leetcode Course Schedule

题目

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.

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

click to show more hints.

Hints:
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.
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
Topological sort could also be done via BFS.
题目来源：https://leetcode.com/problems/course-schedule/

分析

本题目是判断图中是否含有环。题目的提示是个坑，会超时。用一个队列存储入度为0的节点，出队列的时候将此节点指向的节点入度减一，如果减一后入度为0则入队列。直到队列为空后，如果还有节点入度不为0，则存在环。

代码

class Solution {public:    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {        queue<int> help;//存放入度为0的节点        vector<int> courses(numCourses, 0);//记录各个节点入度        for(auto edge : prerequisites){            courses[edge.second]++;        }        for(int i = 0; i < numCourses; i++){            if(courses[i] == 0)                help.push(i);        }        while(!help.empty()){            int i = help.front();            help.pop();            for(auto edge : prerequisites){                if(edge.first == i){                    courses[edge.second]--;                    if(courses[edge.second] == 0)                        help.push(edge.second);                }            }        }        for(auto c : courses)            if(c != 0)                return false;        return true;    }};