題目出處
難度
medium
個人範例程式碼
class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
if not prerequisites: # no course
return True
return self.bfs(numCourses, prerequisites)
def bfs(self, num_courses, prerequisites):
# init
pre_courses, next_courses = self.get_table(num_courses, prerequisites)
start_courses = []
learned_course = []
for course, pre in pre_courses.items():
if not pre:
start_courses.append(course)
queue = collections.deque(start_courses)
visited = set()
while queue:
course = queue.popleft()
learned_course.append(course)
visited.add(course)
for next_course in next_courses[course]:
pre_courses[next_course].remove(course)
if not pre_courses[next_course] and next_course not in visited: # no pre-courses
queue.append(next_course)
else:
return num_courses == len(learned_course)
def get_table(self, num_courses, prerequisites):
pre_courses = {x:[] for x in range(num_courses)} # for record studied
next_courses = {x:[] for x in range(num_courses)} # for search
for course in prerequisites:
pre_courses[course[1]].append(course[0])
next_courses[course[0]].append(course[1])
return pre_courses, next_courses
算法說明
這題考的是 BFS 中的拓樸排序法,個人認為比較難的是在 graph 操作的部分,其他的概念相對來說還好。
另外處理 start_courses ,去尋找 indegree 為 0 的 node
最近在練習程式碼本身就可以自解釋的 Coding style,可以嘗試直接閱讀程式碼理解
input handling
如果沒有 prerequisites,則必 True (無課程或無先修課)
Boundary conditions
當 queue 為空時,如果還沒有辦法上完全部的課程,則為 False
(可能 Graph 有 loop)