题目链接

210. 课程表 II

思路

在 BFS 过程中将所有入度为0的点放入结果集中，如果最终结果集中点的数目和课程数一样，则说明这个结果集可行。

代码

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] indegree = new int[numCourses];
        int[] result = new int[numCourses];
        List<List<Integer>> adjacency = new ArrayList<>();
        Queue<Integer> queue = new LinkedList<>();

        // set adjacency
        for(int i = 0; i < numCourses; i++){
            adjacency.add(new ArrayList<>());
        }
        // set indegree
        // if [1, 0] then let 0 -> 1
        for(int[] nodes : prerequisites){
            adjacency.get(nodes[1]).add(nodes[0]);
            indegree[nodes[0]]++;
        }
        // add nodes into queue which indegree = 0
        for(int i = 0; i < numCourses; i++){
            if(indegree[i] == 0){
                queue.add(i);
            }
        }

        int count = 0;
        while(!queue.isEmpty()){
            int currentNode = queue.remove();
            result[count] = currentNode;
            count++;

            // every adjacency's indegree reduction
            for(Integer nextNode : adjacency.get(currentNode)){
                indegree[nextNode]--;
                if(indegree[nextNode] == 0){
                    queue.add(nextNode);
                }
            }
        }

        if(count == numCourses){
            return result;
        }

        return new int[0];
    }
}