LeetCode 210. Course Schedule II

原题链接在这里：https://leetcode.com/problems/course-schedule-ii/

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].

题解：

用BFS based topological sort.  若是最后res的size 没有加到 numCourses, 说明没有可行方案，返回空的array.

若是可行，就把res转化成array.

Time Complexity: O(V + E). Space: O(V).

AC Java:

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        List<Integer> res = new ArrayList<Integer>();
        List<List<Integer>> graph = new ArrayList<List<Integer>>();
        for(int i = 0; i<numCourses; i++){
            graph.add(new ArrayList<Integer>());
        }
        for(int [] edge : prerequisites){
            graph.get(edge[1]).add(edge[0]);
        }
        
        int [] inDegree = new int[numCourses];
        for(int [] edge : prerequisites){
            inDegree[edge[0]]++;
        }
        
        LinkedList<Integer> que = new LinkedList<Integer>();
        for(int i = 0; i<inDegree.length; i++){
            if(inDegree[i] == 0){
                que.add(i);
            }
        }
        
        while(!que.isEmpty()){
            int source = que.poll();
            res.add(source);
            
            for(int dest : graph.get(source)){
                inDegree[dest]--;
                if(inDegree[dest] == 0){
                    que.add(dest);
                }
            }
        }
        
        if(res.size() != numCourses){
            return new int[0];
        }
        int [] resArr = new int[numCourses];
        for(int i = 0; i<numCourses; i++){
            resArr[i] = res.get(i);
        }
        return resArr;
    }
}

类似Course Schedule.

跟上 Course Schedule III.