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

这道题目是典型的图的拓扑排序题目，这里先简要介绍一下拓扑排序算法的实现步骤：

1. 初始化图邻接矩阵(u依赖于于v，则matrix[u][v]==1)和每个节点的入度数组(u依赖于v，则indegree[v]++)。
2. 从入度数组中选取一个没有前驱节点（即indegree[v]==0）的顶点输出。
3. 从图中删除该顶点和所有以它为头节点的弧。(即所有matrix[v][otherV]==1的，indegre[otherV]–)。
4. 重复2，3步骤，直到全部顶点均已输出，或者图中不存在无前驱的顶点为止.(此时说明图中有环)。

所以，这道题目就是检测最后构造的图中是否有环即可(即检测indegree数组是否有顶点的值大于0)。

AC Code

ac代码如下所示：

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        boolean canFinish = true;
        int[] indegree = new int[numCourses];
        int[][] matrix = new int[numCourses][numCourses];

        // initial data
        for (int i = 0; i < prerequisites.length; i ++) {
            if (matrix[prerequisites[i][0]][prerequisites[i][1]] == 1) continue;
            indegree[prerequisites[i][1]] ++;
            matrix[prerequisites[i][0]][prerequisites[i][1]] = 1;
        }

        // topological sort
        for (int i = 0; i < numCourses; i ++) {
            for (int j = 0; j < numCourses; j ++) {
                if (indegree[j] == 0) {
                    indegree[j] --;

                    // delete node which start is j
                    for (int k = 0; k < numCourses; k ++) {
                        if (matrix[j][k] == 1) {
                            indegree[k] --;
                            matrix[j][k] = 0;
                        }
                    }

                    break;
                }
            }
        }

        // output result
        for (int i = 0; i < numCourses; i ++) {
            if (indegree[i] > 0) {
                canFinish = false;
                break;
            }
        }   

        return canFinish;
    }
}