leetcode 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:
- 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 may 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.
原文地址:https://www.hrwhisper.me/leetcode-reconstruct-itinerary/
更多題解可以查看: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;
}
};
出處:細語呢喃 > leetcode Reconstruct Itinerary
原文地址:https://www.hrwhisper.me/leetcode-reconstruct-itinerary/