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.
看到這道題的想法就是畫圖璧榄，然后求環(huán)，有環(huán)的話就返回false。
至于具體怎么做…想了想結(jié)果是不會做…
答案如下扎附，還有有點(diǎn)懵仔粥，之后再好好想一想乓搬！
class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<vector<int>> matrix(numCourses);
vector<int> in(numCourses, 0);
int n = prerequisites.size();
for (int i=0; i<n; i++)
matrix[prerequisites[i].second].push_back(prerequisites[i].first);
for (int i=0; i<numCourses; i++) {
for (auto it=matrix[i].begin(); it!=matrix[i].end(); it++) {
in[*it]++;
}
}
for (int i=0; i<numCourses; i++) {
int j;
for (j=0; j<numCourses && in[j]!=0; j++);
if (j == numCourses)
return false;
in[j] = -1;
for (auto it=matrix[j].begin(); it!=matrix[j].end(); it++)
in[*it]--;
}
return true;
}
};
