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社区首页 >专栏 >LeetCode 0207 - Course Schedule

LeetCode 0207 - Course Schedule

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Reck Zhang
发布2021-08-11 12:09:20
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发布2021-08-11 12:09:20
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文章被收录于专栏:Reck Zhang

Course Schedule

Desicription

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 1:

代码语言:javascript
复制
Input: 2, [[1,0]] 
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

代码语言:javascript
复制
Input: 2, [[1,0],[0,1]]
Output: false
Explanation: 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 about how a graph is represented.
  2. You may assume that there are no duplicate edges in the input prerequisites.

Solution

代码语言:javascript
复制
class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<unordered_set<int>> graph(numCourses);
        for(auto it : prerequisites) {
            graph[it.second].insert(it.first);
        }
        vector<int> degrees(numCourses, 0);
        for(auto it : graph) {
            for(auto index : it) {
                degrees[index]++;
            }
        }
        for(int i = 0; i < numCourses; i++) {
            int index = 0;
            for(; index < numCourses; index++) {
                if(degrees[index] == 0) {
                    break;
                }
            }
            if(index == numCourses) {
                return false;
            }
            degrees[index] = -1;
            for(auto it : graph[index]) {
                degrees[it]--;
            }
        }
        return true;
    }
};
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目录
  • Course Schedule
    • Desicription
      • Solution
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