LeetCode in Net

210. Course Schedule II

Medium

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] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

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:

Solution

using System.Collections.Generic;

public class Solution {
    public int[] FindOrder(int numCourses, int[][] prerequisites) {
        var graph = new Dictionary<int, List<int>>();
        for (int i = 0; i < numCourses; i++) {
            graph[i] = new List<int>();
        }
        foreach (var classes in prerequisites) {
            graph[classes[0]].Add(classes[1]);
        }
        var output = new List<int>();
        var visited = new Dictionary<int, bool>();
        foreach (var c in graph.Keys) {
            if (Dfs(c, graph, visited, output)) {
                return new int[0];
            }
        }
        return output.ToArray();
    }

    private bool Dfs(int course, Dictionary<int, List<int>> graph, Dictionary<int, bool> visited, List<int> output) {
        if (visited.ContainsKey(course)) {
            return visited[course];
        }
        visited[course] = true;
        foreach (var c in graph[course]) {
            if (Dfs(c, graph, visited, output)) {
                return true;
            }
        }
        visited[course] = false;
        output.Add(course);
        return false;
    }
}
pFad - Phonifier reborn

Pfad - The Proxy pFad of © 2024 Garber Painting. All rights reserved.

Note: This service is not intended for secure transactions such as banking, social media, email, or purchasing. Use at your own risk. We assume no liability whatsoever for broken pages.


Alternative Proxies:

Alternative Proxy

pFad Proxy

pFad v3 Proxy

pFad v4 Proxy