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Course Schedule II

描述

TODO

分析

这题只需要在上一题Course Schedule的基础上,收集结果即可,在元素被弹出队列的时候,将其加入结果数组中。

代码

BFS

# TODO

DFS

// Course Schedule
// Time complexity: O(E+V)
// Space complexity: O(E+V)
class Solution {
public:
bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
vector<vector<int>> graph(numCourses, vector<int>()); // // adjacent list
vector<int> visited(numCourses); // 0, not visited; 1, visited; -1, cyclic
for (auto p : prerequisites) {
graph[p[1]].push_back(p[0]);
}
for (int i = 0; i < numCourses; ++i) {
if (!canFinishDFS(graph, visited, i)) return false;
}
return true;
}
bool canFinishDFS(vector<vector<int>>& graph, vector<int>& visited, int i) {
if (visited[i] == -1) return false;
if (visited[i] == 1) return true;
visited[i] = -1;
for (auto neighbor : graph[i]) {
if (!canFinishDFS(graph, visited, neighbor)) return false;
}
visited[i] = 1;
return true;
}
};

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