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.Example 1:
Input: 2, [[1,0]]
Output: [0,1]
Explanation: 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] .
Example 2:
Input: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [0,2,1,3]
Explanation: 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.
You may assume that there are no duplicate edges in the input prerequisites.
难度：medium
题目：现有标记为 0 到 n - 1 的课程需要学习。某些课程有预备课程需要学习，如果学习课程 0 就要先学习课程 1，表示形式如 [0, 1]. 给定所有课程及其预备课程关系，返回可以学习完所有课程的顺序。这种顺序有许多种，你只需要返回其一即可，如果无法完成所有课程学习，则返回空。
思路：
找出 0 入度结点
然后逐个删除其边，重复 1.
Runtime: 5 ms, faster than 98.86% of Java online submissions for Course Schedule II.Memory Usage: 43.8 MB, less than 66.87% of Java online submissions for Course Schedule II.
class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
// (Array -> List) to store graph
List<Integer>[] nodes = new ArrayList[numCourses];
for (int i = 0; i < numCourses; i++) {
nodes[i] = new ArrayList<Integer>();
}
// count in degree
int[] inDegree = new int[numCourses];
for (int i = 0; i < prerequisites.length; i++) {
(nodes[prerequisites[i][0]]).add(prerequisites[i][1]);
inDegree[prerequisites[i][1]] += 1;
}
// count zero in degree
int zeroInDegreeCount = 0;
List<Integer> zeroInDegreeList = new ArrayList<>();
for (int i = 0; i < inDegree.length; i++) {
if (inDegree[i] <= 0) {
zeroInDegreeList.add(i);
zeroInDegreeCount++;
}
}
// bfs
for (int i = 0; i < zeroInDegreeList.size(); i++) {
for (Integer node : nodes[zeroInDegreeList.get(i)]) {
if (–inDegree[node] <= 0) {
zeroInDegreeList.add(node);
zeroInDegreeCount++;
}
}
}
if (zeroInDegreeCount == numCourses) {
int[] result = new int[numCourses];
for (int i = 0; i < numCourses; i++) {
result[numCourses – 1 – i] = zeroInDegreeList.get(i);
}
return result;
}
return new int[0];
}
}