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:

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.

Tips:

题意：给定一些飞机票，让你从中找出一个序列可以全部用完。要求如下：

1.从JFK出发。

2.如果有多个解，输出字典序最小的。

思路：

一直DFS直到arrival的pq取完，然后加最后一个城市，addFirst。

All the airports are vertices and tickets are directed edges. Then all these tickets form a directed graph.

The graph must be Eulerian since we know that a Eulerian path exists.

Thus, start from "JFK", we can apply the Hierholzer's algorithm to find a Eulerian path in the graph which is a valid reconstruction.

Since the problem asks for lexical order smallest solution, we can put the neighbors in a min-heap. In this way, we always visit the smallest possible neighbor first in our trip.

Code：

Recursive：

public class Solution {
    public List<String> findItinerary(String[][] tickets) {
        LinkedList<String> result = new LinkedList<>();
        if (tickets == null || tickets.length == 0) {
            return result;
        }
        HashMap<String, PriorityQueue<String>> map = new HashMap<>();
        for (String[] ticket : tickets) {
            /*if (!map.containsKey(ticket[0])) {
                map.put(ticket[0], new PriorityQueue<String>());
            }*/
            map.putIfAbsent(ticket[0], new PriorityQueue<>());
            map.get(ticket[0]).add(ticket[1]);
        }
        dfs(result, map, "JFK");
        return result;
    }

    private void dfs(LinkedList<String> result, HashMap<String, PriorityQueue<String>> map, String current) {
        while (map.containsKey(current) && !map.get(current).isEmpty()) {
            dfs(result, map, map.get(current).poll());            
        }
        result.addFirst(current);
    }
}

Stack:

public List<String> findItinerary(String[][] tickets) {
    Map<String, PriorityQueue<String>> targets = new HashMap<>();
    for (String[] ticket : tickets)
        targets.computeIfAbsent(ticket[0], k -> new PriorityQueue()).add(ticket[1]);
    List<String> route = new LinkedList();
    Stack<String> stack = new Stack<>();
    stack.push("JFK");
    while (!stack.empty()) {
        while (targets.containsKey(stack.peek()) && !targets.get(stack.peek()).isEmpty())
            stack.push(targets.get(stack.peek()).poll());
        route.add(0, stack.pop());
    }
    return route;
}