[LeetCode-JAVA] Course Schedule

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;
    }
}