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题目：
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：首先要里解题意 一些课程有先修课程--这些课程必须先学，类似下面的图片，这个就是典型的有向图的拓扑排序
步骤1：找到一个入度为零的顶点
步骤2：从图中删除这个顶点
步骤3：循环1、2 如果所有顶点都走过了 那么就是一个可以完成的图
其次还要了解题中算法的输入 numCourse为要选的课的总数 而prerequisite矩阵是先决条件矩阵（开始一直以为是邻接矩阵）
所以，第一步要先将给出的矩阵，换算出邻接矩阵(也可以是邻接链表)，然后再步步进行。
代码：
public class Solution { public boolean canFinish(int numCourses, int[][] prerequisites) { int[] indegree = new int[numCourses]; int[][] matrix = new int[numCourses][numCourses]; // [i][j] i为j的先决条件 i->j Stack<Integer> stack = new Stack<Integer>(); for(int i = 0 ; i < prerequisites.length ; i++){ if(matrix[prerequisites[i][1]][prerequisites[i][0]] == 1) continue;//输入有重复 indegree[prerequisites[i][0]]++; //入度加一 matrix[prerequisites[i][1]][prerequisites[i][0]] = 1; //p中[j]<-[i] } for(int i = 0 ; i < numCourses ; i++){ if(indegree[i] == 0) //将入度为零的压入栈 stack.push(i); } while(!stack.isEmpty()){ int temp = stack.pop(); //删掉入度为零的点所有连出的线 for(int i = 0 ; i < numCourses ; i++){ //每删掉一个 将对应的入度减一 if(matrix[temp][i] == 1){ matrix[temp][i] = 0; indegree[i]--; //如果减去后i对应的入度为0 则将其入栈 if(indegree[i] == 0) stack.push(i); } } } for(int i = 0 ; i < numCourses ; i++){ //判断是否有入度不为零的点 if(indegree[i] > 0) return false; } return true; } }
[LeetCode-JAVA] Course Schedule
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原文地址：http://www.cnblogs.com/TinyBobo/p/4544860.html
