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.
拓扑排序解决：(BFS,DFS也可以，下次试试)
public boolean canFinish(int numCourses, int[][] prerequisites) {
int[][] matrix = new int[numCourses][numCourses]; //构建图
int[] indegree = new int[numCourses]; //入度
for(int i = 0; i < prerequisites.length; i++){
int ready = prerequisites[i][0];
int pre = prerequisites[i][1];
matrix[pre][ready] = 1;
indegree[ready]++;
}
int count = 0;
Queue<Integer> queue = new LinkedList();
for(int i = 0; i < numCourses; i++){
if(indegree[i] == 0){
queue.offer(i);
}
}
while(!queue.isEmpty()){
int s = queue.poll();
count++;
for(int i = 0; i < numCourses; i++){
if(matrix[s][i] != 0){
if(--indegree[i] == 0){
queue.offer(i);
}
}
matrix[s][i] = 0;
}
}
return count==numCourses;
}