There are a total of n course 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 nunber 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: 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.
class Solution {
public boolean canFinish(int numCourses, int[][] prerequisites) {
// Construct graph and find starting root
int[][] adjMatrix = new int[numCourses][numCourses];
int[] indegree = new int[numCourses];
for (int[] pair : prerequisites) {
adjMatrix[pair[0]][pair[1]] = 1; // course -> prereq
indegree[pair[1]]++; // one more course depends on it
}
// BFS: start with all roots (indegree = 0)
Queue<Integer> queue = new LinkedList();
for (int i = 0; i < indegree.length; i++) {
if (indegree[i] == 0) queue.offer(i);
}
// Standard BFS:
int count = 0;
while (! queue.isEmpty()) {
int course = queue.poll();
count++;
for (int i = 0; i < numCourses; i++) {
// node i now has no incoming edges, add to queue
if (adjMatrix[course][i] == 1 && --indegree[i] == 0) {
queue.offer(i);
}
}
}
return count == numCourses; // visited all courses without circle
}
}