Question

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].

Code

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] degree = new int[numCourses];
        int m = prerequisites.length;
        
        for (int i = 0; i < m; i++) {
            degree[prerequisites[i][0]]++;
        }
        
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < degree.length; i++) {
            if (degree[i] == 0) queue.offer(i);
        }
        
        int[] result = new int[numCourses];
        int count = 0, total = queue.size();
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int pre = queue.poll();
                result[count++] = pre;
                for (int j = 0; j < m; j++) {
                    if (prerequisites[j][1] == pre) {
                        int post = prerequisites[j][0];
                        degree[post]--;
                        if (degree[post] == 0) {
                            queue.offer(post);
                            total++;
                        }
                    }
                }
            }
        }
        if (total != numCourses) return new int[0];
        return result;
    }
}

Solution

與Course Schedule思路相同，拓?fù)渑判颉FS時(shí)按序?qū)⒚恳粚拥捻旤c(diǎn)加入結(jié)果集中。同時(shí)與Course Schedule相同，需要注意圖中是否有環(huán)。