[LeetCode-JAVA] Course Schedule II

题目：

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.

For example:

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]

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].

思路：按照Course Schedule 的思路进行，将stack换成queue，每一次进行一次记录，会显示MLE，估计是邻接矩阵占用的空间太大了，因此要改用邻接链表的方式进行存储。

代码：

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        List<Set<Integer>> posts = new ArrayList<Set<Integer>>();   //初始化邻接链表
        for(int i = 0 ; i < numCourses ; i++){
            posts.add(new HashSet<Integer>());
        }
        
        for(int i = 0 ; i < prerequisites.length ; i++){  //赋值
            posts.get(prerequisites[i][1]).add(prerequisites[i][0]);
        }
        
        int[] indegrees = new int[numCourses];
        for (int i = 0; i < numCourses; i++) {  
            for (int x : posts.get(i)) {  
                indegrees[x]++;  
            }  
        }
        
        LinkedList<Integer> queue = new LinkedList<Integer>();  
        for (int i = 0; i < numCourses; i++) {  
            if (indegrees[i] == 0) {  
                queue.add(i);  
            }  
        }
        
        int[] res = new int[numCourses];  
        int count = 0;  
        while (!queue.isEmpty()) {  
            int cur = queue.poll();  
            for (int x : posts.get(cur)) {  
                indegrees[x]--;  
                if (indegrees[x] == 0) {  
                    queue.add(x);  
                }  
            }  
            res[count++] = cur;  
        }  
          
        if (count == numCourses) return res;  
        return new int[0];  
        
        
    }
}

参考链接：http://blog.csdn.net/ljiabin/article/details/45847019