leetcode solution: Reconstruct Itinerary (332)

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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.

解法1：建图 + DFS

首先是建图的思想，这题看起来就要用DFS，可是没有图可以操作，对于这类的问题要想到用一个Hashmap来构造图的边的关系。
这里有一个小的trick是用到了PriorityQueue， 因为最后的结果是要排序后较小的，那么我们构建hashmap的时候就要把每一个start对应的destination排好序，我们就可以用PQ来存储。
而题目有一个隐含的条件就是一定有解，那么如果我们在搜索碰到一个没有对应destination的城市的时候，他一定是要求的最后一个string。
所以我们在写DFS的时候，要把res.add(cur)放在while语句之后。

Java

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public class Solution {
    public List<String> findItinerary(String[][] tickets) {
        
        // Construct the graph, using HashMap
        HashMap<String, PriorityQueue<String>> map = new HashMap<String, PriorityQueue<String>>();
        
        for (String[] pair: tickets) {
            if (map.containsKey(pair[0])) {
                map.get(pair[0]).add(pair[1]);
            } else {
                map.put(pair[0], new PriorityQueue<String>());
                map.get(pair[0]).add(pair[1]);
            }
        }
        
        // DFS the map and construct the itinerary
        List<String> res = new ArrayList<String>();
        dfs(map, "JFK", res);
        Collections.reverse(res);
        return res;
        
    }
    
    void dfs(HashMap<String, PriorityQueue<String>> map, String cur, List<String> res) {
        while (map.containsKey(cur) && !map.get(cur).isEmpty()) {
            dfs(map, map.get(cur).poll(), res);
        }
        res.add(cur);
        
    }
}