LeetCode 207.Course Schedule

description:
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.

这个题目就是一个典型的topological sorting problems
注意题目中的入度和删除元素之间的关系

public class Solution {    public boolean canFinish(int numCourses, int[][] prerequisites) {        if (numCourses == 0) {            return false;        }        List[] edges = new List[numCourses];        int[] indegree = new int[numCourses];        for (int i = 0; i < numCourses; i++) {            edges[i] = new ArrayList<Integer>();        }        for (int i = 0; i < prerequisites.length; i++) {            indegree[prerequisites[i][0]]++;            edges[prerequisites[i][1]].add(prerequisites[i][0]);        }        Queue queue = new LinkedList<>();        for (int i = 0; i < numCourses; i++) {            if (indegree[i] == 0) {                queue.offer(i);            }        }        int count = 0;        while (!queue.isEmpty()) {            int course = (int)queue.poll();            count++;            int size = edges[course].size();            for (int i = 0; i < size; i++) {                int pointer = (int)edges[course].get(i);                indegree[pointer]--;                if (indegree[pointer] == 0) {                    queue.offer(pointer);                }            }        }        return count == numCourses;    }}