LeetCode: 210. Course Schedule II

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本文深入解析了LeetCode上的课程调度II问题，通过拓扑排序算法实现了课程先修关系的合理安排，确保所有课程都能按先决条件顺利完成。文章详细介绍了如何将边表示的图转化为节点入度表示的邻接表，利用队列进行课程的顺序处理，最终得到一个有效的课程学习顺序。

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题目： 210. Course Schedule II(https://leetcode.com/problems/course-schedule-ii/description/)

public class Solution {

	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]]++;

		Queue<Integer> queue = new LinkedList<Integer>();
		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];
	}
}