[LeetCode] 207. Course Schedule

There are a total of numCourses courses you have to take, labeled from 0 to numCourses-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, is it possible for you to finish all courses?

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0, and to take course 0 you should
             also have finished course 1. So it is impossible.

Constraints:

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.
1 <= numCourses <= 10^5

课程表。

你这个学期必须选修 numCourses 门课程，记为 0 到 numCourses - 1 。

在选修某些课程之前需要一些先修课程。先修课程按数组 prerequisites 给出，其中 prerequisites[i] = [ai, bi] ，表示如果要学习课程 ai 则必须先学习课程 bi 。

例如，先修课程对 [0, 1] 表示：想要学习课程 0 ，你需要先完成课程 1 。
请你判断是否可能完成所有课程的学习？如果可以，返回 true ；否则，返回 false 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/course-schedule
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典型的拓扑排序题目。拓扑排序又称之为有向无环图（DAG）。拓扑排序的定义在这里就不赘述了。这个题目有多种做法，此处我解释一下BFS的思路。首先为所有课程创建一个入度表indegree，遍历所有的课程。所有课程的prerequisites都以一个 pair[0, 1] 表示（0的先修课程是1），所以遍历所有的 pair[0]，建立所有课程的入度表。此时再遍历这个入度表，将入度表中所有为0（没有先修课程）的元素加入一个队列。遍历这个队列，每弹出一个元素 cur，就减去一门课 res--，去 prerequisites 中看是否有其他课程 pair[0] 的先修课程是 cur，若有，则在入度表里面找到这个 pair[0] 并减1。如果这个 pair[0] 也被减到入度为0了，则把他加入队列继续查找，看这个 pair[0] 是否是别的课的先修课。最后会跳出while循环，跳出的时候判断是不是所有课程 res 都遍历完了。

时间O(V + E)

空间O(n)

Java实现

 1 class Solution {
 2     public boolean canFinish(int numCourses, int[][] prerequisites) {
 3         int[] indegree = new int[numCourses];
 4         int res = numCourses;
 5         for (int[] pair : prerequisites) {
 6             indegree[pair[0]]++;
 7         }
 8 
 9         // 把所有入度为0的节点加入队列
10         // 入度为0说明没有先修课程
11         Queue<Integer> queue = new LinkedList<>();
12         for (int i = 0; i < indegree.length; i++) {
13             if (indegree[i] == 0) {
14                 queue.offer(i);
15             }
16         }
17 
18         // 遍历所有入度为0的课程
19         while (!queue.isEmpty()) {
20             int pre = queue.poll();
21             res--;
22             // 看是否有其他课程的先修课程是pre
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                     }
29                 }
30             }
31         }
32         return res == 0;
33     }
34 }