210. Course Schedule II

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.

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].
Note:

  1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
  2. You may assume that there are no duplicate edges in the input prerequisites.

一刷
思路同207

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        List<Integer> seq = new ArrayList<>();
        int[][] topo = new int[numCourses][numCourses];
        int[] indegree = new int[numCourses];
        for(int i=0; i<prerequisites.length; i++){
            int pre = prerequisites[i][1];
            int ready = prerequisites[i][0];
            if(topo[pre][ready] == 0){
                indegree[ready]++;
            }
            topo[pre][ready] = 1;
        }
        
        
        Queue<Integer> wl = new LinkedList();
        for(int i=0; i<numCourses; i++){
            if(indegree[i] == 0) wl.offer(i);
        }
        
        while(!wl.isEmpty()){
            int course = wl.poll();
            seq.add(course);
            for(int i=0; i<numCourses; i++){
                if(topo[course][i]!=0){
                    indegree[i]--;
                    if(indegree[i] == 0) wl.offer(i);
                }
            }
        }
        
        if(seq.size()!=numCourses) return new int[0];
        
        int[] res = new int[seq.size()];
        for(int i=0; i<seq.size(); i++){
            res[i] = seq.get(i);
        }
        return res;
    }
}
最后编辑于
©著作权归作者所有,转载或内容合作请联系作者
平台声明:文章内容(如有图片或视频亦包括在内)由作者上传并发布,文章内容仅代表作者本人观点,简书系信息发布平台,仅提供信息存储服务。

推荐阅读更多精彩内容