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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;
}
}