leetcode 207. 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.

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.

用BFS的做法，用一个in degree来衡量有没有环的出现

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        int[] in = new int[numCourses];
        for(int[] pair : prerequisites){
            in[pair[0]]++;
        }
        
        Queue<Integer> queue = new LinkedList<>(); 
        for(int i=0; i<numCourses; i++){
            if(in[i]==0) queue.offer(i);
        }
        
        while(!queue.isEmpty()){
            int tmp = queue.poll();
            numCourses--;
            for(int[] pair : prerequisites){
                if(pair[1]==tmp && --in[pair[0]]==0)
                    queue.offer(pair[0]);
            }
        }
        return numCourses==0;
    }
}

可以使用一个List数组来保存每个节点的出度，减少while中循环遍历的次数

DFS的做法

public class Solution {
        public boolean canFinish(int numCourses, int[][] prerequisites) {
            ArrayList[] graph = new ArrayList[numCourses];
            for(int i=0;i<numCourses;i++)
                graph[i] = new ArrayList();
                
            boolean[] visited = new boolean[numCourses];
            for(int i=0; i<prerequisites.length;i++){
                graph[prerequisites[i][1]].add(prerequisites[i][0]);
            }

            for(int i=0; i<numCourses; i++){
                if(!dfs(graph,visited,i))
                    return false;
            }
            return true;
        }

        private boolean dfs(ArrayList[] graph, boolean[] visited, int course){
            if(visited[course])
                return false;
            else
                visited[course] = true;;

            for(int i=0; i<graph[course].size();i++){
                if(!dfs(graph,visited,(int)graph[course].get(i)))
                    return false;
            }
            visited[course] = false;
            return true;
        }
    }