[分析]
在题I代码基础上增加记录拓扑遍历顺序即可。
在题I代码基础上增加记录拓扑遍历顺序即可。
public class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
if (prerequisites == null)
return new int[0];
int[] order = new int[numCourses];
List<Set<Integer>> graph = new ArrayList<Set<Integer>>();
int[] indegree = new int[numCourses];
for (int i = 0; i < numCourses; i++)
graph.add(new HashSet<Integer>());
// build graph
for (int i = 0; i < prerequisites.length; i++) {
if (graph.get(prerequisites[i][1]).add(prerequisites[i][0]))
indegree[prerequisites[i][0]]++;
}
LinkedList<Integer> queue = new LinkedList<Integer>();
for (int i = 0; i < numCourses; i++) {
if (indegree[i] == 0)
queue.offer(i);
}
int k = 0;
while (!queue.isEmpty()) {
int v = queue.poll();
order[k++] = v;
for (Integer w : graph.get(v)) {
if (--indegree[w] == 0)
queue.offer(w);
}
}
return k == numCourses ? order : new int[0];
}
}