Course Schedule II

Course Schedule II

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, 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.

思路：

　　dfs

我的代码：

public class Solution {
     public int[] findOrder(int numCourses, int[][] prerequisites) {
        if(prerequisites==null || prerequisites.length==0) {
            int[] a = new int[numCourses];
            for(int i=0;i<numCourses;i++) {
                a[i]=i;
            }
            return a;
        }
        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(graph, visited, visiting, i)) return new int[0];
        }
        int[] nums = new int[numCourses];
        int i = 0;
        for(int count=rst.size()-1; count>=0; count--)
            nums[i++] = rst.get(count);
        return nums;
    }
    private List<Integer> rst = new ArrayList<Integer>();
    public boolean helper(HashSet<Integer>[] graph, boolean[] visited, boolean[] visiting, int cur)
    {
        if(visiting[cur]) return false;
        visiting[cur] = true;
        for(Integer neighbor: graph[cur])
        {
            if(visited[neighbor]) continue;
            if(!helper(graph, visited, visiting, neighbor)) return false;  
        }
        visited[cur] = true;
        visiting[cur] = false;
        rst.add(cur);
        return true;
    }
}

 

学习之处：

• 什么叫一个节点访问结束了，只有当这个节点的所有邻居都访问完了，才算结束了，此时visited[cur] = true;