Leetcode 210 Course Schedule II

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, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

Solution 1: build the graph, then DFS to judge if there is loop. If not topo sort.

 public class Solution {  
   public int[] findOrder(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];  
     Deque<Integer> stack=new LinkedList<>();  
     for (int v=0; v<V; v++)  
       if (!marked[v]) dfs(v,adj,V,marked,onStack,hasCycle,stack);  
     if (hasCycle[0]) return new int[0];  
     int[] res=new int[V];  
     for (int i=0; i<V; i++) res[i]=stack.pop();  
     return res;  
   }  
   private void dfs(int v, List<Integer>[] adj, int V, boolean[] marked, boolean[] onStack, boolean[] hasCycle, Deque<Integer> stack) {  
     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,stack);  
       else if(onStack[w]) hasCycle[0]=true;  
     }  
     stack.push(v);  
     onStack[v]=false;  
   }  
 }