332. Reconstruct Itinerary

最新推荐文章于 2025-05-29 22:51:43 发布

qq_34229391

最新推荐文章于 2025-05-29 22:51:43 发布

阅读量161

点赞数 1

分类专栏： LeetCode Data Structure 文章标签： leetcode euler path

本文链接：https://blog.youkuaiyun.com/qq_34229391/article/details/82556246

版权

LeetCode 同时被 2 个专栏收录

23 篇文章

订阅专栏

Data Structure

3 篇文章

订阅专栏

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:

Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]

Example 2:

Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].
             But it is larger in lexical order.

要将所有的机票全部用完，这是让我们在由机票组成的无向图中寻找一条最小的欧拉路径，下面使用的求解欧拉路径的一笔画算法，因为题目中说明一定有解，也就是说给出的数据中一定有欧拉路径，那么就放心大胆的去做好了，其中为了获取lexical最小的路径，使用multiset将地点的名称进行自动排序。

void dfs(unordered_map<string, multiset<string>>& atlas, vector<string>& result, string depature)
{
    while(atlas[depature].size())
    {
        string temp = *(atlas[depature].begin());
        atlas[depature].erase(atlas[depature].begin());
        dfs(atlas, result, temp);
    }
    result.insert(result.begin(), depature);
}
vector<string> findItinerary(vector<pair<string, string>> tickets)
{
    unordered_map<string, multiset<string>> atlas;
    vector<string> result;
    for(int i = 0; i < tickets.size(); i++)
        atlas[tickets[i].first].insert(tickets[i].second);
    dfs(atlas, result, "JFK");
    return result;
}