【LeetCode】Reconstruct Itinerary（332）

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:

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:
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;
    }
}