LeetCode题解–207. Course Schedule
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链接
LeetCode题目:https://leetcode.com/problems/course-schedule/
难度:Medium
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?
题目大意是选n门课程,某些课程可能有先修课程,试问能否成功选择全部课程。
分析
这题是一道典型的拓扑排序问题,拓扑排序的定义是:将有向图中的顶点以线性方式进行排序,对于任何连接自顶点u到顶点v的有向边uv,在最后的排序结果中,顶点u总是在顶点v的前面。
典型的实现算法有Kahn算法和基于DFS的算法,Kahn算法相比后者的好处是不需要检测图为DAG,所以我选择了Kahn算法。
做法是维护一个入度为0的点的集合(这里选择了队列),每次从集合中取出一个点n,依次删除从n到m的边,如果删后m的入度为0,则把点m也放入这个集合。最后集合为空而且不存在边时,证明选课成功。时间复杂度是O(V+E),其中V为边数,E为点数。
代码
class Solution {public: bool canFinish(int numCourses, vector<pair<int, int>> &prerequisites) { edgeNum = prerequisites.size(); inDegrees = vector<int>(numCourses, 0); for (int i = 0; i < numCourses; i++) { edges.push_back(vector<int>()); } for (auto pre:prerequisites) { edges[pre.second].push_back(pre.first); inDegrees[pre.first]++; } for (int i = 0; i < inDegrees.size(); i++) { if (inDegrees[i] == 0) points.push(i); } while (points.size()) { int n = points.front(); points.pop(); for (auto m:edges[n]) { edgeNum--; inDegrees[m]--; if (!inDegrees[m]) points.push(m); } } return edgeNum == 0; }private: int edgeNum; vector<int> inDegrees; vector<vector<int>> edges; queue<int> points;};
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