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:

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.

public class Solution {    public List<String> findItinerary(String[][] tickets) {        for (String[] ticket : tickets)            targets.computeIfAbsent(ticket[0], k -> new PriorityQueue()).add(ticket[1]);        visit("JFK");        return route;    }        Map<String, PriorityQueue<String>> targets = new HashMap<>();    List<String> route = new LinkedList();        void visit(String airport) {        while(targets.containsKey(airport) && !targets.get(airport).isEmpty())            visit(targets.get(airport).poll());        route.add(0, airport);    }}

总结：priorityqueue实现小顶堆

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;}

Example:

From JFK we first visit JFK -> A -> C -> D -> A. There we're stuck, so we write down A as the end of the route and retreat back to D. There we see the unused ticket to B and follow it: D -> B -> C -> JFK -> D. Then we're stuck again, retreat and write down the airports while doing so: Write down D before the already written A, then JFK before the D, etc. When we're back from our cycle at D, the written route is D -> B -> C -> JFK -> D -> A. Then we retreat further along the original path, prepending C, A and finally JFK to the route, ending up with the route JFK -> A -> C -> D -> B -> C -> JFK -> D -> A.