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_332.java
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_332.java
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package com.fishercoder.solutions;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
/**
* 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.
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.
*/
public class _332 {
public static class Solution1 {
/** credit: https://discuss.leetcode.com/topic/36383/share-my-solution */
public List<String> findItinerary(String[][] tickets) {
Map<String, PriorityQueue<String>> flights = new HashMap<>();
LinkedList<String> path = new LinkedList<>();
for (String[] ticket : tickets) {
flights.putIfAbsent(ticket[0], new PriorityQueue<>());
flights.get(ticket[0]).add(ticket[1]);
}
dfs("JFK", flights, path);
return path;
}
public void dfs(String departure, Map<String, PriorityQueue<String>> flights,
LinkedList path) {
PriorityQueue<String> arrivals = flights.get(departure);
while (arrivals != null && !arrivals.isEmpty()) {
dfs(arrivals.poll(), flights, path);
}
path.addFirst(departure);
}
}
}