leetcode-207-Course Schedule

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, is it possible for you to finish all courses?

Example 1:

Input: 2, [[1,0]] 
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.
Example 2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0, and to take course 0 you should
             also have finished course 1. So it is impossible.

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.

BFS

BFS uses the indegrees of each node. We will first try to find a node with 0 indegree. If we fail to do so, there must be a cycle in the graph and we return false. Otherwise we set its indegree to be -1 to prevent from visiting it again and reduce the indegrees of its neighbors by 1. This process will be repeated for n (number of nodes) times.

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        graph g = buildGraph(numCourses, prerequisites);
        vector<int> degrees = computeIndegrees(g);
        for (int i = 0; i < numCourses; i++) {
            int j = 0;
            for (; j < numCourses; j++) {
                if (!degrees[j]) {
                    break;
                }
            }
            if (j == numCourses) {
                return false;
            }
            degrees[j]--;
            for (int v : g[j]) {
                degrees[v]--;
            }
        }
        return true;
    }
private:
    typedef vector<vector<int>> graph;
    
    graph buildGraph(int numCourses, vector<pair<int, int>>& prerequisites) {
        graph g(numCourses);
        for (auto p : prerequisites) {
            g[p.second].push_back(p.first);
        }
        return g;
    }
    
    vector<int> computeIndegrees(graph& g) {
        vector<int> degrees(g.size(), 0);
        for (auto adj : g) {
            for (int v : adj) {
                degrees[v]++;
            }
        }
        return degrees;
    }
};

DFS

For DFS, in each visit, we start from a node and keep visiting its neighbors, if at a time we return to a visited node, there is a cycle. Otherwise, start again from another unvisited node and repeat this process. We use todo and done for nodes to visit and visited nodes.

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        graph g = buildGraph(numCourses, prerequisites);
        vector<bool> todo(numCourses, false), done(numCourses, false);
        for (int i = 0; i < numCourses; i++) {
            if (!done[i] && !acyclic(g, todo, done, i)) {
                return false;
            }
        }
        return true;
    }
private:
    typedef vector<vector<int>> graph;
    
    graph buildGraph(int numCourses, vector<pair<int, int>>& prerequisites) {
        graph g(numCourses);
        for (auto p : prerequisites) {
            g[p.second].push_back(p.first);
        }
        return g;
    }
    
    bool acyclic(graph& g, vector<bool>& todo, vector<bool>& done, int node) {
        if (todo[node]) {
            return false;
        }
        if (done[node]) {
            return true;
        }
        todo[node] = done[node] = true;
        for (int v : g[node]) {
            if (!acyclic(g, todo, done, v)) {
                return false;
            }
        }
        todo[node] = false;
        return true;
    }
};