原题
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.
分析
两个办法Kahn算法和DFS。
这里参考https://discuss.leetcode.com/topic/17273/18-22-lines-c-bfs-dfs-solutions 建图办法,如果用二维数组的话会超时。
Kahn算法
大体思路是,找到所有入度为0的的点假设在集合S中,然后每次从S中选取一个点V,将它所有的临接的点的入度减1,如果减为0了则将该点添加到集合S中,而刚刚选取的那个点V则从S中移除。
最后当S为空时,也就是当没有入度为0的点时,还存在边,则说明有环。
typedef vector<unordered_set<int >> GRAPH;
class Solution
{
private:
void getDegree(vector<int>°ree,const GRAPH &graph)
{
for(int i=0;i<graph.size();++i)
{
for(auto it=graph[i].begin();it!=graph[i].end();++it)
{
int j=*it;
++degree[j];
}
}
}
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites)
{
GRAPH graph(numCourses);
int length = prerequisites.size();
//init graph
for( int k =0;k<length;++k)
{
int i=prerequisites[k].first;
int j=prerequisites[k].second;
graph[j].insert(i);
}
vector<int>degree(numCourses);
getDegree(degree,graph);
for(int i=0;i<numCourses;++i)
{
int j=0;
for(;j<numCourses;++j)
{
if(!degree[j])
break;
}
if(j==numCourses)
return false;
degree[j]=-1;//移出S
for(auto it=graph[j].begin();it!=graph[j].end();++it)
{
--degree[*it];
}
}
return true;
}
};
比较接近维基百科伪代码的方式
typedef vector<unordered_set<int >> GRAPH;
class Solution
{
private:
void getDegree(vector<int>°ree,const GRAPH &graph)
{
for(int i=0;i<graph.size();++i)
{
for(auto it=graph[i].begin();it!=graph[i].end();++it)
{
int j=*it;
++degree[j];
}
}
}
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites)
{
GRAPH graph(numCourses);
int length = prerequisites.size();
//init graph
for( int k =0;k<length;++k)
{
int i=prerequisites[k].first;
int j=prerequisites[k].second;
graph[j].insert(i);
}
vector<int>degree(numCourses);
getDegree(degree,graph);
for(int i=0;i<numCourses;++i)
{
int j=0;
for(;j<numCourses;++j)
{
if(!degree[j])
break;
}
if(j==numCourses)
return false;
degree[j]=-1;
for(auto it=graph[j].begin();it!=graph[j].end();++it)
{
--degree[*it];
}
}
return true;
}
};
DFS
如果在一条dfs路径中找到相同的节点(已经访问过的节点),则说明有环。这里需要维护两个数组,一个数组用来记录所有节点中已经访问过的节点,一个数组用来记录当前dfs路径中访问过节点,对当前路径中所访问的节点,我们需要进行回溯,否则将只有一条dfs路径。
typedef vector<unordered_set<int >> GRAPH;
class Solution {
private:
void dfs(GRAPH&graph,int i,vector<int>&visited,vector<int>&allvisited,bool &flag)
{
visited[i]=1;
allvisited[i]=1;
for(auto it=graph[i].begin();it!=graph[i].end();++it)
{
int j = *it;
if(!visited[j])
{
dfs(graph,j,visited,allvisited,flag);
}
else
{
flag=true;//has a circle
return ;
}
}
visited[i]=0;
}
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
GRAPH graph(numCourses);
int length = prerequisites.size();
//init graph
for( int k =0;k<length;++k)
{
int i=prerequisites[k].first;
int j=prerequisites[k].second;
graph[j].insert(i);
}
vector<int >visited(numCourses);
vector<int> allvisited(numCourses);
bool flag = false;
for(int i=0;i<graph.size();++i)
{
if(!allvisited[i])
dfs(graph,i,visited,allvisited,flag);
}
if(flag)
return false;
else
return true;
}
};