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.


  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.


DFS+backtracking. At first I did not realize all tickets have to be used. Reading the question thoroughly is very important.

public class Solution {
    Map<String, List<String>> g = new HashMap<String, List<String>>();
    List<String> sol = new ArrayList<String>();
    int size;
    public List<String> findItinerary(String[][] tickets) {
        for (int i = 0; i < tickets.length; i++) {
            String from = tickets[i][0];
            String to = tickets[i][1];
            if (!g.containsKey(from)) {
                List<String> opts = new ArrayList<String>();
                g.put(from, opts);
            else {
                List<String> opts = g.get(from);
                g.put(from, opts);
        size = tickets.length + 1;
        return sol;
    private boolean find(List<String> current) {
        if (current.size() == size) return true;
        String start = current.get(current.size() - 1);
        List<String> opts = g.get(start);
        if (opts == null || opts.size() == 0) {
            return false;
        for (int i = 0; i < opts.size(); i++) {
            String opt = opts.get(i);
            if (find(current)) return true;
            opts.add(i, opt);
            current.remove(current.size() - 1);
        return false;

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