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.

Solution 1: Directed graph loop question. Build the graph and then DFS.

 public class Solution {  
   public boolean canFinish(int numCourses, int[][] prerequisites) {  
     int V=numCourses;  
     List<Integer>[] adj=new List[V];  
     for (int v=0; v<V; v++) adj[v]=new LinkedList<>();  
     for (int i=0; i<prerequisites.length; i++) adj[prerequisites[i][1]].add(prerequisites[i][0]);  
     boolean[] marked=new boolean[V];  
     boolean[] onStack=new boolean[V];  
     boolean[] hasCycle=new boolean[1];  
     for (int v=0; v<V; v++)  
       if(!marked[v]) dfs(v,adj,V,marked,onStack,hasCycle);  
     return !hasCycle[0];  
   }  
   private void dfs(int v, List<Integer>[] adj, int V, boolean[] marked, boolean[] onStack, boolean[] hasCycle) {  
     marked[v]=true;  
     onStack[v]=true;  
     for (int w: adj[v]) {  
       if (hasCycle[0]) return;  
       else if (!marked[w]) dfs(w,adj,V,marked,onStack,hasCycle);  
       else if (onStack[w]) hasCycle[0]=true;  
     }  
     onStack[v]=false;  
   }  
 }