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207. Course Schedule.cpp
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There are a total of numCourses courses you have to take,
labeled from 0 to numCourses-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:
Input: numCourses = 2, prerequisites = [[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:
Input: numCourses = 2, prerequisites = [[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.
Constraints:
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.
1 <= numCourses <= 10^5
class Solution {
public:
vector <bool> vis, dfsVis;
bool dfs(int src, vector<int> adj[]) {
vis[src] = true;
dfsVis[src] = true;
for (auto &v : adj[src]) {
if (!vis[v]) {
if (dfs(v, adj)) return true; // has cycle
} else if (dfsVis[v])
return true; // has cycle
}
dfsVis[src] = false;
return false; // no cycle
}
bool canFinish(int numCourses, vector<vector<int>> &preReq) {
vector<int> adj[numCourses];
for (int i = 0; i < preReq.size(); i++) {
adj[preReq[i][0]].push_back(preReq[i][1]);
}
vis.resize(numCourses, false);
dfsVis.resize(numCourses, false);
for (int i = 0; i < numCourses; i++) {
if (!vis[i])
if (dfs(i, adj)) {
// cycle exists, coureses cannot be taken
return false;
}
}
// cycle doesn't exist, courses can be taken
return true;
}
};