[LeetCode] 210. Course Schedule II

Posted 2021-03-06 aaronliu1991

课程表二。题意跟207题几乎一样，要求返回你为了学完所有课程所安排的学习顺序。可能会有多个正确的顺序，你只要返回一种就可以了。如果不可能完成所有课程，返回一个空数组。例子，

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] .

思路依然是拓扑排序。这一题我依然给出BFS的解法。同版本一一样，需要创建一个数组记录每个课程的入度，同时多维护一个指针K = 0和一个记录结果的数组res。遍历入度表，将入度为0的课程放入队列；并且每当加入一个入度为0的课程，K就++，原因是这些课程是没有先修课程的，可以放在一开始修。再遍历队列（记住队列里面都是没有先修课程的课），每弹出一个cur，就去检查这个cur是否是prerequisites某一个别的课程的先修课程，若是，则将这个别的课程的入度--；这里跟版本一一样，当有任何一个课程的入度被减至0，则将这个课程加入队列，同时也加入res。

时间O(V + E)

空间O(n)

 1 class Solution {
 2     public int[] findOrder(int numCourses, int[][] prerequisites) {
 3         int[] indegree = new int[numCourses];
 4         int[] res = new int[numCourses];
 5         // res用的指针
 6         int k = 0;
 7         // 取得所有的入度
 8         for (int[] pair : prerequisites) {
 9             indegree[pair[0]]++;
10         }
11 
12         Queue<Integer> queue = new LinkedList<>();
13         for (int i = 0; i < indegree.length; i++) {
14             if (indegree[i] == 0) {
15                 queue.offer(i);
16                 res[k] = i;
17                 k++;
18             }
19         }
20 
21         while (!queue.isEmpty()) {
22             int pre = queue.poll();
23             for (int[] pair : prerequisites) {
24                 if (pair[1] == pre) {
25                     indegree[pair[0]]--;
26                     if (indegree[pair[0]] == 0) {
27                         queue.offer(pair[0]);
28                         res[k] = pair[0];
29                         k++;
30                     }
31                 }
32             }
33         }
34         return (k == numCourses) ? res : new int[0];
35     }
36 }