[LeetCode] 210. Course Schedule II
There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.
- For example, the pair
[0, 1], indicates that to take course0you have to first take course1.
Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.
Example 1:
Input: numCourses = 2, prerequisites = [[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: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]] Output: [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].
Example 3:
Input: numCourses = 1, prerequisites = [] Output: [0]
Constraints:
1 <= numCourses <= 20000 <= prerequisites.length <= numCourses * (numCourses - 1)prerequisites[i].length == 20 <= ai, bi < numCoursesai != bi- All the pairs
[ai, bi]are distinct.
课程表二。
现在你总共有 numCourses 门课需要选，记为 0 到 numCourses - 1。给你一个数组 prerequisites ，其中 prerequisites[i] = [ai, bi] ，表示在选修课程 ai 前 必须 先选修 bi 。
例如，想要学习课程 0 ，你需要先完成课程 1 ，我们用一个匹配来表示：[0,1] 。
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 任意一种 就可以了。如果不可能完成所有课程，返回 一个空数组 。来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/course-schedule-ii
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题意跟207题一样，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。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。最后判断 K 是否跟课程总数一样，若是则说明所有的课程应该被遍历完了，返回课程顺序；若不是则说明课程安排里面有环，无法安排一个合理的顺序。
时间O(V + E)
空间O(n)
Java实现
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 }
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