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.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

click to show more hints.

Solution:

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        if (numCourses <= 1)
            return true;

        int[] visited = new int[numCourses];
        HashMap<Integer, List<Integer>> childMap = new HashMap<Integer, List<Integer>>();
        for (int[] pair : prerequisites) {
            int course1 = pair[0], course2 = pair[1];
            if (childMap.containsKey(course1)) {
                childMap.get(course1).add(course2);
            } else {
                List<Integer> childList = new ArrayList<Integer>();
                childList.add(course2);
                childMap.put(course1, childList);
            }
        }

        
        for (int i=0;i<numCourses;i++)
            if (visited[i]==0){
                if (!traverse(visited,i,childMap))  return false;
            }

        return true;
    }
    
    public boolean traverse(int[] visited, int curNode, HashMap<Integer,List<Integer>> childMap){
        visited[curNode] = 1;
        if (childMap.containsKey(curNode)){
            List<Integer> childs = childMap.get(curNode);
            for (int child : childs){
                if (visited[child]==1) return false;
                if (visited[child]==2) continue;
                if (!traverse(visited,child,childMap)) return false;
            }
        }
        visited[curNode] = 2;
        return true;
    }

}