207. Course Schedule

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LeetCode

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Problem

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?

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:

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. 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.

Code

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        int[] indegree = new int[numCourses];
        int res = numCourses;

        // 找入度为0，表示可以开始学习的课
        for(int[] pair : prerequisites){
            // [1,0] => 0->1
            indegree[pair[0]]++;
        }

        Queue<Integer> queue = new LinkedList<>();

        // 入度为0的加入，然后bfs
        for(int i = 0; i < indegree.length; i++){
            if(indegree[i] == 0){
                // 当前这个课已经可以学了
                queue.offer(i);
            }
        }

         while(!queue.isEmpty()){
            int curr = queue.poll();

            for(int[] pair : prerequisites){
                // 如果这门课已经可以学习了，就不加入queue
                if(indegree[pair[0]] == 0){
                    continue;
                }

                if(pair[1] == curr){
                    indegree[pair[0]]--;
                }
                // 这门课原来不能学习，现在可以学习了
                if(indegree[pair[0]] == 0){
                    queue.offer(pair[0]);
                }
            }
        }

        for(int i = 0; i < numCourses; i++){
            if(indegree[i] != 0){
                return false;
            }
        }

        return true;
    }
}