卖萌的弱渣

I am stupid, I am hungry.

Course Schedule 2

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.

Example:

1

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]

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:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices.

Solution

Java

(Course-Schedule2.java) download
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public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] ret = new int[numCourses];
        // 数组链表表示DAG: 课程x的前导课都在graph[x]中
        ArrayList<Integer>[] graph = new ArrayList[numCourses];
        // 入度数组
        int[] indegree = new int[numCourses];
        // 初始化indegree 和　graph
        for(int i=0; i<prerequisites.length; i++){
            if(graph[prerequisites[i][1]]==null)
                graph[prerequisites[i][1]] = new ArrayList<Integer>();
            graph[prerequisites[i][1]].add(prerequisites[i][0]);
            indegree[prerequisites[i][0]]++;
        }
        Queue<Integer> queue = new LinkedList<Integer>();
        // 把所有入度为０的加入queue
        for(int i=0; i<indegree.length; i++){
            if(indegree[i]==0)
                queue.add(i);
        }

        int idx = 0;
        while(!queue.isEmpty()){
            Integer cur = queue.poll();
            ret[idx++] = cur;
            // 这个课程没有前导了
            if(graph[cur]==null) continue;
            for (Integer nextIdx : graph[cur]){
                // 不管是不是０，indegree[next]--都会执行
                if(--indegree[nextIdx]==0)
                    queue.offer(nextIdx);

            }
        }
        return idx != numCourses ? new int[0] : ret;
    }
}