332. Reconstruct Itinerary

Description

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.

Solution

DFS

用基础的dfs就可以解决这道题，但没有想到难住我的是用什么数据结构表示input...

由于要保证lexical order，比较直观的想法是每个toList用PriorityQueue来存储，但实现时发现递归还原PriorityQueue有坑，因为每次poll出来的都是最小值啊，没办法做到便利一个PriorityQueue。

看了些解决方案，最终采用ArrayList来存储toList，对于每个ArrayList都用Collections.sort来排一下序就好了呀。还原一个ArrayList也方便，调用add(insertIndex, element)即可。

class Solution {
    public List findItinerary(String[][] tickets) {
        Map> map = new HashMap<>();
        for (String[] ticket : tickets) {
            String from = ticket[0];
            String to = ticket[1];
            if (!map.containsKey(from)) {
                map.put(from, new ArrayList<>());
            }
            map.get(from).add(to);
        }
        
        for (Map.Entry> entry : map.entrySet()) {
            Collections.sort(entry.getValue());    // sort each list
        }
        
        List itinerary = new LinkedList<>();
        itinerary.add("JFK");
        dfsFindItinerary("JFK", map, tickets.length + 1, itinerary);
        return itinerary;
    }
    
    public boolean dfsFindItinerary(
        String from, Map> map
        , int count, List itinerary) {
        
        if (itinerary.size() == count) {
            return true;
        } 
        
        List toList = map.get(from);
        if (toList == null) {
            return false;
        }
        
        for (int i = 0; i < toList.size(); ++i) {
            String to = toList.get(i);
            toList.remove(i);
            itinerary.add(to);
            
            if (dfsFindItinerary(to, map, count, itinerary)) {
                return true;
            }
            
            itinerary.remove(itinerary.size() - 1);
            toList.add(i, to);
        }
        return false;
    }
}