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LeetCode 207. Course Schedule

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Question

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, is it possible for you to finish all courses?

For example: 

1
2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible. 

1
2, [[1,0],[0,1]]

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: 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. click to show more hints.

Hints: This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort. Topological sort could also be done via BFS.

Explanation

这道题只要看可不可能，那么我们只要将graph理清楚，找到每个先修课的个数，依次放入队列就好了。

Code

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public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        int[][] matrix = new int[numCourses][numCourses];
        int[] degree = new int[numCourses];
        for (int i=0; i< prerequisites.length; i++) {
            int ready = prerequisites[i][0];
            int pre = prerequisites[i][1];
            if (matrix[pre][ready] == 0) {
                degree[ready] ++;
            }
            matrix[pre][ready] = 1;
        }
        int count = 0;
        Queue<Integer> queue = new LinkedList();
        for (int i=0; i< degree.length; i++) {
            if(degree[i] == 0) queue.offer(i);
        }
        while (!queue.isEmpty()) {
            int course = queue.poll();
            count ++;
            for (int i=0; i<numCourses;i++) {
                if (matrix[course][i] != 0) {
                    if (--degree[i] == 0) {
                        queue.offer(i);
                    }
                }
            }
        }
        return count ==numCourses;
    }
}