vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<vector<int>> graph(numCourses);
vector<int> indegrees(numCourses, 0);
queue<int> q;
for (const auto& pair : prerequisites) {
graph[pair.second].emplace_back(pair.first); // must take second before first
++indegrees[pair.first];
}
for (int n = 0; n < indegrees.size(); ++n) {
if (indegrees[n] == 0) {
q.push(n);
}
}
vector<int> result;
result.reserve(numCourses);
int visited = 0;
while (!q.empty()) {
int n = q.front();
q.pop();
++visited;
result.emplace_back(n);
for (int m : graph[n]) {
if (--indegrees[m] == 0) {
q.push(m);
}
}
}
if (numCourses != visited) {
result.clear();
}
return result;
}
string alienOrder(vector<string>& words) {
//find the order, apparently it is a topologcal sorting problem
//1. make graph
//2. top sorting with the graph,
//question is how to avoid cycle?
vector<unordered_set<int> > graph(26);
vector<int> indegrees(26);
vector<int> result;
int row_size = words.size();
//make graph
int stop = '$' - 'a';
int count = 0;
//init graph for every existed letter!!!
//the letter could have no pre or post
//don't forget
for(int i=0;i<words.size();i++){
for(int j=0;j<words[i].size();j++){
int add = words[i][j] - 'a';
if(graph[add].empty())
{
graph[add].insert(stop);
count++;
}
}
}
for(int i = 0; i < row_size - 1; i++)
{
int len1 = words[i].size(), len2 = words[i+1].size();
for(int j = 0; j < min(len1, len2); j++)
{
int u = words[i][j] - 'a';
int v = words[i+1][j] - 'a';
if(u != v)
{
if(graph[u].find(v) == graph[u].end())
{
graph[u].insert(v);
indegrees[v]++;
}
break; //important!!!
}
}
}
//topological sorting
queue<int> q;
for(int k = 0; k < indegrees.size(); k++)
{
if(indegrees[k] == 0 && !graph[k].empty())
{
q.push(k);
}
}
//int cnt = 0;
while(!q.empty())
{
int n = q.front();
q.pop();
//if(graph[n].empty()) continue;
result.push_back(n);
for(auto neigh : graph[n])
{
if(neigh != stop && (--indegrees[neigh] == 0))
{
q.push(neigh);
}
}
count--;
}
if(count != 0) return "";
//convert the results to string
string s;
for(auto v : result)
{
s += (v + 'a');
}
return s;
}
