# Concise JAVA solution based on BFS with comments

• ``````public int[] findOrder(int numCourses, int[][] prerequisites) {
if (numCourses == 0) return null;
// Convert graph presentation from edges to indegree of adjacent list.
int indegree[] = new int[numCourses], order[] = new int[numCourses], index = 0;
for (int i = 0; i < prerequisites.length; i++) // Indegree - how many prerequisites are needed.
indegree[prerequisites[i][0]]++;

for (int i = 0; i < numCourses; i++)
if (indegree[i] == 0) {
// Add the course to the order because it has no prerequisites.
order[index++] = i;
queue.offer(i);
}

// How many courses don't need prerequisites.
while (!queue.isEmpty()) {
int prerequisite = queue.poll(); // Already finished this prerequisite course.
for (int i = 0; i < prerequisites.length; i++)  {
if (prerequisites[i][1] == prerequisite) {
indegree[prerequisites[i][0]]--;
if (indegree[prerequisites[i][0]] == 0) {
// If indegree is zero, then add the course to the order.
order[index++] = prerequisites[i][0];
queue.offer(prerequisites[i][0]);
}
}
}
}

return (index == numCourses) ? order : new int[0];
}``````

• I really like your solution, neat and clean, helps me to fully understand the problem, well done, thank you.

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