[LeetCode 332] Reconstruct Itinerary

最新推荐文章于 2021-01-10 13:04:45 发布

原创最新推荐文章于 2021-01-10 13:04:45 发布 · 2.8k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#DFS #leetcode

Leetcode 同时被 2 个专栏收录

287 篇文章

订阅专栏

Java

106 篇文章

订阅专栏

本文介绍了一种使用深度优先搜索（DFS）算法解决构建机票行程的问题，从JFK出发，按照机票上的出发地和目的地顺序，实现最优的行程路径。通过遍历机票列表并按字母顺序排序目的地机场，确保行程的最小词典顺序。

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.

Solution:

DFS

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

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