332. Reconstruct Itinerary

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:
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.

---

Code

public class Solution {
    public List findItinerary(String[][] tickets) {
        Map> map = new HashMap<>();
        
        for (String[] ticket: tickets) {
            List dests = map.get(ticket[0]);
            if (dests == null) {
                dests = new ArrayList<>();
                dests.add(ticket[1]);
                map.put(ticket[0], dests);
            } else {
                dests.add(ticket[1]);
            }
        }
        
        for (List dests: map.values()) {
            Collections.sort(dests);
        }
        
        List result = new ArrayList<>();
        result.add("JFK");
        
        dfs(result, new ArrayList(), map, "JFK", tickets.length);
        
        return result;
    }
    
    public void dfs(List result, List record, Map> map, String src, int length) {
        if (result.size() > 1) return;
        if (record.size() == length) {
            result.addAll(record);
            return;
        }
        
        List dests = map.get(src);
        
        if (dests != null && dests.size() > 0) {
            for (int i = 0; i < dests.size(); i++) {
                String dest = dests.remove(i);
                record.add(dest);
                dfs(result, record, map, dest, length);
                dests.add(i, dest);
                record.remove(record.size() - 1);
            }
        }
    }
}

Solution

使用DFS。

We know the start point "JFK" and total itinerary length. Use map to store the tickets connect src and dest. Map>, the list should be in ascending order. use dfs to go through the map, find the smallest lexical order