Leetcode: 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.

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。

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

Better solution: Adjacent Matrix

public boolean canFinish(int numCourses, int[][] prerequisites) {
    int[][] matrix = new int[numCourses][numCourses]; // i -> j
    int[] indegree = new int[numCourses];
    
    for (int i=0; i<prerequisites.length; i++) {
        int ready = prerequisites[i][0];
        int pre = prerequisites[i][1];
        if (matrix[pre][ready] == 0)
            indegree[ready]++; //duplicate case
        matrix[pre][ready] = 1;
    }
    
    int count = 0;
    Queue<Integer> queue = new LinkedList();
    for (int i=0; i<indegree.length; i++) {
        if (indegree[i] == 0) queue.offer(i);
    }
    while (!queue.isEmpty()) {
        int course = queue.poll();
        count++;
        for (int i=0; i<numCourses; i++) {
            if (matrix[course][i] != 0) {
                if (--indegree[i] == 0)
                    queue.offer(i);
            }
        }
    }
    return count == numCourses;