LeetCode-Course Schedule-解题报告

原题链接 https://leetcode.com/problems/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.

其实就是拓扑排序。

我使用的dfs，当然我用一个数组保存了每个节点的入度。每次输出入度为0的节点。

class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<vector<int> >graph(numCourses);
vector<int>out(numCourses, 0);
int num = 0;
vector<bool>vis(numCourses, 0);
for (auto &p : prerequisites)
graph[p.second].push_back(p.first), out[p.first]++;
dfs(graph, num, vis, out);
return num == numCourses;
}
void dfs(vector<vector<int> >& g, int& num, vector<bool>& vis, vector<int>& out)
{
if (num == g.size())return;
for (int i = 0; i < g.size(); ++i)
{
if (!vis[i] && out[i] == 0)
{
num++;
vis[i] = true;
for (int j = 0; j < g[i].size(); ++j)
--out[g[i][j]];
dfs(g, num, vis, out);
}
}
}
};