Leetcode 207 Course Schedule
Leetcode 207 Course Schedule
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.
- You may assume that there are no duplicate edges in the input prerequisites.
拓扑排序的裸题。
统计所有点的入度，将入度为0的点拿出来，它可以到达点的度数减1，反复这个过程。
如果所有点都被拿出来，则该有向图无环，如果还有点没被拿出来且图中没有入度为0的点，则图中有环。
也可以采用遍历方法，理解拓扑排序就很容易写出遍历方法了。
class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<int> temp(numCourses, 0);
vector<vector<int>> mp(numCourses, temp);
vector<int> in(numCourses, 0);
for(int i = 0; i < prerequisites.size(); i++)
{
mp[prerequisites[i].second][prerequisites[i].first] = 1;
in[prerequisites[i].first]++;
}
int cnt = 0;
vector<int> vis(numCourses, 0);
while(++cnt <= numCourses)
{
int index = -1;
for(int i = 0; i < numCourses; i++)
if(in[i] == 0 && !vis[i])
{
vis[i] = 1;
index = i;
break;
}
if(index == -1) return false;
for(int i = 0; i < numCourses; i++) if(mp[index][i])in[i]--;
}
return true;
}
};腾讯云开发者
