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. 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. 2.
    All airports are represented by three capital letters (IATA code).
  3. 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
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class Solution {
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public List<String> findItinerary(String[][] tickets) {
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List<String> route = new ArrayList<>();
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if (tickets == null || tickets.length == 0) {
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return route;
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}
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Map<String, PriorityQueue<String>> map = new HashMap<>();
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for (String[] ticket : tickets) {
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map.putIfAbsent(ticket[0], new PriorityQueue<>());
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map.get(ticket[0]).add(ticket[1]);
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}
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dfs("JFK", map, route);
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return route;
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}
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public void dfs(String airport, Map<String, PriorityQueue<String>> map, List<String> route) {
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while (map.containsKey(airport) && !map.get(airport).isEmpty()) {
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String nextAirport = map.get(airport).remove();
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dfs(nextAirport, map, route);
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}
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route.add(0, airport);
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}
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}
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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)