LeetCode 207. Course Schedule

Posted 2022-12-08 逗逗飞

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.

问题描述

给定n门课程，编号为0~n-1，然后某些课有其先修课程，求在给定n门课的先后关系中，能否完成所有的课程？

例如：输入2 表示两门课程，[[1,0]]表示，在修课程1之前，必须修过课程0。

解题思路

其实就是一道简单的 拓扑排序问题。有几个点需要注意： 1、如何存储图的结构——本题解法用二维vector（连接表） 2、入度数组degree 统计 3、是否遍历过某个课程——标记数组（标记数组可以另开一个vis数组，但是本题中可以用degree数组标记为-1表示已遍历该课程） 解题思路：就是求入度为0 的课程，放入队列中，然后将与该课程关联的课程的入度-1，如果关联课程入度为0，放入队列 代码如下：

bool canFinish(int numCourses, vector<pair<int, int> >& prerequisites) 
	queue<int> Q;
	int degree[numCourses];
	memset(degree,0,sizeof(degree));
	vector<vector<int> > Graph(numCourses);
	vector<pair<int, int> >::iterator iter;
	// 首先存储图结构 
	for (iter = prerequisites.begin(); iter != prerequisites.end(); ++iter)  
		++degree[iter->first];
		Graph[iter->second].push_back(iter->first);
	
	// 判断入度为0的点
	for (int i = 0; i < numCourses; ++i) 
		if (degree[i] == 0) 
			Q.push(i);
			degree[i] = -1;
		
	
	// 拓扑 
	while(!Q.empty()) 
		int x = Q.front();
		Q.pop();
		for(int i = 0; i < Graph[x].size(); ++i) 
			--degree[Graph[x][i]];
			if (degree[Graph[x][i]] == 0) 
				Q.push(Graph[x][i]);
				degree[Graph[x][i]] = -1;
			
		
	
	// 判断是否遍历了所有点 
	for(int i = 0; i < numCourses; ++i) 	
		if(degree[i] != -1)
			return false;
	
	return true;

LeetCode 运行结果：
至此：本题解答完毕。 如果您有更优秀的想法，欢迎一起交流！