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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

深度遍历解题。HashMap存储航盘信息，按照起点索引，Priorityqueue存储终点保证按照词典升序排列。之后深度遍历每个终点Priorityqueue的peek，之后反响添加终点，res.add(0, airpot)。代码如下：

public class Solution {    List<String> res = new LinkedList<String>();    HashMap<String, PriorityQueue<String>> map = new HashMap<String, PriorityQueue<String>>();        public List<String> findItinerary(String[][] tickets) {        for (String[] ticket: tickets) {            if (!map.containsKey(ticket[0])) {                map.put(ticket[0], new PriorityQueue<String>());            }            map.get(ticket[0]).offer(ticket[1]);        }        helper("JFK");        return res;    }        private void helper(String airpot) {        while (map.containsKey(airpot) && !map.get(airpot).isEmpty()) {            helper(map.get(airpot).poll());        }        res.add(0, airpot);    }}

迭代法。代码如下：

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;}