leedcode: Course Schedule

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

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more abouthow a graph is represented.

这道题主要是判断有向图是否有环，用的是DFS的思想 这个题目用DFS比较合适，因为DFS一条龙深入下去，如果碰到之前访问过的，就是有环

class Solution {public:    bool dfsVisit(vector<vector<int>*>& auxArray, int i, vector<int>& visit) {                visit[i] = 1;        if (auxArray[i] != NULL) {            for (int j=0; j<auxArray[i]->size(); j++) {                if (visit[auxArray[i]->at(j)] == 1) {                    return false;                }                if (dfsVisit(auxArray, auxArray[i]->at(j), visit) == false) {                    return false;                }            }        }        visit[i] = 2;        return true;    }        bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {                if ((numCourses > 0) && (prerequisites.size() == 0))            return true;                    vector<vector<int>* > auxArray(numCourses, NULL);                for (int i=0; i<prerequisites.size(); i++) {            if (auxArray[prerequisites[i].first] == NULL) {                auxArray[prerequisites[i].first] = new vector<int>;            }                        auxArray[prerequisites[i].first]->push_back(prerequisites[i].second);        }                        vector<int> visit(numCourses, 0);        bool ret = true;                for (int i=0; i<auxArray.size(); i++) {            if (dfsVisit(auxArray, i, visit) == false) {                ret = false;                break;            }        }                for (int i=0; i<auxArray.size(); i++) {            if (auxArray[i] != NULL)                delete auxArray[i];        }                return ret;    }};