207. Course Schedule
2017-02-18 20:46
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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:
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.
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.
解题思路:一刷没ac。
一个拓扑排序的思路是将所有课程看成是节点,依赖关系看作是边,先找到度为0的节点,从图中移除,然后继续找堵为0的节点,顺次移除。看是否能够满足全部移除。
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:
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.
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.
解题思路:一刷没ac。
一个拓扑排序的思路是将所有课程看成是节点,依赖关系看作是边,先找到度为0的节点,从图中移除,然后继续找堵为0的节点,顺次移除。看是否能够满足全部移除。
public class Solution { public boolean canFinish(int numCourses, int[][] prerequisites) { int[][] matrix = new int[numCourses][numCourses]; int[] indegree = new int[numCourses]; int count = 0; for (int i = 0; i < prerequisites.length; i++) { int ready = prerequisites[i][0]; int pre = prerequisites[i][1]; if (matrix[pre][ready] == 0) indegree[ready]++; matrix[pre][ready] = 1; } LinkedList<Integer> queue = new LinkedList<Integer>(); for (int i = 0; i < numCourses; i++) { if (indegree[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 (--indegree[i] == 0) queue.offer(i); } } } return count == numCourses; } }
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