332 Reconstruct Itinerary

332. Reconstruct Itinerary

1. Question

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 fromJFK. Thus, the itinerary must begin withJFK.

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. Implementation

(1) DFS

class Solution {
    public List<String> findItinerary(String[][] tickets) {
        List<String> route = new ArrayList<>();

        if (tickets == null || tickets.length == 0) {
            return route;
        }

        Map<String, PriorityQueue<String>> map = new HashMap<>();

        for (String[] ticket : tickets) {
            map.putIfAbsent(ticket[0], new PriorityQueue<>());
            map.get(ticket[0]).add(ticket[1]);
        }

        dfs("JFK", map, route);
        return route;
    }

    public void dfs(String airport, Map<String, PriorityQueue<String>> map, List<String> route) {
        while (map.containsKey(airport) && !map.get(airport).isEmpty()) {
            String nextAirport = map.get(airport).remove();
            dfs(nextAirport, map, route);
        }
        route.add(0, airport);
    }
}

3. Time & Space Complexity

DFS: 时间复杂度O(n), n为所有的机场个数，空间复杂度O(n)