Leetcode 207. Course Schedule

题目

There are a total of n course 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 nunber of courses and a list of prerequisite pairs, is it possible for you to finish all courses?

Example 1:

Input: 2, [[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: 2, [[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.

Note: 1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented 2. You may assume that there are no duplicate edges in the input prerequisites.

思路

这道题本质上考察的是Topological Sort. Prerequisites表示了一个Dircted Graph，每个节点为一节课，连线为(Course -> Prereq)。我们想 在这个图上找到某种上课顺序，使得前后课程不会互为Prerequisite；即我们要对图像进行拓扑排序，并判断这个图像中有没有循环。

我们可以采用BFS来解决这个问题。首先，我们要记录所有节点的in-degree。当一个节点的in-degree为0时，我们知道它不被任何其他课程所依赖。因此， 我们可以把它和它的所有edges从图中移除（更新与它相邻的node的in-degree）。然后，我们不断重复这个过程，持续将新出现的in-degree为0的节点 放入BFS的queue中，进行移除操作并更新它邻居的in-degree。如果我们发现如果还有节点未被访问，且不存在in-degree为0的节点，我们可以断定图像中一定有环。 如果我们能够在没有环的情况下访问完所有的节点，说明我们找到了图像的拓扑排序。

解答

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {

        // Construct graph and find starting root
        int[][] adjMatrix = new int[numCourses][numCourses];
        int[] indegree = new int[numCourses];
        for (int[] pair : prerequisites) {
            adjMatrix[pair[0]][pair[1]] = 1; // course -> prereq
            indegree[pair[1]]++; // one more course depends on it
        }

        // BFS: start with all roots (indegree = 0)
        Queue<Integer> queue = new LinkedList();
        for (int i = 0; i < indegree.length; i++) {
            if (indegree[i] == 0) queue.offer(i);
        }

        // Standard BFS:
        int count = 0;
        while (! queue.isEmpty()) {
            int course = queue.poll();
            count++;
            for (int i = 0; i < numCourses; i++) {
                // node i now has no incoming edges, add to queue
                if (adjMatrix[course][i] == 1 && --indegree[i] == 0) {
                    queue.offer(i);
                }
            }
        }

        return count == numCourses; // visited all courses without circle
    }
}

Complexity Analysis

• Time Complexity: O(n). 建立Graph和Indegree数组都需要遍历所有元素，同时BFS也会访问每个节点，总共时间为O(n).

• Space Complexity: O(n). 我们使用了额外的二位数组、队列来储存所有的节点和边。

拓展

我们以上用BFS实现了拓扑排序。你能用DFS实现它吗？我们从没被访问过的节点开始，连续访问它的所有相邻节点。如果访问到之前访问过的节点，则有循环； 如果没有，我们从另一个未被访问过的节点开始继续深度优先搜索。

总结

对拓扑排序、BFS、DFS这类算法，我们需要具备基本的熟练度。清晰判断题目考察方向后，能够很快给出实现。

Reference

• jianchao-li's solution
• justjiayu's solution