Reconstruct Itinerary

时间:2021-01-26 17:03:51

Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.

Note:

  1. If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string. For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
  2. All airports are represented by three capital letters (IATA code).
  3. You may assume all tickets form at least one valid itinerary.

Example 1:
tickets = [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Return ["JFK", "MUC", "LHR", "SFO", "SJC"].

Example 2:
tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Return ["JFK","ATL","JFK","SFO","ATL","SFO"].
Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"]. But it is larger in lexical order.

重建行程单,在图中找一条路径,能经过所有的边。

参考:http://www.cnblogs.com/grandyang/p/5183210.html

class Solution{
public:
vector<string> findItinerary(vector<pair<string,string>> tickets){\
unordered_map<string,multiset<string>> m;
for(auto t : tickets){
m[t.first].insert(t.second);
}
vector<string> res;
dfs(m,"JFK",res);
return vector<string> (res.rbegin(),res.rend());
} void dfs(unordered_map<string,multiset<string>>& m,string s,vector<string>& res){
while(m[s].size()){
string t = *m[s].begin();
m[s].erase(m[s].begin());
dfs(m,t,res);
}
res.push_back(s);
}
};