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, return the ordering of courses you should take to finish all courses.
There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
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 the correct course order is [0,1]
4, [[1,0],[2,0],[3,1],[3,2]]
There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].
public class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
//本题是求拓扑图，验证图是否有环，从叶子节点开始求（出度为0的点），如果所有点都能遍历到，则满足条件
//需要辅助数组，数组的下标代表课程编号，数组的值代表出度
//queue保存的是出度为0的点，每次向里面添加需要计算个数count，每次从queue中弹出时，保存结果到list。最后倒序输出结果
int []flag=new int[numCourses];
int [] res=new int[numCourses];
List<Integer> list=new ArrayList<Integer>();
for(int i=0;i<prerequisites.length;i++){
flag[prerequisites[i][1]]++;
}
LinkedList<Integer> queue=new LinkedList<Integer>();
int count=0;
for(int i=0;i<numCourses;i++){
if(flag[i]==0){//出度为0
queue.add(i);
count++;
}
}
while(!queue.isEmpty()){
int k=queue.remove();
list.add(k);
for(int i=0;i<prerequisites.length;i++){
if(k==prerequisites[i][0]){
int l=prerequisites[i][1];
flag[l]--;
if(flag[l]==0){
count++;
queue.add(l);
}
}
}
}
for(int i=0;i<list.size();i++){
res[i]=list.get(list.size()-1-i);
}
if(count==numCourses)
return res;
else return new int[0];
}
}