LeetCode 210 Course Schedule II

LeetCode

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [a<sub>i</sub>, b<sub>i</sub>] indicates that you must take course b<sub>i</sub> first if you want to take course a<sub>i</sub>.

For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: [0,1]
Explanation: 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].

Example 2:

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: 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].

Example 3:

Input: numCourses = 1, prerequisites = []
Output: [0]

Constraints: 1 <= numCourses <= 2000
`0 <= prerequisites.length <= numCourses (numCourses - 1)*prerequisites[i].length == 2*0 <= a<sub>i</sub>, b<sub>i</sub> < numCourses*a<sub>i</sub> != b<sub>i</sub>* All the pairs[a<sub>i</sub>, b<sub>i</sub>]` are distinct.

题目大意：

课程有先修课要求，求修课的顺序

算法思路：

拓扑排序的经典题

注意事项：

1. 注意题目要求课程可能存在循环，记得第四部侦测循环

Python代码：

def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
	in_degree = [0] * numCourses
	graph = [[] for _ in range(numCourses)]
	for li in prerequisites:
		in_degree[li[0]] += 1
		graph[li[1]].append(li[0])
	queue = collections.deque([i for i in range(len(in_degree)) if in_degree[i] == 0])
	res = []
	while queue:
		node = queue.popleft()
		res.append(node)
		for neighbor in graph[node]:
			in_degree[neighbor] -= 1
			if in_degree[neighbor] == 0:
				queue.append(neighbor)
	return res if numCourses == len(res) else []

Java代码：

public int[] findOrder(int numCourses, int[][] prerequisites) {
	ArrayList<ArrayList<Integer>> graph = new ArrayList<ArrayList<Integer>>();
	ArrayList<Integer> res = new ArrayList<>();
	for(int i=0;i<numCourses;i++)
		graph.add(new ArrayList<Integer>());
	int[] inDegree = new int[numCourses];
	//populate inDegree & convert to graph
	for(int i=0;i<prerequisites.length;i++){
		//[0,1] means 1->0
		inDegree[prerequisites[i][0]]++;
		graph.get(prerequisites[i][1]).add(prerequisites[i][0]);
	}
	Queue<Integer> q = new LinkedList<Integer>();
	for(int i=0;i<inDegree.length;i++){
		if(inDegree[i]==0)
			q.offer(i);
	}
	int count = 0;
	while(!q.isEmpty()){
		Integer v = q.poll();
		res.add(v);
		count++;
		for(int neighbor : graph.get(v)){
			if(--inDegree[neighbor]==0)
				q.add(neighbor);
		}
	}
	if(count != numCourses)
		res.clear();

	return res.stream().mapToInt(i->i).toArray();
}

算法分析：

时间复杂度为O(n)，空间复杂度O(n)