[LeetCode] 210. Course Schedule II Java

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, 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].

题意及分析：给出一个课程序列，有些课程需要先序课程，求一个满足条件的课程排序顺序。可以使用拓扑排序，每次删除一个入度为0的点，用一个list保存每次删除的点（入度为0的点），因为我这里是把先序课程放在前面的，所以最后需要将list反转一下保存到数组中输出。总体和200题类似。

代码：

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] map=new int[numCourses];
        for(int i=0;i<prerequisites.length;i++){        //计算每个点的入度
            map[prerequisites[i][0]]++;
        }
        Queue<Integer> queue = new LinkedList<>();  //记录入度为0的点
        for(int i=0;i<map.length;i++){
            if(map[i]==0) queue.add(i);
        }
        List<Integer> res = new ArrayList<>();
        int count = queue.size();   //初始的入度为o的点
        while(!queue.isEmpty()){
            int temp = queue.poll();
            res.add(temp);
            for(int i=0;i<prerequisites.length;i++){
                if(temp== prerequisites[i][1]){
                    int t  = prerequisites[i][0];
                    map[t]--;
                    if(map[t]==0){
                        queue.add(t);
                        count++;
                    }
                }
            }
        }
        if(count!=numCourses){
            int[] a = new int[0];
            return a;
        }else {
            int[] a = new int[res.size()];
            for(int i=res.size()-1;i>=0;i--){       //这里需要反转一下
                a[i] = res.get(i);
            }
            return a;
        }
    }
}