leetcode Reconstruct Itinerary

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本文介绍了一个使用DFS算法解决机票行程重建的问题，通过遍历给定的机票信息，找到从JFK机场出发的最小字典序行程路径。

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Reconstruct Itinerary

题目详情：

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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

解题方法：

使用dfs算法，从JFK开始遍历每个节点。要注意的是这里要求按照字典序访问节点，并且可能会有重复的节点即有两张相同的车票。所以这里使用map<string, multiset<string>>数据结构存储图是最简单的方式。

最开始我选择的方法是遍历到一个节点就将其加入到结果vector中，但是这样的缺点是当一个节点的子节点没有邻居时，不能先访问它，这样处理起来就有点麻烦。后来参考 discuss中的算法，按序访问节点的每一个邻居，并将其删除，当节点没有邻居时，再将其加入到结果vector中。最后反转结果vector。

代码详情：

class Solution {
public:
vector<string> findItinerary(vector<pair<string, string>> tickets) {
for (auto ticket : tickets)
targets[ticket.first].insert(ticket.second);
visit("JFK");
return vector<string>(route.rbegin(), route.rend());
}

map<string, multiset<string>> targets;
vector<string> route;

void visit(string airport) {
while (targets[airport].size()) {
string next = *targets[airport].begin();
targets[airport].erase(targets[airport].begin());
visit(next);
}
route.push_back(airport);
}
};