Course Schedule II

The same as I:

public class Solution {
    /**
     * @param numCourses a total of n courses
     * @param prerequisites a list of prerequisite pairs
     * @return the course order
     */
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        // Write your code here
        List[] edges = new ArrayList[numCourses];
        int[] degree = new int[numCourses];
        int[] rst = new int[numCourses];
        
        for (int i = 0; i < numCourses; i++) {
            edges[i] = new ArrayList<Integer>();
        }
        
        for (int i = 0; i < prerequisites.length; i++) {
            degree[prerequisites[i][0]]++;
            edges[prerequisites[i][1]].add(prerequisites[i][0]);
        }
        
        Queue<Integer> q = new LinkedList<Integer>();
        for (int i = 0; i < numCourses; i++) {
            if (degree[i] == 0) {
                q.offer(i);    
            }
        }
        int index = 0;
        while(!q.isEmpty()) {
            int course = q.poll();
            rst[index++] = course;
            int n = edges[course].size();
            for (int i = 0; i < n; i++) {
                int nextCourse = (int)edges[course].get(i);
                degree[nextCourse]--;
                if (degree[nextCourse] == 0) {
                    q.offer(nextCourse);
                }
            }
        }
        
        if (index == numCourses) {
            return rst;
        }
        
        return new int[0];
    }
}

 

转载于:https://www.cnblogs.com/codingEskimo/p/7007010.html