[LeetCode 210] 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 1:

Input: 2, [[1,0]] 
Output: `[0,1]`
Explanation: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] .`

Example 2:

**Input:** 4, [[1,0],[2,0],[3,1],[3,2]]
**Output:** `[0,1,2,3] or [0,2,1,3]`
**Explanation:** 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:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

SOLUTION1: DFS VS SOLUTION2: BFS + INDEGREE

区别只在于需要记录路径

class Solution {
    /*************************** Solution 1 DFS ************************************
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        //1. Generate nodeVsNeighbors map
        Map> nodeVsNeighbors = new HashMap<> ();
        for (int i = 0; i < prerequisites.length; i++) {
            int course = prerequisites[i][0];
            int preCourse = prerequisites[i][1];
            
            List courses = nodeVsNeighbors.getOrDefault (preCourse, new ArrayList ());
            courses.add (course);
            nodeVsNeighbors.put (preCourse, courses);
        }
        
        
        //2. 
        boolean[] visited = new boolean[numCourses];
        boolean[] recStack = new boolean [numCourses];
        Stack tracker = new Stack<> ();
        
        for (int i = 0; i < numCourses; i++) {
            if (!visited[i]) {
                if (findOrderHelper (nodeVsNeighbors, visited, recStack, tracker, i))
                    return new int[0];
            }
        }
        
        int[] result = new int[tracker.size ()];
        int index = 0;
        
        while (!tracker.isEmpty ()) {
            result[index++] = tracker.pop ();
        }
        
        return result;
    }
    
    public boolean findOrderHelper (Map> nodeVsNeighbors, boolean[] visited, boolean[] recStack, Stack tracker, int currentIndex) {
        if (recStack[currentIndex])
            return true;
        
        if (visited[currentIndex])
            return false;
        
        visited[currentIndex] = true;
        recStack[currentIndex] = true;
        
        for (int nextCourse : nodeVsNeighbors.getOrDefault (currentIndex, new ArrayList ())) {
            if (findOrderHelper (nodeVsNeighbors, visited, recStack, tracker, nextCourse))
                return true;
        }
        
        recStack[currentIndex] = false;
        tracker.push (currentIndex);
        return false;
    }
    ************************************ Solution 1 END ************************/
    public int[] findOrder(int numCourses, int[][] prerequisites) {
//         if (prerequisites == null || prerequisites.length == 0 || prerequisites[0].length == 0) {
//             return true;
//         }
        
//         if (numCourses == 0) {
//             return true;
//         }
        
        
        //1, get preCourseVsCourses Map  and ZeroDegree map
        Map> preCourseVsCourses = new HashMap<> ();
        Map zeroDegree = new HashMap <> ();
        
        for (int i = 0; i < prerequisites.length; i++) {
            int preCourse = prerequisites[i][1];
            int course = prerequisites[i][0];
            
            List courses = preCourseVsCourses.getOrDefault (preCourse, new ArrayList ());
            courses.add (course);
            preCourseVsCourses.put (preCourse, courses);
            
            int degree = zeroDegree.getOrDefault (course, 0);
            zeroDegree.put (course, degree + 1);
        }
        
        //2. Put all node with zero degree into a queue
        Queue queue1 = new LinkedList<> ();
        for (int i = 0; i < numCourses; i++) {
            if (!zeroDegree.containsKey (i) || zeroDegree.get (i) == 0) {
                queue1.offer (i);
            }
        }
        
        //3. try to find if there is a cycle or can visit all
        Queue queue2 = new LinkedList<> ();
        int countOfZeroDegreeCourse = 0;
        int[] result = new int[numCourses];
        int index = 0;
        
        while (!queue1.isEmpty ()) {
            int currentCourse = queue1.poll ();
            countOfZeroDegreeCourse ++;
            result [index ++] = currentCourse;
                
            for (int nextCourse : preCourseVsCourses.getOrDefault (currentCourse, new ArrayList ())) {
                if (zeroDegree.containsKey (nextCourse)) {
                    int degree = zeroDegree.get (nextCourse);
                    zeroDegree.put (nextCourse, degree - 1);
                    
                    if (degree - 1 == 0) {
                        queue2.offer (nextCourse);
                    }
                }
            }
            
            if (queue1.isEmpty ()) {
                queue1 = queue2;
                queue2 = new LinkedList ();
            }
        }
        
        return countOfZeroDegreeCourse == numCourses ? result : new int[0];
    }
}