Leetcode 210 Course Schedule II

本文介绍了一种解决课程先修顺序问题的算法实现。通过构建图结构来表示课程及其依赖关系,并利用拓扑排序来确定有效的课程学习顺序。适用于存在课程先修条件的场景。

here 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, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

同 207 Course Schedule

每次取出点的时候放入数组最后输出。

require 'set'
def can_finish(num_courses, prerequisites)
  graph, neighbour = Hash.new{|hsh,key| hsh[key] = Set.new}, Hash.new{|hsh,key| hsh[key] = Set.new}
  prerequisites.each do |x,y|
    graph[x] << y
    neighbour[y] << x
  end
  zero_degree, count = [], 0
  ans =[]
  num_courses.times {|x| zero_degree << x if graph[x].empty?}
  while not zero_degree.empty?
    node = zero_degree.pop
    ans << node
    count += 1
    neighbour[node].each do |x|
      graph[x] -= [node]
      zero_degree << x if graph[x].empty?
    end
  end
  return ans if count == num_courses
  []
end

 

转载于:https://www.cnblogs.com/lilixu/p/4570878.html

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