标签：
Course Schedule
问题：
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?
思路：
拓扑排序，dfs
我的代码：
public class Solution { public boolean canFinish(int numCourses, int[][] prerequisites) { if(numCourses<=0 || prerequisites==null || prerequisites.length==0) return true; HashSet<Integer>[] graph = new HashSet[numCourses]; for(int i=0; i<numCourses; i++) graph[i] = new HashSet<Integer>(); boolean[] visited = new boolean[numCourses]; boolean[] visiting = 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(visited[i]) continue; if(helper(visited, visiting, graph, i) == false) return false; } return true; } public boolean helper(boolean[] visited, boolean[] visiting, HashSet<Integer>[] graph, int j) { if(visiting[j]) return false; visiting[j] = true; for(Integer neighbor: graph[j]) { if(visited[neighbor]) continue; if(helper(visited, visiting, graph, neighbor) == false) return false; } visiting[j] = false; visited[j] = true; return true; } }
学习之处:
这道题有思路，但是一直不会写拓扑排序，主要的纠结在于如何判断是否又环，这个人的思路很赞，通过visited判断节点是否访问过，visiting用于判断是否有环。学习了。
标签：
原文地址：http://www.cnblogs.com/sunshisonghit/p/4485006.html
