README_EN.md

210. Course Schedule II

中文文档

Description

There are a total of n courses you have to take labelled from 0 to n - 1.

Some courses may have prerequisites, for example, if prerequisites[i] = [ai, bi] this means you must take the course bi before the course ai.

Given the total number of courses numCourses and a list of the prerequisite pairs, 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 <= ai, bi < numCourses
• ai != bi
• All the pairs [ai, bi] are distinct.

Solutions

Python3

class Solution:
    def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
        edges = collections.defaultdict(list)
        indegree = [0] * numCourses
        for i, j in prerequisites:
            edges[j].append(i)
            indegree[i] += 1
        q = collections.deque()
        for i in range(numCourses):
            if indegree[i] == 0:
                q.append(i)
        res = []
        while q:
            i = q.popleft()
            res.append(i)
            for j in edges[i]:
                indegree[j] -= 1
                if indegree[j] == 0:
                    q.append(j)
        return res if len(res) == numCourses else []

Java

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        List<Integer>[] edges = new List[numCourses];
        for (int i = 0; i < numCourses; ++i) {
            edges[i] = new ArrayList<>();
        }
        int[] indegree = new int[numCourses];
        for (int[] p : prerequisites) {
            edges[p[1]].add(p[0]);
            ++indegree[p[0]];
        }
        Queue<Integer> q = new LinkedList<>();
        for (int i = 0; i < numCourses; ++i) {
            if (indegree[i] == 0) {
                q.offer(i);
            }
        }
        int[] res = new int[numCourses];
        int cnt = 0;
        while (!q.isEmpty()) {
            int i = q.poll();
            res[cnt++] = i;
            for (int j : edges[i]) {
                --indegree[j];
                if (indegree[j] == 0) {
                    q.offer(j);
                }
            }
        }
        return cnt == numCourses ? res : new int[0];
    }
}

C++

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> edges(numCourses);
        vector<int> indegree(numCourses);
        for (auto p : prerequisites)
        {
            edges[p[1]].push_back(p[0]);
            ++indegree[p[0]];
        }
        queue<int> q;
        for (int i = 0; i < numCourses; ++i)
        {
            if (indegree[i] == 0) q.push(i);
        }
        vector<int> res;
        while (!q.empty())
        {
            int i = q.front();
            q.pop();
            res.push_back(i);
            for (int j : edges[i])
            {
                --indegree[j];
                if (indegree[j] == 0) q.push(j);
            }
        }
        return res.size() == numCourses ? res : vector<int>();
    }
};

Go

func findOrder(numCourses int, prerequisites [][]int) []int {
	edges := make([][]int, numCourses)
	indegree := make([]int, numCourses)
	for _, p := range prerequisites {
		edges[p[1]] = append(edges[p[1]], p[0])
		indegree[p[0]]++
	}
	var q []int
	for i := 0; i < numCourses; i++ {
		if indegree[i] == 0 {
			q = append(q, i)
		}
	}
	var res []int
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		res = append(res, i)
		for _, j := range edges[i] {
			indegree[j]--
			if indegree[j] == 0 {
				q = append(q, j)
			}
		}
	}
	if len(res) == numCourses {
		return res
	}
	return []int{}
}

C#

using System.Collections.Generic;

public class Solution {
    public int[] FindOrder(int numCourses, int[][] prerequisites) {
        var indegree = new int[numCourses];
        var edgeCount = prerequisites.Length;
        var edge = new List<int>[numCourses];
        for (var i = 0; i < edgeCount; ++i)
        {
            var child = prerequisites[i][0];
            var parent = prerequisites[i][1];
            if (edge[parent] == null)
            {
                edge[parent] = new List<int>();
            }
            edge[parent].Add(child);
            ++indegree[child];
        }

        var queue = new Queue<int>();
        for (var i = 0; i < numCourses; ++i)
        {
            if (indegree[i] == 0) queue.Enqueue(i);
        }

        var result = new int[numCourses];
        var count = 0;
        while (queue.Count > 0)
        {
            var node = queue.Dequeue();
            result[count++] = node;
            if (edge[node] != null)
            {
                foreach (var next in edge[node])
                {
                    if (--indegree[next] == 0)
                    {
                        queue.Enqueue(next);
                    }
                }
            }
        }
        return count == numCourses ? result : new int[0];
    }
}

...