Course Schedule

Course Schedule

问题：

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?

思路：

　　拓扑排序，dfs

我的代码：

public class Solution {

    public boolean canFinish(int numCourses, int[][] prerequisites) {

        if(numCourses<=0 || prerequisites==null || prerequisites.length==0) return true;

        HashSet<Integer>[] graph = new HashSet[numCourses];

        for(int i=0; i<numCourses; i++)

            graph[i] = new HashSet<Integer>();

        boolean[] visited = new boolean[numCourses];

        boolean[] visiting = new boolean[numCourses];

        

        for(int i=0; i<prerequisites.length; i++)

        {

            graph[prerequisites[i][1]].add(prerequisites[i][0]);

        }

        for(int i=0; i<numCourses; i++)

        {

            if(visited[i]) continue;

            if(helper(visited, visiting, graph, i) == false)    return false;

        }

        return true;

    }

    public boolean helper(boolean[] visited, boolean[] visiting, HashSet<Integer>[] graph, int j)

    {

        if(visiting[j]) return false;

        visiting[j] = true;

        

        for(Integer neighbor: graph[j])

        {

            if(visited[neighbor]) continue;

            if(helper(visited, visiting, graph, neighbor) == false) return false;

        }

        

        visiting[j] = false;

        visited[j] = true;

        return true;

    }

    

    

}

View Code

学习之处:

　　这道题有思路，但是一直不会写拓扑排序，主要的纠结在于如何判断是否又环，这个人的思路很赞，通过visited判断节点是否访问过，visiting用于判断是否有环。学习了。