210. Course Schedule II

很像图的问题 实际上只用考虑indegree 当indegree等于0的时候 就可以被放进queue 进行bfs 最后能放进n个course就可以返回

 

 

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        if(numCourses == 0) return new int[0];
        HashMap<Integer, List<Integer>> graph = new HashMap<>();
        int[] indegree = new int[numCourses];
        for(int[] pre : prerequisites){
            indegree[pre[0]]++;
            if(!graph.containsKey(pre[1])){
                graph.put(pre[1], new ArrayList<>());
            }
            graph.get(pre[1]).add(pre[0]);
        }
        Queue<Integer> queue = new LinkedList<>();
        List<Integer> res = new ArrayList<>();
        for(int i = 0; i < numCourses; i++){
            if(indegree[i] == 0){
                queue.offer(i);
                res.add(i);
            }
        }

        int size = queue.size();
        while(!queue.isEmpty()){
            for(int i = 0; i < size; i++){
                int course = queue.poll();
                if(graph.containsKey(course)){
                    List<Integer> pairs = graph.get(course);
                    for(int pair : pairs){
                        indegree[pair]--;
                        if(indegree[pair] == 0){
                            queue.offer(pair);
                            res.add(pair);
                        }
                    }
                }
                
            }
            size = queue.size();
        }
        int[] arr = new int[numCourses];
        if(res.size() == numCourses){
            for(int i = 0; i < numCourses; i++){
                arr[i] = res.get(i);
            }
            return arr;
        }
        return new int[0];    
    }
    
}