[leetcode]332. Reconstruct Itinerary

题目链接：https://leetcode.com/problems/reconstruct-itinerary/#/description

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.

class Solution {public:    vector<string> findItinerary(vector<pair<string, string>> tickets) {        for(auto ticket:tickets)            targets[ticket.first].insert(ticket.second);        visit("JFK");        return vector<string>(route.rbegin(),route.rend());    }private:    map<string,multiset<string>> targets;    vector<string> route;    void visit(string airport)    {        while(targets[airport].size())        {            string next=*targets[airport].begin();            targets[airport].erase(targets[airport].begin());            visit(next);        }        route.push_back(airport);    }};

class Solution {public:    vector<string> findItinerary(vector<pair<string, string>> tickets) {        // Each node (airport) contains a set of outgoing edges (destination).        unordered_map<string, multiset<string>> graph;        // We are always appending the deepest node to the itinerary,        // so will need to reverse the itinerary in the end.        vector<string> itinerary;        if (tickets.size() == 0){            return itinerary;        }        // Construct the node and assign outgoing edges        for (pair<string, string> eachTicket : tickets){            graph[eachTicket.first].insert(eachTicket.second);        }        stack<string> dfs;        dfs.push("JFK");        while (!dfs.empty())        {            string topAirport = dfs.top();            if (graph[topAirport].empty())            {                // If there is no more outgoing edges, append to itinerary                // Two cases:                // 1. If it searchs the terminal end first, it will simply get                //    added to the itinerary first as it should, and the proper route                //    will still be traversed since its entry is still on the stack.                // 2. If it search the proper route first, the dead end route will also                //    get added to the itinerary first.                itinerary.push_back(topAirport);                dfs.pop();            }            else            {                // Otherwise push the outgoing edge to the dfs stack and                // remove it from the node.                dfs.push(*(graph[topAirport].begin()));                graph[topAirport].erase(graph[topAirport].begin());            }        }        // Reverse the itinerary.        reverse(itinerary.begin(), itinerary.end());        return itinerary;    }};