Description
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?
Example:
Input: 2, [[1,0]]
Output: true
There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.
Input: 2, [[1,0],[0,1]]
Output: false
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.
Code
判断是否有环,可以用拓扑排序求解。话不多说,直接上代码。
class Solution {
private:
vector< vector<int> > V;
vector<int> vis; // -1访问过一次,0没访问过,1访问完了
bool dfs(int u) {
int v, i;
vis[u] = -1;
for (i=0; i<V[u].size(); i++) {
v = V[u][i];
if (!vis[v]) dfs(v);
if (vis[v] == -1) return false;
}
vis[u] = 1;
return true;
}
public:
bool canFinish(int n, vector< pair<int, int> >& edges) {
int i;
V.resize(n);
vis.resize(n);
for(int i = 0; i < edges.size(); ++i)
V[edges[i].first].push_back(edges[i].second);
for (i=0; i<n; i++) vis[i] = 0;
for (i=0; i<n; i++)
if (!vis[i]) {
if(!dfs(i)) return false;
}
return true;
}
};
Summary
1.拓扑排序有两种实现,一种是找入度为0的点;一种是DFS,DFS更高效一些。
2.如何检查环?
基于入度:如果找不到入度为0的点,但是依旧有为被访问过的节点,那就有环;
基于DFS:DFS有两种实现,递归版需要存储“黑白灰”三种状态,也就是我上面的做法;非递归版则可以判断邻居节点是否已经出现在栈中。