leetcode 207 Course Schedule
z
 2019-04-01 09:28:23  

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:

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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:

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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.
  • 等价于判断图中是否存在环
  • 对于图的表述，可以使用邻接矩阵和入度来表示
  • 对所有入度为0的课程，学习他。
  • 学习完一个课程之后，将以该课程为prerequisite的课程的入度都减一
  • 如果存在一个课程入度为0，将其加入待学习的set中
  • 每学习一个课程，count一下它
  • 最后比较count的数目是否为所有课程的数目
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class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        if (prerequisites.length == 0 || prerequisites[0].length == 0) return true;
        LinkedList<Integer>[] neighbors = new LinkedList[numCourses];
        for (int i=0; i<numCourses; i++) neighbors[i] = new LinkedList<Integer>();
        int[] indegree = new int[numCourses];
        int n = 0;
        for (int[] crs: prerequisites) {
            int take = crs[0];
            int pre = crs[1];
            if (neighbors[pre] == null) {
                neighbors[pre] = new LinkedList<Integer>();
            }
            neighbors[pre].add(take);
            indegree[take]++;
        }
        
        Queue<Integer> q = new LinkedList<>();
        
        for (int i=0; i<numCourses; i++) {
            if (indegree[i] == 0) {
                q.offer(i);
            }
        }
        
        while (!q.isEmpty()) {
            int crs = q.poll();
            n ++;
            for (int pre: neighbors[crs]) {
                indegree[pre] --;
                if (indegree[pre] == 0) {
                    q.offer(pre);
                }
            }
        }
        return n == numCourses;
    }
}