[LeetCode 332] 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 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.
Solution:
DFS

We know the start point "JFK" and total itinerary length. Use map to store the tickets connect src and dest. Map<String, List<String>>, the list should be in ascending order. use dfs to go through the map, find the smallest
lexical order 

public List<String> findItinerary(String[][] tickets) {
Map<String, List<String>> itineraryMap = new HashMap<>();
for(String[] ticket : tickets) {
List<String> dests = itineraryMap.get(ticket[0]);
if(dests == null) {
dests = new ArrayList<>();
dests.add(ticket[1]);
itineraryMap.put(ticket[0], dests);
} else{
dests.add(ticket[1]);
}
}
for(List<String> dests : itineraryMap.values()) {
Collections.sort(dests);
}
List<String> res = new ArrayList<>();
res.add("JFK");
dfs(res, new ArrayList<String>(), itineraryMap, "JFK", tickets.length);
return res;
}

private void dfs(List<String> res, List<String> cur, Map<String, List<String>> itineraryMap, String src, int len) {
if(res.size() >1) {
return;
}
if(cur.size() ==len) {
res.addAll(cur);
return;
}
List<String> dests = itineraryMap.get(src);
if(dests != null && dests.size() >0) {
for(int i=0;i<dests.size();i++) {
String des = dests.remove(i);
cur.add(des);
dfs(res, cur, itineraryMap, des, len);
dests.add(i, des);
cur.remove(cur.size()-1);
}
}
}