leetcode题解Java | 210. Course Schedule II

题目：https://leetcode.com/problems/course-schedule-ii/#/description

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, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

Note:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

分析：

这道题的本质是考拓扑排序。所以用一个队列来做，每次让入度为零的点入队，出队时，与此点相连的点的入度减一，如果这时这个点的入度也为零了，就入队。做到队列为空为止。

Java实现：

public class Solution {   public int[] findOrder(int numCourses, int[][] prerequisites){List<Integer> edges[] = new ArrayList[numCourses];boolean[] visit = new boolean[numCourses];int[] in_degree = new int[numCourses];int[] ans = new int[numCourses];int n=numCourses;for(int i=0; i<numCourses; ++i)edges[i] = new ArrayList<>();for(int i=0; i<prerequisites.length; ++i){int v1 = prerequisites[i][0];int v2 = prerequisites[i][1];edges[v1].add(v2);++in_degree[v2];}Queue<Integer> q = new LinkedList<>();for(int i=0; i<numCourses; ++i)if(in_degree[i] == 0)q.add(i);while(q.isEmpty()!=true){int v = q.poll();visit[v]=true;ans[--n] = v;for(int i:edges[v]){--in_degree[i];if(in_degree[i]==0)q.add(i);}}for(int i=0; i<numCourses; ++i)if(visit[i]==false){ans = new int[0];return ans;}return ans;}}