항해99 2주차 - 코스스케쥴
Today I learned
2022/01/20
회고록
1/20
항해 99, 알고리즘 1주차
교재 : 파이썬 알고리즘 인터뷰
12장 그래프
1. 이론
2. 문제
There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.
For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.
Return true if you can finish all courses. Otherwise, return false.
Example 1:
Input: numCourses = 2, prerequisites = [[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: numCourses = 2, prerequisites = [[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.
Constraints:
1 <= numCourses <= 105
0 <= prerequisites.length <= 5000
prerequisites[i].length == 2
0 <= ai, bi < numCourses
All the pairs prerequisites[i] are unique.
https://leetcode.com/problems/course-schedule/
3. MySol
- Recursive DFS
from typing import List
import collections
class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
graph = collections.defaultdict(list)
for x, y in prerequisites:
graph[x].append(y)
traced = set()
def dfs(i):
if i in traced:
return False
traced.add(i)
for y in graph[i]:
if not dfs(y):
return False
traced.remove(i)
return True
for x in list(graph):
if not dfs(x):
return False
return True
if __name__ == '__main__':
courses = [[0,1],[1,0]]
num = 3
sol = Solution()
result = sol.canFinish(3, courses)
print(f'result : {result}')
4. 배운 점
- 해당 문제는 이해하게 되면 DFS및 재귀함수에 대해 이해하는데 굉장히 큰 도움이 되는 문제이다.
5. 총평
재귀, DFS 훈련
Author And Source
이 문제에 관하여(항해99 2주차 - 코스스케쥴), 우리는 이곳에서 더 많은 자료를 발견하고 링크를 클릭하여 보았다 https://velog.io/@jsw4215/항해99-2주차-코스스케쥴저자 귀속: 원작자 정보가 원작자 URL에 포함되어 있으며 저작권은 원작자 소유입니다.
우수한 개발자 콘텐츠 발견에 전념 (Collection and Share based on the CC Protocol.)