Reconstruct Itinerary

问题来源

问题描述

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.

问题分析

显然，这是一道与有向图路径相关的题目。题目的大意是从给出的一堆有向边组合而成的图中找出一条从顶点JFK 到达其他顶点的字母序最小的路径。我们暂时不管字母序的问题，仅考虑如何选取一条路径。显然，DFS是一个较优的选择。以JFK为起点深度遍历整张图，注意到图中是有环的，所以我们还需做多的一步操作便是删除已遍历的边避免进入死循环。至于字母序的问题，我们只需要要每个结点的子结点按照字典序排好，遍历时从字典序最小的结点开始遍历即可。每次将当前结点的子结点遍历完成后再将当前结点压入路径数组中，最后将路径数组反转即可得到答案。

解决代码

class Solution {  public:    vector<string> findItinerary(vector<pair<string, string>> tickets) {        map<string, multiset<string>> edges;        for (auto loc : tickets) {          edges[loc.first].insert(loc.second);        }        vector<string> res;        if (tickets.size() == 0) return res;        dfs(res, "JFK", edges);        reverse(res.begin(), res.end());        return res;    }    void dfs(std::vector<string>& r, string loc, map<string, multiset<string>> & edges) {      while (edges[loc].size()) {        auto tmp = *(edges[loc].begin());        edges[loc].erase(edges[loc].begin());        dfs(r, tmp, edges);        }      r.push_back(loc);    }};