链接：https://leetcode.com/problems/course-schedule-ii/
难度：medium

解题思路，优先输出没有前置课程的课，用degrees数组来记录课i的前置课程的数量degrees[i]。当其前置课程数量被选出后，课程的degrees[i] --。剩下就是广搜的思路，每次入队列时只考虑degrees[i] == 0的课程。若最后课程数量不等于numCourses，则说明至少有一门课有循环依赖。

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        
        int[] degrees = new int[numCourses];
        for(int i = 0; i < prerequisites.length; i ++) {
            degrees[prerequisites[i][0]] ++;
        }
        Queue<Integer> queue = new LinkedList<>();
        addEle(queue, degrees, numCourses);
        
        int res[] = new int[numCourses];
        int visited[] = new int[numCourses];
        int i = 0;
        while(queue.peek() != null) {
            int target = queue.poll();
            if(visited[target] == 1) {
                continue;
            }
            res[i ++] = target;
            visited[target] = 1;
            adjustDegrees(prerequisites, degrees, target);
            addEle(queue, degrees, numCourses);
            
        }
        
        if(i < numCourses) return new int[]{};
        return res;
        
    }
    
    private void adjustDegrees(int[][] pr,int[] degrees, int target) {
        for(int i = 0; i < pr.length; i ++) {
            if(pr[i][1] == target) {
                degrees[pr[i][0]] --;
                degrees[pr[i][1]] --;
            }
        }
    }
    
    private void addEle(Queue<Integer> queue, int[] degrees, int numCourses) {
        for(int i = 0; i < numCourses; i ++) {
            if(degrees[i] == 0) {
                queue.offer(i);
            }
        }
    }
    
   
}

Runtime: 38 ms, faster than 6.90% of Java online submissions for Course Schedule II.
Memory Usage: 44.6 MB, less than 93.90% of Java online submissions for Course Schedule II.