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.

思路：
宽度优先搜索，用一个数组存储每个课程的前置课程数量，用一个二维list存储每个课程可解锁的所有课程。
宽搜时，每搜索到一个课程，就把他能解锁的所有子课程的前置课程数减1，如果某个子课程前置课程数归0，就把它加入到队列中。
最后比较宽搜到的课程数是否与课程总数相等。

public boolean canFinish(int numCourses, int[][] prerequisites) {
    //tip 后面的是前置条件

    int[] preCount = new int[numCourses];
    List<List<Integer>> unlock = new ArrayList<>();
    for (int i = 0; i < numCourses; i++) {
        unlock.add(new ArrayList<>());
    }

    for (int i = 0; i < prerequisites.length; i++) {
        int[] pre = prerequisites[i];
        preCount[pre[0]] += 1;
        unlock.get(pre[1]).add(pre[0]);
    }

    //use queue to search and count
    int cnt = 0;
    Queue<Integer> q = new LinkedList<>();
    for (int i = 0; i < numCourses; i++) {
        if (preCount[i] == 0) {
            q.offer(i);
        }
    }
    while (!q.isEmpty()) {
        int num = q.poll();
        cnt++;
        List<Integer> subs = unlock.get(num);
        for (int i = 0; i < subs.size(); i++) {
            int next = subs.get(i);
            preCount[next] -= 1;
            if (preCount[next] == 0) {
                q.offer(next);
            }
        }
    }

    return cnt == numCourses;
}