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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.
思路：即判断该图是否是有向无环图，或是否存在拓扑排序。
public class Solution { public boolean canFinish(int numCourses, int[][] prerequisites) { int[] degree = new int[numCourses];//每个定点的入度 Stack<Integer> stack = new Stack<Integer>();//存放入度为0的顶点 List<Integer>[] adjList = new ArrayList[numCourses];//邻接表 for(int q=0;q<numCourses;q++) adjList[q] = new ArrayList<Integer>(); //初始化邻接表 for(int p=0;p<prerequisites.length;p++) { adjList[prerequisites[p][1]].add(prerequisites[p][0]); } //初始化每个顶点入度 for(int i=0;i<numCourses;i++) { for(Integer vertex:adjList[i]) degree[vertex]++; } //将入度为0的顶点入栈 for(int j=0;j<numCourses;j++) { if(degree[j]==0) stack.push(j); } int count = 0;//当前拓扑排序顶点个数 while(!stack.empty()) { count++; int v = stack.pop(); for(Integer i:adjList[v]) { if(--degree[i]==0) stack.push(i); } } if(count!=numCourses) return false; else return true; } }
当然也可以通过DFS或者BFS求解
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原文地址：http://www.cnblogs.com/mrpod2g/p/4485902.html
