(numCourses): if not self.checkLoop(graph, visited, i): return False return True def checkLoop(self, graph: dict, visited: dict, course: int) -> bool: if visited[course] == 1: return True if visited[course] == -1: return False visited[course] = -1 for neig in graph[course]: if not self.checkLoop(graph, visited, neig): return False visited[course] = 1 return True
4. Test cases:
1) [(0,1),(2,0),(1,2)], => False