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.

首先根据题意构造一个图，然后用DFS判断图中是否存在环，如果存在环就不能完成课程。代码如下：

public class Solution {
    boolean[] onStack;
    boolean[] isVisited;
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        if(prerequisites == null || prerequisites.length == 0) return true;
        List<Integer>[] graph = new List[numCourses];
        onStack = new boolean[numCourses];
        isVisited = new boolean[numCourses];
        for(int i = 0; i < prerequisites.length; i++) {
            int key = prerequisites[i][1];
            int value = prerequisites[i][0];
            if(graph[key] == null) graph[key] = new ArrayList<Integer>();
            graph[key].add(value);
        }
        for(int i = 0; i < numCourses; i++) {
            if(isVisited[i] == false && hasCycle(i, graph) == true) 
                return false;
        }
        return true;
    }
    
    public boolean hasCycle(int i, List<Integer>[] graph) {
        boolean cycle = false;
        onStack[i] = true;
        isVisited[i] = true;
        if(graph[i] != null) {
            for(int j : graph[i]) {
                if(isVisited[j] == false) {
                    cycle = hasCycle(j, graph);
                    if(cycle == true) 
                        break;
                } else if(onStack[j] == true) {
                    cycle = true;
                    break;
                }
            }
        }
        onStack[i] = false;
        return cycle;
    }
}