Course Schedule

题目原网址为https://leetcode.com/problems/course-schedule/description/

一.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:

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.

click to show more hints.

二.我的解法

class Solution {public:bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {        vector<list<int>> graph(numCourses);for (int i = 0; i < prerequisites.size(); i++) {graph[prerequisites[i].second].push_back(prerequisites[i].first);}queue<int> source;vector<int> degree(numCourses);for (int i = 0; i < prerequisites.size(); i++) {degree[prerequisites[i].first]++;}for (int i = 0; i < numCourses; i++) {if (!degree[i]) {source.push(i);}}while (!source.empty()) {int course = source.front();source.pop();for (auto i = graph[course].begin(); i != graph[course].end(); i++) {degree[*i]--;if (!degree[*i]) {source.push(*i);}}}for (int i = 0; i < numCourses; i++) {if (degree[i] > 0) {return false;}}return true;}};

三.解法分析

把每一个课程看做一个节点，如果课程a要先于课程b上，那么有一条a到b的有向边，那么这个问题就可以变成判断图是否有环的问题，这里我用了

拓扑排序来判读图是否有环，如果可以完成拓扑排序，那么就没有环，否则就有。我这里拓扑排序的实现思路如下：

就是每次选取入度为0的点从图上去掉，假设为点a，那么所有的点b满足a到b有边的点入度减一，如此不断地进行下去，如果所有的点都从图中删去，

那么图就没有环，否则就有环