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 很類(lèi)似。用兩個(gè)HashMap表示入度和Graph. 建立queue來(lái)遍歷節(jié)點(diǎn)，先把入度為零的點(diǎn)offer到queue里面，再一個(gè)一個(gè)poll出來(lái)遍歷。用courseRemaining來(lái)記錄還需要遍歷的節(jié)點(diǎn)。每次遍歷某個(gè)節(jié)點(diǎn)，要使得它的neighbors的入度減一。（就是當(dāng)前節(jié)點(diǎn)指向的點(diǎn)，在這里就是以當(dāng)前節(jié)點(diǎn)作為先修課的課程）如果neighbors當(dāng)中有節(jié)點(diǎn)在入度減一后入度為零，要將其加入到queue里面，同時(shí)courseRemaining自減。

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<Integer, Integer> indegree = new HashMap<>();
        Map<Integer, Set<Integer>> neighbors = new HashMap<>();
        Queue<Integer> 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<Integer>());
        }
        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;
    }
}