LeetCode 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?

For example:

2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

2, [[1,0],[0,1]]

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.

click to show more hints.

Hints:

1. This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
2. There are several ways to represent a graph. For example, the input prerequisites is a graph represented by a list of edges. Is this graph representation appropriate?
3. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
4. Topological sort could also be done via BFS.

class Node {
public:
    int in;
    vector<int> req;
    Node():in(0){}
};

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<Node> nodes(numCourses);
        int len = prerequisites.size();
        for (int i=0; i<len; i++) {
            nodes[prerequisites[i][0]].req.push_back(prerequisites[i][1]);
            nodes[prerequisites[i][1]].in++;
        }
        queue<int> zeros;
        for (int i = 0; i<numCourses; i++) {
            if (nodes[i].in == 0 && nodes[i].req.size() != 0) {
                zeros.push(i);
            }
        }
        while (!zeros.empty()) {
            int first = zeros.front();
            zeros.pop();
            
            nodes[first].in = -1;
            for (int k=0; k<nodes[first].req.size(); k++) {
                if (--nodes[nodes[first].req[k]].in == 0) {
                    zeros.push(nodes[first].req[k]);
                }
            }
        }
        for (int i=0; i<numCourses; i++) {
            if (nodes[i].in > 0 && nodes[i].req.size() != 0) return false;
        }
        return true;
    }
};

273ms感觉时间复杂度不行，有待改进