[LeetCode] (medium) 332. Reconstruct Itinerary

https://leetcode.com/problems/reconstruct-itinerary/

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:

Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]

Example 2:

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.

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不想多说什么，优化点在于消除一些冗余操作： 

class Solution {
public:
    unordered_map<string, multiset<string>> M;
    vector<string> result, tem;
    
    int ticket_cnt = 0;
    bool found = false;
    
    vector<string> findItinerary(vector<pair<string, string>> tickets) {
        ticket_cnt = tickets.size();
        
        // cout << ("ATL" < "SFO") << endl;
        
        for(int i = 0; i < tickets.size(); ++i){
            M[tickets[i].first].insert(tickets[i].second);
        }
        
        // result.push_back("JFK");
        dfs(0, "JFK");
        
        return vector<string>(result.rbegin(), result.rend());
        
    }
              
    void dfs(int cur_num, string cur_node){
        if(found) return;
        if(cur_num == ticket_cnt){
            found = true;
        }
        while(!M[cur_node].empty()){
            string nxt = *(M[cur_node].begin());
            M[cur_node].erase(M[cur_node].begin());
            // result.push_back(nxt);
            dfs(cur_num+1, nxt);
        }
        result.push_back(cur_node);
    }
};