leetcode #207

最新推荐文章于 2025-02-16 22:37:03 发布

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#作业

本文探讨了如何通过构建有向图并运用拓扑排序的方法来解决课程依赖问题，即判断是否能完成所有课程的学习任务。具体地，文章介绍了一种广度优先搜索（BFS）的实现方式，用于检测课程依赖图中是否存在环路。

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.

Hints:

This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
Topological sort could also be done via BFS.

题目分析：

给定一个课程表，该课程表中要上某些课程必须先上其他课程，问能否上完所有课程。

题目解析：

将该问题转化为有向图，课程当作顶点，有向边从一个顶点指向另一个定点表示上完该课程才能上下一个课程。题目问题转化为该有向图中是否有圈。可以采用拓扑排序的思路。这里提供广搜版本的代码。

代码：

from queue import Queue

class Solution:
    def canFinish(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: bool
        """
        graph = [[] for _ in range(numCourses)]
        queue = Queue()
        arr = []
        for edge in prerequisites:
            graph[edge[1]].append(edge[0])
        predecessor_count = [0] * numCourses
        for i in range(len(graph)):
            for vertex in graph[i]:
                predecessor_count[vertex] += 1

        for i in range(len(predecessor_count)):
            if predecessor_count[i] == 0:
                queue.put(i)
                arr.append(i)
        while not queue.empty():
            q = queue.get()
            for i in graph[q]:
                predecessor_count[i] -= 1
                if predecessor_count[i] == 0:
                    queue.put(i)
                    arr.append(i)
        return True if len(arr)==numCourses else False