Problem

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

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

For example:

1	2, [[1,0]]

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]

1	4, [[1,0],[2,0],[3,1],[3,2]]

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].

Note:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.

题目大意：

给你一些课程和它的先修课程，让你输出修课顺序。如果无法修完所有课程，返回空数组。

Idea

Topological sorting

拓扑排序

Solution 1: Topological Sorting

Time complexity: O(V+E)

Space complexity: O(V+E)

C++

// Author: Huahua

// Runtime: 16 ms

class Solution {

public:

vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {

vector<vector<int>> graph(numCourses);

for(const auto& p : prerequisites)

graph[p.second].push_back(p.first);

// states: 0 = unkonwn, 1 == visiting, 2 = visited

vector<int> v(numCourses, 0);

vector<int> ans;

for (int i = 0; i < numCourses; ++i)

if (dfs(i, graph, v, ans)) return {};

std::reverse(ans.begin(), ans.end());

return ans;

}

private:

bool dfs(int cur, vector<vector<int>>& graph, vector<int>& v, vector<int>& ans) {

if (v[cur] == 1) return true;

if (v[cur] == 2) return false;

v[cur] = 1;

for (const int t : graph[cur])

if (dfs(t, graph, v, ans)) return true;

v[cur] = 2;

ans.push_back(cur);

return false;

}

};

Java

// Author: Huahua

// Runtime: 83 ms

class Solution {

public int[] findOrder(int numCourses, int[][] prerequisites) {

ArrayList<ArrayList<Integer>> graph = new ArrayList<>();

for (int i = 0; i < numCourses; ++i)

graph.add(new ArrayList<Integer>());

for (int i = 0; i < prerequisites.length; ++i) {

int course = prerequisites[i][0];

int prerequisite = prerequisites[i][1];

graph.get(course).add(prerequisite);

}

int[] visited = new int[numCourses];

List<Integer> ans = new ArrayList<Integer>();

Integer index = numCourses;

for (int i = 0; i < numCourses; ++i)

if (dfs(i, graph, visited, ans)) return new int[0];

return ans.stream().mapToInt(i->i).toArray();

}

private boolean dfs(int curr, ArrayList<ArrayList<Integer>> graph, int[] visited, List<Integer> ans) {

if (visited[curr] == 1) return true;

if (visited[curr] == 2) return false;

visited[curr] = 1;

for (int next : graph.get(curr))

if (dfs(next, graph, visited, ans)) return true;

visited[curr] = 2;

ans.add(curr);

return false;

}

花花酱 LeetCode 210. Course Schedule II

Problem

Idea

Solution 1: Topological Sorting

C++

Java

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