lintcode616- Course Schedule II- medium

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

Given n = 2, prerequisites = [[1,0]]
Return [0,1]

Given n = 4, prerequisites = [1,0],[2,0],[3,1],[3,2]]
Return [0,1,2,3] or [0,2,1,3]

 

和前面一题一模一样的做法，只不过直接把得到的拓扑序返回即可。还有ArrayList.add()改为数组里的添加只要用一个int pointer辅助即可。

 

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
        int[] order = new int[numCourses];
        int pointer = 0;
        int[] indegrees = new int[numCourses];
        List[] edges = new List[numCourses];
        
        for (int i = 0; i < numCourses; i++) {
            indegrees[i] = 0;
            edges[i] = new ArrayList<Integer>();
        }
        
        for (int i = 0; i < prerequisites.length; i++) {
            int c1 = prerequisites[i][0];
            int c2 = prerequisites[i][1];
            indegrees[c1]++;
            edges[c2].add(c1);
        }
        
        Queue<Integer> queue = new LinkedList<Integer>();
        Set<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < numCourses; i++) {
            if (indegrees[i] == 0) {
                queue.offer(i);
                set.add(i);
            }
        }
        while (!queue.isEmpty()) {
            int course = queue.poll();
            order[pointer++] = course;
            for (int i = 0; i < edges[course].size(); i++) {
                int next = (int) edges[course].get(i);
                indegrees[next]--;
                if (indegrees[next] == 0) {
                    queue.offer(next);
                    set.add(next);
                }
            }
        }
        
        if (pointer == numCourses) {
            return order;
        }
        return new int[0];
    }
}