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:
["JFK", "LGA"]
has a smaller lexical order than ["JFK", "LGB"]
. 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.
原文地址:https://www.hrwhisper.me/leetcode-reconstruct-itinerary/
更多题解可以查看:https://www.hrwhisper.me/leetcode-algorithm-solution/
class Solution { public: vector<string> findItinerary(vector<pair<string, string>> tickets) { unordered_map<string, map<string,int>> m; for (const auto &p : tickets) { m[p.first][p.second]++; } string start = "JFK"; vector<string> ans; ans.push_back(start); dfs(start, ans, tickets.size()+ 1, m); return ans; } bool dfs(const string & cur,vector<string> &ans,const int &n,unordered_map<string, map<string, int>> &m) { if (ans.size() == n) return true; for (auto ticket = m[cur].begin(); ticket != m[cur].end(); ticket++) { //map<string, int>::iterator if (ticket->second != 0) { ticket->second--; ans.push_back(ticket->first); if (dfs(ticket->first, ans, n, m)) return true; ans.pop_back(); ticket->second++; } } return false; } };
出处:细语呢喃 > leetcode Reconstruct Itinerary
原文地址:https://www.hrwhisper.me/leetcode-reconstruct-itinerary/