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:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

click to show more hints.

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.

Solution

Topological sort BFS, time O(n + e), space O(n)

对于一个有向图来说，如果其中存在环，这个图不存在拓扑排序。反之如果没有环，则一定有拓扑排序。

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        List<Integer>[] graph = new ArrayList[numCourses];   // i -> j
        int[] indegree = new int[numCourses];
        
        for (int i = 0; i < numCourses; ++i) {
            graph[i] = new ArrayList<>();
        }
        
        for (int[] prerequisite : prerequisites) {
            int course = prerequisite[0];
            int pre = prerequisite[1];
            graph[pre].add(course);
            ++indegree[course];
        }
        
        Queue<Integer> 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 = queue.poll();
            ++count;
            
            for (Integer i : graph[course]) {
                if (--indegree[i] == 0) {
                    queue.offer(i);
                }
            }
        }
        
        return count == numCourses;
    }
}

DFS

从graph的每个节点开始，如果全都没有遇到cycle，则可以finish。
注意这里遇到一个坑，原本想用Arrays.fill(graph, new ArrayList<>())初始化graph，后来发现graph中的每个元素都指向同一个ArrayList对象...还是乖乖地一个一个实例化吧。

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        List<Integer>[] graph = new ArrayList[numCourses];   // i -> j
        for (int i = 0; i < numCourses; ++i) {
            graph[i] = new ArrayList<>();
        }
        
        for (int[] prerequisite : prerequisites) {
            graph[prerequisite[1]].add(prerequisite[0]);
        }
        
        boolean[] visited = new boolean[numCourses];
        for (int i = 0; i < numCourses; ++i) {
            if (!dfs(i, graph, visited)) {
                return false;
            }
        }
        
        return true;
    }
    
    // return false if cycle detected, means cannot finish
    public boolean dfs(int i, List<Integer>[] graph, boolean[] visited) {
        if (visited[i]) {
            return false;
        }
        
        visited[i] = true;
        
        for (int j : graph[i]) {
            if (!dfs(j, graph, visited)) {
                return false;
            }
        }
        
        visited[i] = false;
        return true;
    }
}