第五周 拓扑排序

算法题目 ： Course Schedule          

算法题目描述： 

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 abouthow a graph is represented
2. You may assume that there are no duplicate edges in the input prerequisites.

算法分析：

这就是典型的拓扑排序的问题了，问我们是否可以依次输出这些课并且不形成环。我们就可以用拓扑排序依次输出度为0的值，并用一个队列存入度为0的节点，然后依次出对，如果最后存在度不为0的节点，就说明存在环。应该就是这样。

算法代码（C++):

class Solution {public:    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {        vector<vector<int>> graph(numCourses, vector<int>(0));        vector<int> inDegree(numCourses, 0);        for (auto u : prerequisites) {            graph[u[1]].push_back(u[0]);            ++inDegree[u[0]];        }        queue<int> que;        for (int i = 0; i < numCourses; ++i) {            if (inDegree[i] == 0) que.push(i);        }        while (!que.empty()) {            int u = que.front();            que.pop();            for (auto v : graph[u]) {                --inDegree[v];                if (inDegree[v] == 0) que.push(v);            }        }        for (int i = 0; i < numCourses; ++i) {            if (inDegree[i] != 0) return false;        }        return true;    }};