210. Course Schedule II

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:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

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BFS算法 检索前度为0的点加入队列 依次让以此为前度的点前度减一 然后再把新找到的前度为0的点加入队列

public class Solution {      public int[] findOrder(int numCourses, int[][] prerequisites) {          List<Set<Integer>> adjLists = new ArrayList<Set<Integer>>();          for (int i = 0; i < numCourses; i++) {              adjLists.add(new HashSet<Integer>());          }                    for (int i = 0; i < prerequisites.length; i++) {              adjLists.get(prerequisites[i][1]).add(prerequisites[i][0]);          }                    int[] indegrees = new int[numCourses];          for (int i = 0; i < numCourses; i++) {              for (int x : adjLists.get(i)) {                  indegrees[x]++;              }          }                    Queue<Integer> queue = new LinkedList<Integer>();  //这一步比我快，不用重复搜索哪些数前向为0，直接把前向0的加入队列        for (int i = 0; i < numCourses; i++) {              if (indegrees[i] == 0) {                  queue.offer(i);              }          }                    int[] res = new int[numCourses];          int count = 0;          while (!queue.isEmpty()) {              int cur = queue.poll();              for (int x : adjLists.get(cur)) {                  indegrees[x]--;                  if (indegrees[x] == 0) {                      queue.offer(x);  //这一步比我快，不用重复搜索哪些数前向为0，直接把前向0的加入队列                }              }              res[count++] = cur;          }                    if (count == numCourses) return res;          return new int[0];      }  }