【Leetcode】207. Course Schedule

Description:

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:

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.

You may assume that there are no duplicate edges in the input prerequisites.

思路：

本题实际上是对有向无环图（DAG）的判断，解法通常有拓扑排序和DFS（或BFS），这里可以用拓扑排序。拓扑排序的判断方法是找出一个图中入度为0的点，然后删除它，再在剩下的点中找入度为0的点，直到所有的点都删除完毕，如果所有点都能删除掉，那么就完成了一次拓扑排序，这时候说明图中无环，否则则说明有环。

下面是使用C++的实现过程：

class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<vector<int>> graph(numCourses);
vector<int> indegree(numCourses,0);//用于记录入度
for(int i=0;i<prerequisites.size();i++){
graph[prerequisites[i].second].push_back(prerequisites[i].first);
indegree[prerequisites[i].first]++;
}
queue<int> Q;//用于记录入度为0的点
for(int i = 0;i < numCourses;i++){
if(indegree[i] == 0) Q.push(i);
}
int counter = 0;
while(!Q.empty())
{
int u = Q.front();
Q.pop();
count++;//每删除一个点做一次记录
for(int v=0;v<graph[u].size();v++){
if(--indegree[graph[u][v]]==0) Q.push(graph[u][v]);
}
}
return count == numCourses;
}
};