leetcode Course Schedule

最新推荐文章于 2021-05-17 14:36:15 发布

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#leetcode #算法 #c++ #面试题 #图论

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本博客探讨了如何通过图论中的环检测来解决课程计划问题，具体阐述了如何利用拓扑排序和深度优先搜索算法判断是否存在环，从而决定是否能够完成所有课程。

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:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

click to show more hints.

Hints:
This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
Topological sort could also be done via BFS.
题目来源：https://leetcode.com/problems/course-schedule/

分析

本题目是判断图中是否含有环。题目的提示是个坑，会超时。用一个队列存储入度为0的节点，出队列的时候将此节点指向的节点入度减一，如果减一后入度为0则入队列。直到队列为空后，如果还有节点入度不为0，则存在环。

代码

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        queue<int> help;//存放入度为0的节点
        vector<int> courses(numCourses, 0);//记录各个节点入度
        for(auto edge : prerequisites){
            courses[edge.second]++;
        }
        for(int i = 0; i < numCourses; i++){
            if(courses[i] == 0)
                help.push(i);
        }
        while(!help.empty()){
            int i = help.front();
            help.pop();
            for(auto edge : prerequisites){
                if(edge.first == i){
                    courses[edge.second]--;
                    if(courses[edge.second] == 0)
                        help.push(edge.second);
                }
            }
        }
        for(auto c : courses)
            if(c != 0)
                return false;
        return true;
    }
};