207. Course Schedule

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, is it possible for you to finish all courses?
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 it is possible.
2, [[1,0],[0,1]]
There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.

click to show more hints.
Hints:This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
Topological sort could also be done via BFS.

思路是BFS + Topological sort，跟 Course Schedule ii 很类似。用两个HashMap表示入度和Graph. 建立queue来遍历节点，先把入度为零的点offer到queue里面，再一个一个poll出来遍历。用courseRemaining来记录还需要遍历的节点。每次遍历某个节点，要使得它的neighbors的入度减一。（就是当前节点指向的点，在这里就是以当前节点作为先修课的课程）如果neighbors当中有节点在入度减一后入度为零，要将其加入到queue里面，同时courseRemaining自减。

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        if (numCourses <= 1){
            return true;
        }
        if (prerequisites.length == 0 || prerequisites[0].length == 0){
            return true;
        }
        Map indegree = new HashMap<>();
        Map> neighbors = new HashMap<>();
        Queue queue = new LinkedList<>();
        for (int i = 0; i < prerequisites.length; i++){
           
            if (indegree.containsKey(prerequisites[i][0])){
                indegree.put(prerequisites[i][0], indegree.get(prerequisites[i][0]) + 1);
            } else {
                indegree.put(prerequisites[i][0], 1);
            }
        }
        int coursesRemaining = numCourses;
        for (int i = 0; i < numCourses; i++){
            if (!indegree.containsKey(i)){
                queue.offer(i);
                coursesRemaining--;
            }
            neighbors.put(i, new HashSet());
        }
        for (int i = 0; i < prerequisites.length; i++){
            neighbors.get(prerequisites[i][1]).add(prerequisites[i][0]);
        }

        while (!queue.isEmpty()){
            Integer curt = queue.poll();
            for(Integer nei : neighbors.get(curt)){
                indegree.put(nei, indegree.get(nei) - 1);
                if (indegree.get(nei) == 0){
                    queue.offer(nei);
                    coursesRemaining--;
                }
            }
        }
        // System.out.print(coursesRemaining);
        return coursesRemaining == 0;
    }
}