LeetCode No332. Reconstruct Itinerary

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1. 题目描述

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:
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.

2. 思路

将航线抽象建模为一个图，则题目的意思为找出一条航线其可以由所有的票连接而成，若有多条，则选取字典序最小的那条航线。
航线的起点为”JFK”节点，当一个节点的出度为0时，其必定为航线的终点。
由于邻接边用multiset表示，当前的票已经排序，所以深度遍历时，只要存在边被遍历完，则必然是字典序最小的那条。

3. 代码及复杂度分析

class Solution {public:    vector<string> findItinerary(vector<pair<string, string>> tickets) {        map<string, multiset<string>> targets;        for (auto ticket : tickets)             targets[ticket.first].insert(ticket.second);        vector<string> itenary;        stack<string> s;        s.push("JFK");        while (!s.empty()) {            string curr_airport = s.top();            if (targets[curr_airport].empty()) {                itenary.push_back(curr_airport);                s.pop();            } else {                s.push(*(targets[curr_airport].begin()));                targets[curr_airport].erase(targets[curr_airport].begin());            }        }        reverse(itenary.begin(), itenary.end());        return itenary;    }};

4. 参考链接

https://leetcode.com/discuss/85439/short-iterative-solution-explanation-recursive-backtracking

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