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.

The hint provided by leetcode is:

Hints:

  1. 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.
  2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
  3. Topological sort could also be done via BFS.

 

BFS: 

典型的拓扑排序。原理也很简单，在一个有向图中，每次找到一个没有前驱节点的节点（也就是入度为0的节点），然后把它指向其他节点的边都去掉，重复这个过程（BFS），直到所有节点已被找到，或者没有符合条件的节点（如果图中有环存在）。

回顾一下图的三种表示方式：边表示法（即题目中表示方法），邻接表法，邻接矩阵。用邻接表存储图比较方便寻找入度为0的节点。

 1 public class Solution {
 2     public boolean canFinish(int numCourses, int[][] prerequisites) {
 3         ArrayList<Integer>[] graph = new ArrayList[numCourses];
 4         int[] indegree = new int[numCourses];
 5         
 6         for (int i=0; i<numCourses; i++) {
 7             graph[i] = new ArrayList<Integer>();
 8         }
 9         
10         for (int[] each : prerequisites) {
11             graph[each[0]].add(each[1]);
12             indegree[each[1]]++;
13         }
14         
15         Queue<Integer> queue = new LinkedList<Integer>();
16         for (int i=0; i<indegree.length; i++) {
17             if (indegree[i] == 0) {
18                 queue.offer(i);
19             }
20         }
21         
22         int count = 0;
23         while (!queue.isEmpty()) {
24             int cur = (int)queue.poll();
25             count++;
26             
27             for (int course : graph[cur]) {
28                 indegree[course]--;
29                 if (indegree[course] == 0) {
30                     queue.offer(course);
31                 }
32             }
33         }
34         return count == numCourses;
35     }
36 }

 

Better solution: Adjacent Matrix

 1 public boolean canFinish(int numCourses, int[][] prerequisites) {
 2     int[][] matrix = new int[numCourses][numCourses]; // i -> j
 3     int[] indegree = new int[numCourses];
 4     
 5     for (int i=0; i<prerequisites.length; i++) {
 6         int ready = prerequisites[i][0];
 7         int pre = prerequisites[i][1];
 8         if (matrix[pre][ready] == 0)
 9             indegree[ready]++; //duplicate case
10         matrix[pre][ready] = 1;
11     }
12     
13     int count = 0;
14     Queue<Integer> queue = new LinkedList();
15     for (int i=0; i<indegree.length; i++) {
16         if (indegree[i] == 0) queue.offer(i);
17     }
18     while (!queue.isEmpty()) {
19         int course = queue.poll();
20         count++;
21         for (int i=0; i<numCourses; i++) {
22             if (matrix[course][i] != 0) {
23                 if (--indegree[i] == 0)
24                     queue.offer(i);
25             }
26         }
27     }
28     return count == numCourses;