# 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 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.

## 2. Implementation

**(1) DFS**

```java
class Solution {
    public List<String> findItinerary(String[][] tickets) {
        List<String> route = new ArrayList<>();

        if (tickets == null || tickets.length == 0) {
            return route;
        }

        Map<String, PriorityQueue<String>> map = new HashMap<>();

        for (String[] ticket : tickets) {
            map.putIfAbsent(ticket[0], new PriorityQueue<>());
            map.get(ticket[0]).add(ticket[1]);
        }

        dfs("JFK", map, route);
        return route;
    }

    public void dfs(String airport, Map<String, PriorityQueue<String>> map, List<String> route) {
        while (map.containsKey(airport) && !map.get(airport).isEmpty()) {
            String nextAirport = map.get(airport).remove();
            dfs(nextAirport, map, route);
        }
        route.add(0, airport);
    }
}
```

## 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)