Course_Schedule

题目描述：

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?
(一共有N节课要上，但是上每节课有一个前提，必须上另一节课才能上这节课，判断能不能把所有课都上完.)
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.
    You may assume that there are no duplicate edges in the input prerequisites.

思路:构成有向环就会使课程无法上完，因为课程之间互为条件,所以只需要利用并查集保存各节点的前提节点寻找有无回环即可.

public class Course_Schedule {public static boolean canFinish(int numCourses, int[][] prerequisites) {int root[] = new int[numCourses];for(int i=0;i<numCourses;i++)root[i] = i;for(int i=0;i<prerequisites.length;i++){int prerequisite[] = prerequisites[i];int index = prerequisite[0];int leaf = prerequisite[0];root[index] = prerequisite[1];//若当前节点沿着路径向前面遍历，如果发现前面路径上某点的根节点是该节点那么说明产生了回环.while(root[index]!=index){if(root[index]==leaf){return false;}index = root[index];}}return true;            }public static void main(String[] args) {int numCourses = 2; int[][] prerequisites = {{1,0},{0,1}};System.out.println(canFinish(numCourses,prerequisites));}}