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:

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

 

Approach #2: C++. [graph + dfs]

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        // states: 0: not visit     1: visiting     2: visited
        vector<int> node(numCourses, 0);
        graph = vector<vector<int>>(numCourses);
        
        for (auto prerequisite : prerequisites) {
            graph[prerequisite.second].push_back(prerequisite.first);
        }
        
        for (int i = 0; i < numCourses; ++i) {
            if (dfs(node, i)) return false;
        }
        
        return true;
    }
private:
    vector<vector<int>> graph;
    
    bool dfs(vector<int>& node, int cur) {
        if (node[cur] == 1) return true;
        if (node[cur] == 2) return false;
        
        node[cur] = 1;
        
        for (auto k : graph[cur]) {
            if (dfs(node, k)) return true;
        }
        
        node[cur] = 2;
        return false;
    }
};

　　

Approach #2: Java. [BFS]

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        ArrayList[] graph = new ArrayList[numCourses];
        int[] degree = new int[numCourses];
        Queue queue = new LinkedList();
        int count = 0;
        
        for (int i = 0; i < numCourses; ++i) {
            graph[i] = new ArrayList();
        }
        
        for (int i = 0; i < prerequisites.length; ++i) {
            degree[prerequisites[i][1]]++;
            graph[prerequisites[i][0]].add(prerequisites[i][1]);
        }
        
        for (int i = 0; i < degree.length; ++i) {
            if (degree[i] == 0) {
                queue.add(i);
                count++;
            }
        }
        
        while (queue.size() != 0) {
            int course = (int)queue.poll();
            for (int i = 0; i < graph[course].size(); ++i) {
                int pointer = (int)graph[course].get(i);
                degree[pointer]--;
                if (degree[pointer] == 0) {
                    queue.add(pointer);
                    count++;
                }
            }
        }
        
        if (count == numCourses) return true;
        else return false;
    }
}

　　

Approach #3: Python.

class Solution(object):
    def canFinish(self, n, pres):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: bool
        """
        from collections import deque
        ind = [[] for _ in xrange(n)]
        oud = [0] * n
        for p in pres:
            oud[p[0]] += 1
            ind[p[1]].append(p[0])
        dq = deque()
        
        for i in xrange(n):
            if oud[i] == 0:
                dq.append(i)
        
        k = 0
        
        while dq:
            x = dq.popleft()
            k += 1
            for i in ind[x]:
                oud[i] -= 1
                if oud[i] == 0:
                    dq.append(i)
                    
        return k == n

　　

Analysis:

At the first, we initialize each node with 0(the state represent that this node haven't visited). and we use a two-dimensional vector to store the linked node. graph[a].push_back(b) represent that a is the prerequisites of b.

for every number from 0 to n-1 it will have some (or not) prerequisites int the vector of graph[i]. We travel the elements in graph[i] if there is a node which is visiting (which node state is 2), so we can know there is a circle and it will return false in the answer.

In the processing, If the node equal to 1(which reperesent this node has been visited, in other word it's prerequisites has been satisfied).