1. 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:
- 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.
2. Answer
import java.util.*;
public class Solution {
public List<String> findItinerary(String[][] tickets) {
List<String> result = new ArrayList<String>();
if(tickets == null || tickets.length == 0){
return result;
}
Map<String, ArrayList<String>> graph = new HashMap<String, ArrayList<String>>();
for(int i=0; i<tickets.length; i++){
if(!graph.containsKey(tickets[i][0])){
ArrayList<String> adj = new ArrayList<String>();
adj.add(tickets[i][1]);
graph.put(tickets[i][0], adj);
}else{
ArrayList<String> newadj = graph.get(tickets[i][0]);
newadj.add(tickets[i][1]);
graph.put(tickets[i][0], newadj);
}
}
for(ArrayList<String> a : graph.values()){
Collections.sort(a);
}
Stack<String> stack = new Stack<String>();
stack.push("JFK");
while(!stack.isEmpty()){
while(graph.containsKey(stack.peek()) && !graph.get(stack.peek()).isEmpty()){
stack.push(graph.get(stack.peek()).remove(0));
}
result.add(0,stack.pop());
}
return result;
}
}