Reconstruct Itinerary
Desicription
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:
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”]. All airports are represented by three capital letters (IATA code). You may assume all tickets form at least one valid itinerary. Example 1:
代码语言：javascript复制Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]Example 2:
代码语言：javascript复制Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].
But it is larger in lexical order.Solution
代码语言：javascript复制class Solution {
public:
std::vector<std::string> findItinerary(std::vector<std::vector<std::string>>& tickets) {
auto graph = std::map<std::string, std::multiset<std::string>>();
for(const auto& ticket : tickets) {
graph[ticket[0]].insert(ticket[1]);
}
auto res = std::vector<std::string>();
dfs("JFK", graph, res);
return std::vector<std::string>(res.rbegin(), res.rend());
}
private:
static void dfs(const std::string& current, std::map<std::string, std::multiset<std::string>>& graph, std::vector<std::string>& res) {
while(!graph[current].empty()) {
auto next = *graph[current].begin();
graph[current].erase(graph[current].begin());
dfs(next, graph, res);
}
res.push_back(current);
}
};
